Class 11 Engineering Maths - Relations And Functions

$\displaystyle f\left ( x \right )=1$ and $\phi \left ( x \right )=\sin ^{2}x+\cos ^{2}x.$
Show that both functions are equal?

For

$f(x)=1$
domain:

$x \in (-\infty, \infty)$
Range:

$f(x) = \{1\}$

For

$\phi \left ( x \right )=\sin ^{2}x+\cos ^{2}x=1$
domain:

$x \in (-\infty, \infty)$
Range:

$\phi(x)= \{1\}$
Therefore,

$f(x)$ and

$\phi (x)$ are equal.

State whether the given statement is true or false. If true enter $1$ or else enter $0$ .
$(x, y)$ is an ordered pair, whose first component is $x$ and the second component is $y$ .

For any two elements

$x$ and

$y$ of same or different sets, the ordered pair is given by

$(x,y)$ or

$(y,x)$ .

Also, in an ordered pair

$(x,y)$ ,

$x$ is called the first component and

$y$ is called the second component.

So, the given statement is true.

Hence, the answer is

$1$ .

The cartesian product $A\times A$ has $9$ elements among which are found $(-1, 0)$ and $(0, 1)$ . Find the set $A$ and the remaining elements of $A\times A$ .

$n(A\times A) = 9$

Set

$A =\{-1,0,1\}$ Since

$(-1,0)$ and

$(0,1)$ are elements in

$A\times A$

Remaining elements

$= \{(-1,-1), (-1,1), (0,-1), (0,0), (1,-1), (1,0), (1,1)\}$

If $x$ is real, then find the solution set of $\sqrt{x+1}+\sqrt{x-1}=1$ .

For equation,

$\sqrt{x+1}+ \sqrt{x-1}=1$ ,

$x\geq1\ \& \ x\geq-1$ (non-negative terms inside square roots)

$\Rightarrow x\geq 1$ (intersection of sets)

$\Rightarrow \sqrt{x+1}\geq \sqrt{2}>1$

$\Rightarrow \sqrt{x+1} +\sqrt{x-1}>1$ (square roots of real numbers are non-negative)
This contradicts our initial equation for finding a real solution.

$\therefore$ The equation has no real solution.

Use the elements of set A ={x, y, z} to form all possible ordered pairs.
Find the number of elements in the ordered pair.

All possible ordered pairs

$=(^3C_1) \times (^3C_1)=9$

What is the $y-$ coordinate of point $R$ ?

A point is denoted by

$(x-$ cordinate,

$y -$ cordinate

$)$ .
Hence,

$y-$ coordinate of point

$R(5,1)$ is

$1$ .

State whether each of the following statements are true or false. If the statement is false rewrite the given statement correctly
(i) If $P = \{m, n\}$ and $Q = \{n, m\}$ then $\displaystyle P\times Q = \{(m, n), (n, m)\}$
(ii) If $A$ and $B$ are non-empty sets then $\displaystyle A\times B$ is a non-empty set of ordered pairs $(x, y)$ such that $\displaystyle x\in A$ and $\displaystyle y\in B$
(iii) If $A = \{1, 2\}, B = \{3, 4\}$ then $\displaystyle A\times \left ( B\cap \phi \right )=\phi$

(i) Given

$P = \{m, n\}$ and

$Q = \{n, m\}$ then

$\displaystyle P\times Q = \{(m, n), (m, m), (n, n), (n, m)\}$

So, given value of

$P\times Q$ is incorrect.

Hence, the given statement (i) is false.

(ii) If

$A$ and

$B$ are non-empty sets, then

$\displaystyle A\times B$ is a non-empty set of ordered pairs

$(x, y)$ such that

$\displaystyle x\in A$ and

$\displaystyle y\in B$

Hence, the given statement (ii) is true.

(iii)

$A = \{1,2\}$ and

$B = \{3,4\}$

$A\times (B\cap\phi) = A\times \phi = \phi$

So, (iii) is true.

If $\displaystyle \left ( \frac{x}{3}+1,y-\frac{2}{3} \right )=\left ( \frac{5}{3},\frac{1}{3} \right )$ find the values of x and y

It is given that

$\displaystyle \left ( \frac{x}{3}+1,y-\frac{2}{3} \right )=\left ( \frac{5}{3},\frac{1}{3} \right )$
Since the ordered pairs are equal the corresponding element will also be equal
Therefore

$\displaystyle \frac{x}{3}+1=\frac{5}{3}$ and

$\displaystyle y-\frac{2}{3}=\frac{1}{3}$

$\Rightarrow y =1$

$\Rightarrow \displaystyle \frac{x}{3}+1=\frac{5}{3}$

$\Rightarrow x=2$

$\displaystyle \therefore x = 2$ and

$y = 1$

Find the number of ordered pairs in $R o R$ ?

To obtain the elements of R o R, we proceed as follows. Since

$\displaystyle \left ( 1,4 \right ) \epsilon R\:and\:\left ( 4,5 \right ) \epsilon R$ we have

$\displaystyle \left ( 1,5 \right ) \epsilon$ R o R.Again since

$\displaystyle \left ( 1,4 \right ) \epsilon R\:and\:\left ( 4,6 \right )\epsilon R$ we have (1,6)

$\displaystyle \epsilon$ R o R.Similarly (3,6)

$\displaystyle \epsilon$ R o R since (3,7)

$\displaystyle \epsilon$ R and (7,6)

$\displaystyle \epsilon$
R Hence

$R o R=(1,5), (1,6), (3,6)$

$A = \{1, 2, 3, 5\}$ and $B = \{4, 6, 9\}$ . Define a relation $R$ from $A$ to $B$ by $R = \{(x, y):$ the difference between $x$ and $y$ is odd $\displaystyle x\in A, y\in B\}$ . Write $R$ in roster form

$A = \{1, 2, 3, 5\}$ and

$B = \{4, 6, 9\}$

$R = \{(x, y):\mbox{ the difference between x and y is odd} \ \displaystyle x\in A, y\in B\}$

$\displaystyle \therefore R = \{(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)\}$

Let $A = \{1, 2, 3, 4, 6\}$ and $R$ be the relation on $A$ defined by $\{(a, b): a, \displaystyle b\in A\ , b$ is exactly divisible by $a\}$
(i) Write $R$ in roster form
(ii) Find the domain of $R$
(iii) Find the range of $R$

$A = \{1, 2, 3, 4, 6\}$

$R = \{(a, b): a, \displaystyle b\in A, b \mbox{ is exactly divisible by } a\}$
(i)

$R = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)\}$
(ii) Domain of

$R = \{1, 2, 3, 4, 6\}$
(iii) Range of

$R = \{1, 2, 3, 4, 6\}$

$(x, y)$ is an ordered pair whose components are $x$ and $y$ .

For any two elements

$x$ and

$y$ of same or different sets, the ordered pair is given by

$(x,y)$ or

$(y,x)$ .

Thus,

$(x,y)$ is an ordered pair whose components are

$x$ and

$y$ .

So, the given statement is true.

Hence, the answer is

$1$ .

Let $A$ and $B$ be two sets such that $n(A) = 3$ and $n(B) = 2$ . If $(x, 1), (y, 2), (z, 1)$ are in $\displaystyle A\times B$ find $A$ and $B$ where $x, y$ and $z$ are distinct elements

It is given that

$n(A) = 3$ and

$n(B) = 2$ and

$(x, 1), (y, 2), (z, 1)$ are in

$\displaystyle A\times B$

We know that

$A =$ Set of first elements of the ordered pair elements of

$\displaystyle A\times B$

$B =$ Set of second elements of ordered pair elements of

$\displaystyle A\times B$

$\displaystyle \therefore x, y$ and

$z$ are the elements of

$A$ and

$1$ and

$2$ are the elements of

$B$

Since

$n(A) = 3$ and

$n(B) =2$ it is clear that

$A = \{x, y, z\}$ and

$B = \{1, 2\}$

Suppose $3$ bulbs are selected at random from a lot. Each bulb is tested and classified as defective $(D)$ or non-defective $(N)$ . Write the sample space of this experiment?

Given

$3$ bulbs are to be selected at random from the lot. Each bulb in the lot is tested and classified as defective

$(D)$ or non-defective

$(N)$
The sample space of this experiment is given by,

$S = {DDD, DDN, DND, DNN, NDD, NDN, NND, NNN}$

Let $A =\{x, y, z\}$ and $B = \{1, 2\}$ . Find the number of relations from $A$ to $B$ .

It is given that

$A = \{x, y, z\}$ and

$B = \{1, 2\}$

$\displaystyle \therefore$

$\displaystyle A\times B = \{(x, 1), (x, 2), (y, 1), (y, 2), (z, 1), (z, 2)\}$
Since

$n(\displaystyle A\times B)=6$

The number of subsets of

$\displaystyle A\times B$ is

$\displaystyle 2^{6}$ .

Determine the domain and range of the relation $R$ defined by $R = \{(x, x + 5) : \displaystyle x\in \{0, 1, 2, 3, 4, 5\}\}$

$R = \{(x, x + 5) : \displaystyle x\in \{0, 1, 2, 3, 4, 5\}\}$

$\displaystyle \Rightarrow R = \{(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)\}$

$\displaystyle \therefore$ Domain of

$R = \{0, 1, 2, 3, 4, 5\}$
Range of

$R = \{5, 6, 7, 8, 9, 10\}$

Write the relation $R = \{(x,\displaystyle x^{3}): x\mbox{ is a prime number less than 10}\}$ in roster form

$R = \{(x,\displaystyle x^{3}): x\mbox{ is a prime number less than 10}\}$
The prime numbers less than

$10$ are

$2, 3, 5$ and

$7$

$\displaystyle \therefore R = \{(2, 8), (3, 27), (5, 125), (7, 343)\}$

For the given function F= { (1, 3), (2, 5), (4, 7), (5, 9), (3, 1) }, write the domain and range.

Domain

$=$ {

$1,2,4,5,3$ }

and Range

$=$ {

$3,5,7,9,1$ }

If $k$ is given a real constant $f(x)$ is a real quadratic in ' $x$ such that $f(x + k) = f(-x)$ . The coefficient of $x^2$ is $1$ . Find $f(x)$ .

Let

$f(x)={ x }^{ 2 }+bx+c$

$f(k+x)=f(-x)\\ { (x+k) }^{ 2 }+b(x+k)+c={ (-x) }^{ 2 }+b(-x)+c\\ { x }^{ 2 }+{ k }^{ 2 }+2kx+bx+bk+c={ x }^{ 2 }-bx+c\\ { x }^{ 2 }+(2k+b)x+({ k }^{ 2 }+bk+c)={ x }^{ 2 }-bx+c$

Comparing both sides

$\Rightarrow 2k=-2b\\ \Rightarrow b=-k\\ \Rightarrow f(x)={ x }^{ 2 }-kx+c$

where

$c\in R$

A function $f : [-3, 7) \rightarrow R$ is defined as follows
$f(x)=\left\{\begin{matrix}4x^2-1; & -3 \leq x < 2 \\ 3x-2; & 2\leq x\leq 4 \\ 2x-3; & 4 < x < 7 \end{matrix}\right.$
(i) $f(5)+f(6)$
(ii) $f(1)-f(3)$
(iii) $f(-2)-f(-4)$
(iv) $\dfrac{f(3)+f(1)}{2f(6)-f(1)}$

$(i)$

$f(5)=2(5)-3=7\\f(6)=2(6)-3=9\\\therefore f(5)+f(6)=7+9=16$

$(ii)$

$f(1)=4(1^2)-1=3\\f(3)=3(3)-2=7\\\therefore f(1)-f(3)=3-7=-4$

$(iii)$

$f(-4)$ is not defined because

$-4$ is not in domain.

$(iv)$

$f(3)=3(3)-2=7\\f(1)=4(1^2)-1=3\\f(6)=2(6)-3=9\\ \therefore \dfrac {f(3)+f(1)}{2f(6)-f(1)}=\dfrac {7+3}{18-3}=\dfrac {2}{3}$

Let R be a relation from A = {1,2,3,4} to B = {1,3,5} i.e. (a,b) $\epsilon$ R if a < b then find R o $R^{-1}$

R = {(1,3), (1,5), (2,3), (2,5), (3,5), (4,5))

$\therefore \, R^{-1}$ = {(3,1), (5,1), (3,2), (5,2), (5,3), (5,4)}

$\therefore \, RoR^{-1} \, = \, (3, \, 1) \, \epsilon \, R^{-1} \, and \, (1, \, 5) \, \epsilon \, R$

$\therefore \, (3,5) \, \epsilon \, RoR^{-1}$ etc.

$RoR^{-1}$ = {(3,3), (3,5), (5,3), (5,5)}

If $G=\left\{ 7,8 \right\}$ and $H=\left\{ 5,4,2 \right\}$ , find $H\times G$ .

Given two sets are

$H=\left\{ 5,4,2 \right\}$ and

$G=\left\{ 7,8 \right\}$ ,

$\therefore$

$H\times G=\left\{ \left( 5,7 \right) ,\left( 5,8 \right) ,\left( 4,7 \right) ,\left( 4,8 \right) ,\left( 2,7 \right) ,\left( 2,8 \right) \right\}$

Let $S=\{1,2,3\}$ and a relation $R$ is defined by $aRb$ iff $a=b$ then find $R$ .

Given

$S=\{1,2,3\}$ and a relation

$R$ is defined by

$aRb$ iff

$a=b$ .

Then

$R=\{(1,1),(2,2),(3,3)\}$ since

$1R1,2R2,3R3$ holds.

If $f(x)=\log_e\sec x$ and $\phi(x)=\log_e\tan x$ then prove that $e^{2f(x)}-e^{2\phi(x)}=1$ .

Given,

$f(x)=\log_e\sec x$ and

$\phi(x)=\log_e\tan x$ .

Then

$e^{2f(x)}-e^{2\phi(x)}$

$=e^{2\log_e \sec x}-e^{2\log _e \tan x}$

$=e^{\log_e \sec^2 x}-e^{\log _e \tan^2 x}$

$=\sec^2x-\tan^2x$

$=1$ .

Find the domain of definations of the following functions:
$f(x)=\sqrt {1-\sqrt {1-x^{2}}}$

$\begin{array}{l} For\, domain \\ 1-{ x^{ 2 } }\ge 0 \\ \Rightarrow \left( { 1-x } \right) \left( { 1+x } \right) \ge 0 \\ \Rightarrow \left( { 1-x } \right) \left( { 1+x } \right) \le 0 \\ \Rightarrow x\in \left[ { -1,1 } \right] \\ also \\ 1-\sqrt { 1-{ x^{ 2 } } } \ge 0 \\ \Rightarrow -\sqrt { 1-{ x^{ 2 } } } \ge -1 \\ \Rightarrow \sqrt { 1-{ x^{ 2 } } } \le 1 \\ \Rightarrow 1-{ x^{ 2 } }\le 1 \\ \Rightarrow -{ x^{ 2 } }\le 0 \\ \Rightarrow { x^{ 2 } }\ge 0 \end{array}$

Hence domain of

$f(x)$ is

$[-1,1]$

Let $A = \left\{ {1,2,3,4} \right\}$ and $R$ is a relation on $A$ such that $R = \left\{ {\left( {a,b} \right):a = 2b} \right\}$

Consider,

$A = \left\{ {1,2,3,4} \right\},R = \left\{ {\left( {a,b} \right):a = 2b} \right\}$

$\begin{array}{c}B = 2 \times 1,2 \times 2,2 \times 3,2 \times 4\\ = \left\{ {2,4,6,8} \right\}\end{array}$
and if

$R:A \to \left\{ {1,2} \right\} \to \left\{ {2,4} \right\}$

If $A = \{a, b, c\}$ , $B = \{1, 2\}$ , then find $A \times B , \, B \times A \, and \, A \times A$

$A=\left \{ a,b,c \right \}$

$B=\left \{ 1,2 \right \}$

$A\times B=\left \{ (a,1),(a,2),(b,1),(b,2),(c,1),(c,2) \right \}$

$B\times A=\left \{ (1,a),(1,b),(1,c),(2,a),(2,b),(2,c) \right \}$

$A\times A=\left \{ (a,a),(a,b),(a,c),(b,a),(b,b),(b,c),(c,a),(c,b),(c,c) \right \}$

If the relation $R: A \rightarrow B$ where $A = \{1, 2, 3, 4\}, B = \{1, 3, 5\}$ is defined by $R = \{(x, y): x < y, x \in A, y \in B\}$ then find $R$ and $R^{-1}$ .

Given the relation

$R: A \rightarrow B$ where

$A = \{1, 2, 3, 4\}, B = \{1, 3, 5\}$ is defined by

$R = \{(x, y): x < y, x \in A, y \in B\}$ .

$R=\{(2,1),(3,1),(4,1),(4,3)\}$ .

And

$R^{-1}=\{(1,2),(1,3),(1,4),(3,4)\}$ .

If $P = \{a , b\}$ and $Q = \{x, y, z\}$ , show that $P \times Q \neq Q \times P.$

$P=\left \{ a,b \right \}$

$Q=\left \{ x,y,z \right \}$

$P\times Q=\left \{ (a,x),(a,y),(a,z),(b,x),(b,y),(b,z) \right \}$

$Q\times P=\left \{ (x,a),(x,b),(y,a),(y,b),(z,a),(z,b) \right \}$

By the definition of equality of ordered pairs, the pair

$(a,x)$ is not equal to the pair

$(x,a)$

Therefore,

$P\times Q\neq Q\times P$

Let $A=\{1,-1\}$ . Then find $A\times A$ .

Given,

$A=\{1,-1\}$ .

Then

$A\times A=\{(1.1),(1,-1),(-1,1),(-1,-1)\}$ .

Given $R = \{(x, y): x, y \in W, x^2 + y^2 = 25\}$ , where $W$ is the set of all whole numbers. Find the domain and range of $R$ .

We have

$3^2+4^2=25$ or,

$4^2+3^2=25$ and

$0^2+(5)^2=25$ or,

$5^2+0^2=25$ .

So the domain of

$R$ is

$\{0,3,4,5\}$ . [ Values corresponding to

$x$ for

$x$ being whole number]

And the range is

$\{0,3, 4,5\}$ . [ Values corresponding to

$y$ for

$y$ being whole number]

If $R = \{(x, y)|y = 2x + 7|$ , where $x \in R$ and $-5 \le x \le 5\}$ is relation, find the domain of $R$ .

Given,

$R = \{(x, y)|y = 2x + 7|$ , where

$x \in R$ and

$-5 \le x \le 5\}$ is relation.

Now, the domain of the relation

$R$ is nothing but the set of values of

$x$ and it is

$-5\le x\le 5$ for

$x\in \mathbb{R}$ of

$[-5,5]$ .

If $f(x) = x + (1/x)$ , show that $\{f(x)\}^3 = f(x^3) + 3f(1/x)$

$f(x)=x+\dfrac{1}{x} \Rightarrow \boxed{f(x^3)=x^3+\frac{1}{x^3}}$ ...(1)

$\Rightarrow f(y_x)=\dfrac{1}{x}+\dfrac{1}{(1/x)}\Rightarrow \boxed{x+\dfrac{1}{x}=f(1/x)}$ ...(2)

$(f(x))^3=(x+\dfrac{1}{x})^3$

$=x^3+(\dfrac{1}{x})^3+3(x.\dfrac{1}{x})(x+\dfrac{1}{x})$

$=(x^3+\dfrac{1}{x^3})+3(1)(x+\dfrac{1}{x})$

$\boxed{(f(x))^3=f(x^3)+3f(\dfrac{1}{x})}$ from (1) and (2)

If $A = \{0, 1\}$ and $B = \{1, 2, 3\}$ , show that $A \times B \neq B \times A$

$A=\left \{ 0,1 \right \}$

$B=\left \{ 1,2,3 \right \}$

$A\times B=\left \{ (0,1),(0,2),(0,3),(1,1),(1,2),(1,3) \right \}$

$B\times A=\left \{ (1,0),(2,0),(3,0),(1,1),(2,1),(3,1) \right \}$

By the definition of equality of ordered pairs, the pair

$(0,1)$ in

$A\times B$ is not equal to the pair

$(1,0)$ in

$B\times A$

Therefore,

$A\times B\neq B\times A$

Let $A = \{1, 2, 3\}$ . Find the number of relations from $A$ to $A.$

If there are

$x$ elements in set

$A$ , and

$y$ elements in set

$B$ , then no. of relations from

$A$ to

$B$ is

$n(A\times B)=xy$

Here, Number of elements in set A is

$3$

So, number of relations from

$A$ to

$A$ is

$=3\times 3=9$

The Cartesian product $P\times P$ has $16$ elements among which two elements are $(a,b)$ and $(c,d)$ . Find the set $P$ and the remaining elements of $P\times P$

$P$ has n elements then the cartesian product of

$P\times P$ has

$2^{n}$ elements.

By using this

$P$ has 4 elements and we can see all those 4 in given question;

$P=\{a,b,c,d\}$

If $\dfrac{1}{{\left| a \right|}} > \dfrac{1}{b}$ , then $\left| a \right| < b$ , where a & b are non-zero real numbers.

$\dfrac {1}{|a|} > \dfrac {1}{b}$

Multiplying both sides by

$\left|a\right|$

$1>\dfrac{\left|a\right|}{b}$

Multiply both sides by

$6$

$b>\left|a\right|$

Let A = {1,2} and B = {3,4}. Find the number of relations from A to B.

Given, A = {1,2} and B = {3,4}.

Number of elements in set A =

$n(A)=2$

Number of elements in set B =

$n(B)=2$

Number of relations from A to B =

$2^{n(A) \times n(B)}$

$=2^{2\times 2}=2^4=16$

Find the domain and range of the following real functions:
i) $f(x) = -|x|$
ii) $f(x) = \sqrt{9 - x^2}$

$f(x)=-\left | x \right |$

Range

$f\in (x\leq 0)\ or (-\infty ,0]$

Domain

$x\in (-\infty <x<\infty )\ or (-\infty ,\infty )$

ii)

$f(x)=\sqrt{9-x^2}$

Range

$(0\leq f(x)\leq 3)\ or [0,3]$

Domain

$x(-3\leq x\leq 3 )\ or [-3,3]$

If $f\left( x \right) = {x^3} - \frac{1}{{{x^3}}}$ , prove that $f\left( x \right) + f\left( {\frac{1}{x}} \right) = 0$

$f(x)=x^2-\dfrac {1}{x^3}$

$\Rightarrow \ f\left (\dfrac {1}{x}\right)=\left (\dfrac {1}{x}\right)^3-\dfrac {1}{\left (\dfrac {1}{x}\right)^3} =\dfrac {1}{x^3}-\dfrac {1}{1/x^3}$

$=\dfrac {1}{x^3}-x^3$

$\therefore \ f(x)+f\left (\dfrac {1}{x}\right)=x^3-\dfrac {1}{x^3}+\dfrac {1}{x^3}-x^3$

$\therefore \ \boxed {f(x)+f(1/x) =0}$

Find domain and range of
$R = \{ (a, b) / a \in N, \, \, \, a < 5 \, \, \text{ and} \, \, b = 2\}$

The Relation is,

$R=\left \{ (a,b):a\epsilon N,\; a<5\; and \; b=2 \right \}$

Domain for this relation, is all values of First element of the ordered pairs

$(a,b)$ , that is

$a$

Now,

$a$ is a natural number and

$a<5$

So, domain is the set

$\left \{ 1,2,3,4 \right \}$

Range is all values of the second element of the ordered pairs

$(a,b)$ , that is

$b$

So, range is the set

$\left \{ 2 \right \}$

If $A=\left\{ 3,4 \right\} \quad and\quad B=\left\{ 1,2 \right\}$ then of $A\times B$ is .

A = {3,4}

B = {1,2}

$A\times B =$ {(3,1), (3,2), (4,1), (4,2)}

Determine the domain and the range of the relation $R$ defined by $R=\{ \left( x+1,\quad x+5 \right) :x\in \{ 0,\quad 1,\quad 2,\quad 3,\quad 4,\quad 5\}\}$

$R=\{(x+1), x+5\}$

$D_f\in R\{0, 1, 2, 3, 4, 5\}$

$R_f\in \{(1, 2, 3, 4, 5, 6), (5, 6, 7, 8, 9, 10)\}$ .

1107930_1195924_ans_120d790896b2469ebd2571ea705007ab.jpg

Show that the relation $R$ in the set $Z$ of integers given by
$R=\left\{(a,b):2\ divides\ a-b\right\}$

$2\left \{ (a,b)=2 divides a=b \right \}$

where R is in set 2 of integers

a-a=0

2 divides a-a

$\Rightarrow (a,a)\epsilon R$

$\therefore R$ is reflexive

Let

$(a,a)\epsilon R$

$\therefore 2$ divides

$a-b\Rightarrow a-b=2n$

for same

$n\epsilon z\Rightarrow b-a=2(-n)$

$\Rightarrow$ 2 divides

$b-a=(b,a)\epsilon R$

R is symmetric

Let (a,b) &

$(b,c)\epsilon R$

2 divides a-b & b-c

$\therefore a-b=2n$ &

$b-c=2n_{2}$

for same

$n_{1},n_{2}\epsilon 2$

$=2n_{1}+2n_{2}=a-c=2(n_{1}+n_{2})$

$\therefore 2$ divides a-c

$(a,c)\epsilon R$

$\therefore (a,b)(b,c)\epsilon R\Rightarrow (a,c)\epsilon R$

$\therefore R$ is transitive

So, R is an equivalence relation

1188780_1292737_ans_ca987ac614334202bd9faf4a35de2514.jpg

If $p(x)=ix^2+5x+8$ , then $p{(-1)}$ =

We have,

$p(x)=ix^2+5x+8$

Put

$x=-1$ , we get

$p(-1)=i(-1)^2+5(-1)+8$

$p(-1)=i-5+8$

$p(-1)=i+3$

Hence, this is the answer.

The value of $x$ satisfying the equation $-3(x+6)=24$ ?

Given,

$-3(x+6)=24$

$(x+6)=\dfrac{-24}{3}$

$x+6=-8$

$x=-8-6$

$x=-14$

Let $f:R\rightarrow R$ be defined as $f(x)=10x+7$ . Find the function $g:R\rightarrow R$ such that $gof=fog=I_{R}$ .

1347429_1262716_ans_0d1936bb278b48dd97a7c62d29a4e074.jpg

If $f(x)=\cfrac{\sin{([x]\pi)}}{{x}^{2}+x+1}$ , where $[x]$ denotes the integral part of $x$ . Show that $f(x)$ is a constant function.

$f(x)=\dfrac{\sin([x]\pi )}{x^{2}+x+1}$

$[x]=$ integer

$\therefore [x]\pi =n\pi$

$\therefore \sin n\pi =0$

$\therefore f(x)=0 =$ constant function

If $A=\left\{1,2\right\}$ , form the set $A\times A\times A$ .

$\begin{array}{l} A\times A\times A\times =\left\{ { 1,2 } \right\} \times \left\{ { 1,2 } \right\} \times \left\{ { 1,2 } \right\} \\ =\{ \left( { 1,1,1 } \right) ,\left( { 1,1,2 } \right) ,\left( { 1,2,1 } \right) \\ \left( { 1,2,2 } \right) ,\left( { 2,1,1 } \right) ,\left( { 2,1,2 } \right) \\ \left( { 2,2,1 } \right) ,\left( { 2.2.2 } \right) \} \end{array}$

If $A = \left\{ {1,2} \right\},B = \left\{ {x:x \in N\,\,and\,\,{x^2} - 9 = 0} \right\},$ Find $A \times B.$

Given ,

$A=\left \{ 1,2 \right \}$

$B=\left \{ x:x \in N\,\, and \,\,x^{2}-9=0 \right \}$

Here,

$x^{2}-9=0$

$\Rightarrow x^{2}=9$

$\Rightarrow x=\pm 3\Rightarrow x=3$ or

$x=-3$ which is neglected since

$x\in N$

$\therefore B=\left \{ 3 \right \}$

$A\times B=\left \{ (1,3),(2,3) \right \}$

If $A\times B=\left\{\left(a,1\right),\left(b,3\right),\left(a,3\right),\left(b,1\right),\left(a,2\right),\left(b,2\right)\right\}$ .Find $A$ and $B$

Clearly,

$A$ is the set of all first entries in ordered pairs in

$A\times B$ and

$B$ is the set of all second entries in ordered pairs in

$A\times B$

$\therefore\,A=\left\{a,b\right\}$ and

$B=\left\{1,2,3\right\}$

Find $x$ and $y$ if $\left(x+3,5\right)=\left(6,2x+y\right)$

By the definition of equality of ordered pairs, we obtain

$\left(x+3,5\right)=\left(6,2x+y\right)$

$\Rightarrow\,x+3=6$ and

$5=2x+y$

$\Rightarrow\,x=6-3=3$ and

$y=5-2x$

$\Rightarrow\,x=3$ and

$y=5-2\times 3=5-6=-1$

$\therefore\,x=3,y=-1$

Let $R$ be relation from $Q$ to $Q$ defined by $R=\left\{(a,b):a,b\ \in \ Q\ and\ a-b\ \in \ Z\right\}$ .
Show that :
$(a,b)\ \in \ R$ implies that $(b,a)\ \in \ R$ .

$(a, b)\in R$

$\Rightarrow a-b$ is an integer, then

$b-a$ would also be an integer being

$-(a-b)$

$\Rightarrow b-a$ also

$\in Z\Rightarrow (b, a)\in R$ .

1207985_1392536_ans_4fe96a0bdf324bb5b7a50c24b75e7cd5.jpg

Find the relations between $a$ and $b$ , so tat the function $f$ defined by:
$f\left( x \right)=\begin{cases} ax+1, \\ bx+3, \end{cases}\begin{matrix} if\ x\le 3 \\ if\ x>3 \end{matrix}$
is continuous at $x=3$ .

Given the function

$f(x)$ is continuous at

$x=3$ .

Then we'll have,

$\lim\limits_{x\to 3-}f(x)=$

$\lim\limits_{x\to 3+}f(x)$

or,

$3a+1=3b+3$

or,

$3(a-b)=2$ .

This is the required relation between

$a,b$ .

If $A=\left\{a,b\right\}$ and $B=\left\{1,2,3\right\}$ find $\left(A\times B\right)\cap\left(B\times\,A\right)$

We have

$A=\left\{a,b\right\}$ and

$B=\left\{1,2,3\right\}$

$A\times \,B=\left\{\left(a,1\right),\left(a,2\right),\left(a,3\right),\left(b,1\right),\left(b,2\right),\left(b,3\right)\right\}$

$B\times \,A=\left\{\left(1,a\right),\left(1,b\right),\left(2,a\right),\left(2,b\right),\left(3,a\right),\left(3,b\right)\right\}$

$\left(A\times B\right)\cap\left(B\times\,A\right)$

$=\left\{\left(a,1\right),\left(a,2\right),\left(a,3\right),\left(b,1\right),\left(b,2\right),\left(b,3\right)\right\}\cap\left\{\left(1,a\right),\left(1,b\right),\left(2,a\right),\left(2,b\right),\left(3,a\right),\left(3,b\right)\right\}=\phi$

The cartesian product $A\times A$ has $9$ elements among which are found $\left(-1,0\right)$ and $\left(0,1\right)$ .Find the set $A$ and the remaining elements of $A\times A$

Since

$\left(-1,0\right)\in\,A\times A$ and

$\left(0,1\right)\in A\times A$

$\therefore\,\left(-1,0\right)\in\,A\times A\Rightarrow\,-1,0\in A$

and

$\left(0,1\right)\in\,A\times A\Rightarrow\,0,1\in A$

$\therefore\,-1,0,1\in A$

It is given that

$A\times A$ has

$9$ elements

$\therefore\,A$ has exactly has

$3$ elements.

Hence

$A=\left\{-1,0,1\right\}$

If $A=\left\{1,3,5,6\right\}$ and $B=\left\{2,4\right\}$ .Find $A\times\,B$ and $B\times \,A$

We have

$A=\left\{1,3,5,6\right\}$ and

$B=\left\{2,4\right\}$

$A\times B=\left\{\left(1,2\right),\left(1,4\right),\left(3,2\right),\left(3,4\right),\left(5,2\right),\left(5,4\right),\left(6,2\right),\left(6,4\right)\right\}$

$B\times A=\left\{\left(2,1\right),\left(2,3\right),\left(2,5\right),\left(2,6\right),\left(4,1\right),\left(4,3\right),\left(4,5\right),\left(4,6\right)\right\}$

Let $R$ be relation from $Q$ to $Q$ defined by $R=\left\{(a,b):a,b\ \in \ Q\ and\ a-b\ \in \ Z\right\}$ .
Show that :
$(a,b)\ \in \ R$ and $(b,c)\ \in R$ implies that $(a,c)\ \in \ R$ .

$(a, b)\in R\Rightarrow a-b\in Z$

$(b, c)\in R\Rightarrow b-c\in Z$

$\Rightarrow (a-c)=(a-b)+(b-c)$ being sum of two integers is also a integer

$\Rightarrow (a-c)\in z\Rightarrow (a, c)\in R$ .

1207986_1392538_ans_e2dd86cb2a154fa4a29a6dce3924154f.jpg

Find the values of $a$ and $b$ if $\left(3a-2,b+3\right)=\left(2a-1,3\right)$

By the definition of equality of ordered pairs, we obtain

$\left(3a-2,b+3\right)=\left(2a-1,3\right)$

$\Rightarrow\,3a-2=2a-1$

$\Rightarrow\,3a-2a=-1+2$

$\Rightarrow\,a=1$

Again

$b+3=3\Rightarrow\,b=3-3=0$

$\therefore\,a=1,b=0$

If $A=\left\{a,b\right\}$ and $B=\left\{1,2,3\right\}$ find $A\times A$ and $B\times\,B$

We have

$A=\left\{a,b\right\}$

$A\times \,A=\left\{\left(a,a\right),\left(b,b\right)\right\}$

We have

$B=\left\{1,2,3\right\}$

$B\times \,B=\left\{\left(1,1\right),\left(2,2\right),\left(3,3\right)\right\}$

List the relation R defined by $R=\left \{ (a,b):a\leq b^{3} \right \}$ in Roaster form. $a,b \in N$

Roaster form means representing the elements in an order.

$R=\{(a,b):a\leq b^3\}\\R=\{(1,1),(1,2),(2,2),(1,3),(2,3),......\}$

If $A=\left\{a,b\right\}$ and $B=\left\{1,2,3\right\}$ find $A\times B$ and $B\times\,A$

We have

$A=\left\{a,b\right\}$ and

$B=\left\{1,2,3\right\}$

$A\times \,B=\left\{\left(a,1\right),\left(a,2\right),\left(a,3\right),\left(b,1\right),\left(b,2\right),\left(b,3\right)\right\}$

$B\times \,A=\left\{\left(1,a\right),\left(1,b\right),\left(2,a\right),\left(2,b\right),\left(3,a\right),\left(3,b\right)\right\}$

Let $R=\left\{ \left( a,{ a }^{ 3 } \right) :\text{a is a prime number less than }\ 5 \right\}$ be a relation. Find the range of $R$ .

$R=\{(a,a^3):\,a$ is a prime number less than

$5\}$

Prime number less than

$5$ are

$2$ and

$3$ .

$\therefore$

$a=2,3$

$\therefore$

$R=\{(2,2^3),(3,3^3)\}=\{(2,8),(3,27)\}$

$\therefore$ The range of

$R$ is

$\{8,27\}$

Write the domain of the relation $R$ defined on the set $\mathbb{Z}$ of integers as follows:
$(a,b)\in R\Leftrightarrow {a}^{2}+{b}^{2}=25$

Given the relation defined on

$\mathbb{Z}$ is

$(a,b)\in R\Leftrightarrow {a}^{2}+{b}^{2}=25$ .

Since

$a,b$ are integers then the domain of the relation, i.e. the set of the integers satisfying the given equation, will be

$\left\{ 0,\pm 3,\pm 4,\pm 5 \right\}$ .

If $R=\left\{ (x,y):x+2y=8 \right\}$ is a relation on $N$ , then write the range of $R$ .

Given,

$x+2y=8$

or,

$y=\dfrac{8-x}{2}$

or,

$y=4-\dfrac{x}{2}$

Since the relation is on

$N$ , so

$x$ and

$y$ are both natural .

Thus values possible for

$x$ are

$2,4,6$ .

Which gives possible values of y to be

$3,2,1$ .

Thus Range of

$R$ is

$\{1,2,3\}$ .

If $R=\left\{ (x,y):{ x }^{ 2 }+{ y }^{ 2 }\le 4;x,y\in \mathbb{Z} \right\}$ is a relation on $\mathbb{Z}$ , write the domain of $R$ .

Given the relation

$R=\left\{ (x,y):{ x }^{ 2 }+{ y }^{ 2 }\le 4;x,y\in \mathbb{Z} \right\}$ .

Since

$a,b$ are integers then the domain of the relation, i.e. the set of integers satisfying the given equation will be

$R$ will be

$\left\{ 0,\pm 1,\pm 2 \right\}$ .

Write the identity relation on set $A=\left\{ a,b,c \right\}$ .

Given,

$A=\left\{ a,b,c \right\}$ .

Now the identity relation on

$A$ will be

$\left\{ \left( a,a \right) ,\left( b,b \right) ,\left( c,c \right) \right\}$ .

Find the domain and range of each of the following real value functions:
$f(x)= |x-1|$

We have,

$f(x)=|x-1|$ . Clearly,

$f(x)$ is defined for all

$x\in R$ .

So, domain

$(f)=R$

Also,

$f(x)=|x-1|\ge 0$ for all

$x\in R$ .

So, range

$(f)=[0, \infty]$

There are four men and six women on the city councils. If one council number is selected for a committee at random, how likely that it is a woman

Out of

$4$ men and

$6$ women one person can be chosen in

$^{10}{{C}_{1}}=10$ ways.

The number of ways of selecting

$1$ women out of 6 women

$=\ ^{6}{{C}_{1}}=6$ .

$\therefore$ Required probability

$=\dfrac{6}{10}=\dfrac{3}{5}$

Enter 1 if it is true else enter 0.
For the relation $\displaystyle R = \{ (-1, 1), (1, 1), (-2, 4), (2, 4), (3, 9)\}$ is dom $\displaystyle (R) = \{-2, -1, 1, 2, 3\}$ and range $\displaystyle (R) = \{1, 4,9\}$ .

1909779_1709456_ans_8b4b7f6f26274b4a8f5ae9123e9eeefb.jpeg

Let the relation $R$ be defined in $N$ by $aRb$ if $2a+3b=30$ . Then $R=$ ....

Given that,

$2a+3b=30$

$3b=30-2a$

$b=\dfrac{30-2a}{3}$

For

$a=3,\ b=8$

$a=6,\ b=6$

$a=9,\ b=4$

$a=12,\ b=2$

$R=\left\{ (3,8), (6,6), (9,4), (12,2) \right\}$

Let the relation $R$ be defined on the set $A=\left\{ 1,2,3,4,5 \right\}$ by $R={(a,b):a^2b^2 < 8}.$ Then $R$ is given by ______.

Given

$A=\left\{ 1,2,3,4,5 \right\}$

$R={(a,b):∣a^2−b^2∣ < 8}.$

$R=\left\{(1,1), (1,2), (2,1), (2,2), (2,3), (3,2), (3,3), (4,3), (3,4), (4,4), (5,5)\right\}$

For the set $A=\left\{1,2,3\right\}$ , define a relation $R$ in the set $A$ as follows:
$R=\left\{(1,1),(2,2),(3,3),(1,3)\right\}$
Write the ordered pairs to be added to $R$ to make it the smallest equivalence relation.

$(3,1)$ is the single ordered pair which needs to be added to

$R$ to make it the smallest equivalence relation.

Let $R$ relation defined on the set of natural number $N$ as follows:
$R=\left\{(x,y): x\in \ N, y\in \ N, 2x+y=41\right\}$ . Find the domain and rang of the relation $R$ . Also verify $R$ is reflexive, symmetric and transitive.

Given that,

$R=\left\{(x,y): x\in \ N, y\in \ N, 2x+y=41\right\}$ .

Domain

$=\left\{1,2,3,...20 \right\}$

Range

$=\left\{1,3,5,7... 39\right\}$

$R=\left\{(1,39),(2,37),(3,35),(19,3),(20,1)\right\}$

$R$ is not reflexive as

$(2,2)\notin R$

$21\times 2+2\neq 41$

$(1,39)\in R$ but

$(39,1)\in R (1,39)\in R$ but

$(39,1)\notin R$

So,

$R$ is not transitive

$(11,19)\in R, (19,3)\in R$

But

$(11,3)\in R$

Hence

$R$ is neither reflexive, nor symmetric and nor transitive.

If $R$ be a relation $<$ from $A=\{1,2,3,4\}$ to $B=\{1,3,5\}$ i.e., $(a,b) \in$ R if $a<b$ , then find $ROR^{-1}$

$A=\left\{ 1,2,3,4 \right\}$

$B=\left\{ 1,3,5 \right\}$

$R=\left\{ \left( 1,3 \right) ,\left( 1,5 \right) ,\left( 2,3 \right) ,\left( 2,5 \right) ,\left( 3,5 \right) ,\left( 4,5 \right) \right\}$

${ R }^{ -1 }=\left\{ \left( 3,1 \right) ,\left( 5,1 \right) ,\left( 3,2 \right) ,\left( 5,2 \right) ,\left( 5,3 \right) ,\left( 5,4 \right) \right\}$

For composition

$RO{ R }^{ -1 }$ , we will pick up an element of

${ R }^{ -1 }$ first then of

$R$ .
eg:

$\left( 3,1 \right) \in { R }^{ -1 }$

$\left( 1,3 \right) \in R$

$\Rightarrow \left( 3,3 \right) \in RO{ R }^{ -1 }$

$\therefore RO{ R }^{ -1 }=\left\{ \left( 3,3 \right) ,\left( 3,5 \right) ,\left( 5,3 \right) ,\left( 5,5 \right) \right\}$

Let $A=\left \{ 1, 2, 3, 4 \right \}$ and $B=\left \{ x, y, z \right \}$ . Consider the subset $R=\left \{ (1, x),(1,y),(2,z),(3,x) \right \}$ of $A\times B.$ Is R, a relation from A to B? If yes, find domain and range of R. Draw arrow diagram of R and represent R in a tabular form.

Since every subset of

$A\times B$ is a relation from A to B, therefore, R is also a relation
from A to B.
Domain of

$R=\left \{ 1, 2, 3 \right \}$ Range of R

$=\left \{ x, y, z \right \}$
Arrow diagram of R in the tabular form, we construct a box with the help of

$\left \{ n(A)+2 \right \}$ horizontal lines and

$\left \{ n(B)+2 \right \}$ vertical lines. In this case, we draw 6 horizontal lines and 5 vertical lines to get 20 boxes

$\left \{ n(A)+1 \right \}\times \left \{ n(B)+1 \right \}$ in general). Place R in the extreme left top and elements of A below R in a vertical line. Write elements of B in the first horizontal line as shown.
As

$(1, x)\in R,$ place 1 in the row containing x. As

$(1, z)\notin R,$ place 0 on the row containing 1 and column containing z. Complete the table corresponding to other pairs also. Here, we use the convention that if

$(a, b)\in R,$ we write '1' in the row containing a and column containing b and if

$(a, b)\notin R,$ we write '0' in the corresponding box.

If $f: R\rightarrow R$ defined by $f(x)=x^2+1$ , then the values of $f^{-1}(17)$ are a and b, then a+b =?

Given,

$f(x)=x^2+1$

$f^{-1}(x^2+1)=x$
Clearly for

$x^2+1=17 \Rightarrow x=4,-4$

$\Rightarrow f^{-1}(17)=4,-4$
That is, let

$a=4$ and

$b=-4,$

$\Rightarrow a+b=0$
So ,the correct answer is

$0$

For each set of ordered pairs below, state whether it is a function or not give reasons
(i){(3, 2), (4, 2), (5, 2)}
if a function enter 1 else 0

A function

$f$ from

$X$ to

$Y$ is a subset of the Cartesian product

$X$

$Y$ subject to the following condition: every element of

$X$ is the first component of one and only one ordered pair in the subset. In other words, for every

$x$ in

$X$ , there is exactly one element

$y$ such that the ordered pair

$(x, y)$ is contained in the subset defining the function

$f$ .

And given ordered pairs satisfy all the condition

If $x, y\in \left [ 0, 2\pi \right ]$ , then find the total number of ordered pairs $\left ( x, y \right )$ satisfying the equation $\sin x\cos y=1$

$\sin x\cos y=1$

Possible ways are

$\Rightarrow$

$\sin x=1$ ,

$\cos y=1$ or

$\sin x=-1$ ,

$\cos y=-1$

$\sin x=1$ ,

$\cos y=1$ ; hence,

$x=\pi /2$ ,

$y=0, 2\pi$

$\sin x=-1$ ,

$\cos y=-1$ ; hence,

$x=3\pi /2$ ,

$y=\pi$

Thus, the possible ordered pairs are

$\displaystyle \left ( \frac{\pi }{2}, 0 \right )$ ,

$\displaystyle \left ( \frac{\pi }{2}, 2\pi \right )$ and

$\displaystyle \left ( \frac{3\pi }{2}, \pi \right )$

$\therefore$ Total number of ordered pairs is

$\Rightarrow 3$

Let a relation R be defined by

$\displaystyle R= \left \{ \left ( 4,5 \right ), \left ( 1,4 \right ), \left ( 4,6 \right ), \left ( 7,6 \right ), \left ( 3,7 \right ) \right \}$ then find $\displaystyle R o R$

Given

$R=\{(4,5), (1,4), (4,6), (7,6), (3,7)\}$

$\displaystyle \left ( 1,4 \right )\epsilon R, \left ( 4,5 \right )\epsilon R$

$\therefore \left ( 1,5 \right )\epsilon R o R$

Similarly

$(3,6)$ and

$(1,6)$

$\displaystyle \epsilon R OR.$

$\displaystyle \therefore R o R= \left \{ \left ( 1,5 \right ), \left ( 3,6 \right ), \left ( 1,6 \right ) \right \}.$

Given a relation $\displaystyle R=\left \{ \left ( 1,2 \right ),\left ( 2,3 \right ) \right \}$ on the set of natural numbers,Find the minimum number of ordered pairs should be added so that the enlarged relation is symmetric,transitive and reflexive.

For the relation R to be reflexive, it is necessary that

$\displaystyle \left ( n,n \right )\epsilon R$ for every

$\displaystyle n \epsilon N$ that is,R must have all pairs

$\displaystyle \left ( 1,1 \right ),\left ( 2,2 \right ),\left ( 3,3 \right )$

For R to be symmetric, it must contain the pair (2,1) and (3,2) since the pairs (1,2) and (2,3) are already present.

For R to be transitive,it must contain the pair(1,3) since (1,2) and (2,3) are already there. We must then also include the pair (3,1) for symmetry.

Hence the relation

$\displaystyle R'$ obtained from R by adding a minimum number of ordered pairs to R to make it an equivalence relation is

$\displaystyle R'=\left \{ \left ( 1,2 \right ),\left ( 2,1 \right ),\left ( 2,3 \right ),\left ( 3,2 \right ),\left ( 1,3 \right ),\left ( 3,1 \right ) ,\left ( 1,1 \right ),\left ( 2,2 \right ),\left ( 3,3 \right ) \right \}$

Use elements of set P = {2 , 3} to find the number of all possible ordered pairs.

Possible ordered pairs formed using the set

$P = \{2,3\}$ are

$(2,2), (2,3), (3,3), (3,2)$ .

Let $A$ and $B$ be two sets such that $\displaystyle A\times B$ Consists of $6$ elements. If three elements of $\displaystyle A\times B$ are $\displaystyle \left ( 1,4 \right ),\left ( 2,6 \right ),\left ( 3,6 \right )$ then $\:B\times A$ . is $=\left \{ \left ( a,1 \right ),\left ( 6,1 \right ),\left ( 4,d \right ),\left ( c,2 \right ) ,\left ( 4,3 \right )\left ( b,3 \right )\right \}$ . Find $a+b+c+d$

From the given ordered pairs of

$\displaystyle A\times B$ we conclude that

$\displaystyle A=\left \{ 1,2,3 \right \},B=\left \{ 4,6 \right \}$

Since

$\displaystyle A\times B$ has six elements i.e.

$3\times2$ ,

$A$ and

$B$ are as stated above.

$\displaystyle \therefore A\times B=\left \{ \left ( 1,4 \right ),\left ( 1,6 \right ),\left ( 2,4 \right ),\left ( 2,6 \right ),\left ( 3,4 \right ),\left ( 3,6 \right ) \right \}$

By reversing the above ordered pairs, we get

$\displaystyle B\times A=\left \{ \left ( 4,1 \right ),\left ( 6,1 \right ),\left ( 4,2 \right ),\left ( 6,2 \right ) ,\left ( 4,3 \right )\left ( 6,3 \right )\right \}.$

$\Rightarrow a+b+c+d=18$

Given $A = \{5, 6, 7\}$ and $B = \{3, 4\}$ . Form all possible ordered pairs and write the total number of ordered pairs formed so that both the components are from B.

The elements in

$A=[5,6,7]$ set are totally different from the elements in set

$B= [3,4]$

so there is no way to convert this set into subset lookslike as set

$B$

So, answer is

$0$

Let $A = \{1, 2\}$ and $B = \{2, 3, 4\}$ .
If $B \times B= \{(2,2), (2,3),(2, 4), (3,2), (3,3), (3,4) , (4,2), (4,3), (4,4)\}$ .
Please enter $1$ or else $0$ .

Cartesian product of two sets is obtained by pairing each element of first set with each element of second set.

Hence,

$B \times B= \{(2,2), (2,3),(2, 4), (3,2), (3,3), (3,4) , (4,2), (4,3), (4,4)\}$ .

$B \times A=\{(2,1),(2,2), (3,1), (3,2),(4,1),(4,2)\}$ . If true enter $1$ , or else enter $0$ .

Cartesian product of two sets is found by taking or pairing each element of first set with each element of second set.
Hence,

$B \times A = \{(2,1), (2,2),(3, 1), (3,2), (4,1), (4,2)\}$

Let $A = \{1, 2\}$ and $B = \{2, 3, 4\}$ , then $A \times A = \{ (1,1), (1,2),(2,1), (2,2)\}$ . If true enter $1$ , or else enter $0$ .

Cartesian product of two sets is found by taking or pairing each element of first set with each element of second set.
Hence,

$A \times A = \{ (1,1), (1,2),(2,1), (2,2)\}$ .

Let $A = \{1, 2\}$ and $B = \{2, 3, 4\}$ , then $A \times B = \{ (1,2), (1,3),(1, 4), (2,2), (2,3), (2,4) \}$ . If true enter $1$ , or else enter $0$ .

Cartesian product of two sets is found by taking or pairing each element of first set with each element of second set.

Hence,

$A \times B = \{ (1,2), (1,3),(1, 4), (2,2), (2,3), (2,4) \}$ .

Given $A = \{5, 6, 7\}$ and $B = \{3, 4\}$ . Form all possible ordered pairs and write the total number of ordered pairs formed. So that :
Both the components are from A
Find the total number of such pairs.

Possible ordered pairs with both components from A are

$(5,5), (5,6), (5,7), (6,5), (6,6), (6,7), (7,5), (7,6), (7,7)$

Thus the total number of such ordered pairs is 9.

If $A = \{x, y\}$ and $B = \{y, x\}$ ; then $A \times B = ?$
What are the number of elements of such set?

Cartesian product of two sets is obtained by pairing each element of first set with each element of second set.

Given :

$A = \{x, y\}$ and

$B = \{y, x\}$

Then,

$A \times B = \{ (x,y), (x,x),(y,y), (y,x) \}$

Thus there are

$4$ elements in the above set

If $A =$ { $x$ : $x \,\,\in\,\, W$ , $3 \leq x < 6$ }, $B = \{3, 5, 7\}$ and $C = \{2, 4\}$ ; find : $(A - B) \times C$

$A = \{x : x$

$\in$

$w, 3$

$\le$

$x < 6\}$

Set

$A =$ {

$3,4,5$ }

$B = \{3, 5, 7\}$

$C = \{2, 4\}$

$(A - B)$

$\times$

$C =\{(4, 2), (4,4)\}$

Now,

$A - B$ , removing the elements of B which are in A

$A - B =$ {

$4$ } as

$3, 4$ are common elements of A and B.

Multiplying two sets means, finding all possible sets by multiplying each element of first set with each element of the other.

Hence,

$(A - B) \times C =$ {

$(4,2), (4, 4)$ }

If P = {a, b, c} and Q = {b, c, d} ; find number of elements in $(P \cap Q) \times P$

Given,

P = {a, b, c} and Q = {b, c, d}

Intersection of two sets has the elements which are common in both the sets.

So,

$P \cap Q = \{b,c\}$

Cartesian product of two sets is found by forming ordered pairs.

So,

$(P \cap Q) \times P = \{ (b,a), (b,b), (b,c), (c,a), (c,b), (c,c)\}$ .

$n((P \cap Q) \times P)=6$

Given $A = \{ x \in W : 7 \le x \leq 10 \},$ find $A \times A.$

Given :

$A = \{ x \in W : 7 \le x \leq 10 \},$

$A =\{ 7,8,9,10 \}$

Product of two sets is found by forming ordered pairs of elements of both sets.

So,

$A \times A =$

$\{(7,7), (7,8), (7,9), (7,10),(8,7), (8,8),(8,9), (8,10),(9,7), (9,8) , (9,9), (9, 10),(10,7), (10,8), (10,9), (10,10) \}$

If $M=\{x : x \,\,\epsilon\,\, N, 1 < x \leq 4\}$ and $N = \{y : y\,\,\epsilon\,\,W, y < 3\}$ ;
Find: $n(N$ $\times$ $N$ ).

Cartesian product of two sets is obtained by pairing each element of first set with each element of second set.
Now set

$M = \{2,3,4\}$
and set

$N = \{ 0,1,2 \}$
Now,

$N \times N= \{(0,0),(0,1),(0,2),(1,0),(1,1)(1,2),(2,0),(2,1),(2,2)\}$

Hence,

$n(N\times N)=9$

$(M \cup N)\times (M - N)$

Union of two sets has all the elements of both the sets.

So,

$M \cup N =\{0,1,2,3\}$

$N - M$ means subtracting the common elements of

$M$ and

$N$ from

$N$ .

So,

$N - M = \{3\}$
Cartesian product of two sets is found by forming ordered pairs, where

$x$ -coordinate is from first set and

$y$ -coordinate is from second set.

So,

$(M \cup N) \times (N - M) =\{ (0,3), (1,3), (2,3), (3,3)\}$

If $M=\{x : x \,\,\epsilon\,\, N, 1 < x \leq 4\}$ and $N = \{y : y\,\,\epsilon\,\,W, y < 3\}$ ; find M $\times$ N :
Also find the number of such ordered pairs.

$M=\left\{ x:x\in N,1<x\le 4 \right\}$

Since,

$x\in N$ and

$1<x\le 4$

$\therefore x=2,3,4$

$\therefore M=\left\{ 2,3,4 \right\} \longrightarrow 1$

$N=\left\{ y:y\in W,y<3 \right\}$

$\therefore N=\left\{ 0,1,2 \right\}$ [

$\because y\in$ Whole Number

$\Longrightarrow y$ can be

$0,1,2$ ]

$M\times N=$ Every element of set

$M$ will be paired with each element of set

$N$

$M\times N=\left\{ \left( 2,0 \right) ,\left( 2,1 \right) ,\left( 2,2 \right) ,\left( 3,0 \right) ,\left( 3,1 \right) ,\left( 3,2 \right) ,\left( 4,0 \right) ,\left( 4,1 \right) ,\left( 4.2 \right) \right\}$

Thus the number of such ordered pairs are

$9$ .

$\displaystyle R_{4}= \left \{ \left ( 1,3 \right ), \left ( 2,5 \right ), \left ( 4,7 \right ), \left ( 5,9 \right ), \left ( 3,1 \right ) \right \}$
Is it a mapping?
If it is true enter 1, else enter 0.

$R_{4}$ is a mapping because every element in

$X$ is mapped to elements in

$Y$

Hence, the correct answer is

$1$ .

Given ordered pairs : (5, 4), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6), (6, 7), (8, 4), (8, 5), (8, 6), (8,8).
Use these ordered pairs to find the following relation :
$R_4$ = "is greater than"

For the Relation

${ R }_{ 4 }$ we have to find the ordered pairs in the list where

$x$ co-ordinate is greater than

$y$ co-ordinate.

So,

$(5,4), (6,4) , (6,5), (8,4), (8,5), (8,6)$ have their

$x$ co-ordinate greater than

$y$ co-ordinate.

Hence,

${ R }_{ 4 } = (5,4), (6,4) , (6,5), (8,4), (8,5), (8,6)$
Domain is the set of all

$x$ co-ordinates

$= \{5,6,8\}$
Range is the set of all

$y$ co-ordinates

$= \{4,5,6 \}$ .

Given ordered pairs : (5, 4), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6), (6, 7), (8, 4), (8, 5), (8, 6), (8,8).
Find the ordered pair(s) from above given list which satisfies the following relation.
$R_3$ = "is one less than"

For the relation

${ R }_{ 3}$ we have to find the ordered pairs in the list where

$x$ co-ordinate is one less than

$y$ co-ordinate.

So,

$(5,6), (6,7)$ have their

$x$ co-ordinate one less than

$y$ co-ordinate.

Hence,

${ R }_{ 3 } = \{(5,6), (6,7) \}$
Domain is the set of all

$x$ co-ordinates

$=\{ 5,6 \}$
Range is the set of all

$y$ co-ordinates

$=\{ 6,7\}$ .

Given A = {8, 9, 10, 12}, B = {2, 3, 4, 5} and the relation R from A to B means : "is multiple of" :
Write the domain range of relation R.

When R is a relation from A to B, then the Domain are the elements from the set A and range are the elements from the Set B satisfying the condition of the relation.

Hence, Domain

$= \{ 8,9,10,12 \}$ and Range

$= \{ 2,3,4,5 \}$ as they satisfy the given relation.

Let $\displaystyle A=\left \{ 1,2,3,4,5 \right \},B=\left \{ 1,3,5,7,9,10,17,26,30 \right \}$ Which of the following sets of ordered pairs are (i) relations (ii) functions (iii) neither,from A to B
(a) $\displaystyle R_{1}=\left \{ \left ( 1,3 \right ),\left ( 3,3 \right ),\left ( 5,17 \right ) ,\left ( 1,2 \right )\right \}$
(b) $\displaystyle R_{2}=\left \{ \left ( 3,7 \right ),\left ( 4,5 \right ),\left ( 5,30 \right ) \right \}$
(c) $\displaystyle R_{3}=\left \{ \left ( x,y \right ):x \epsilon A,y \epsilon B,x+y=8 \right \}$
(d) $\displaystyle R_{4}=\left \{ \left ( x,y \right ):x \epsilon A,y\epsilon B,y=x^{2}+1 \right \}$ (e) $\displaystyle R_{5}=\left \{ \left ( x,y \right ):x\epsilon A,y\epsilon B,y=2x-1 \right \}$

$\displaystyle R_{1}\neq R$ as

$\displaystyle \left ( 1,2 \right )\epsilon A\times B,$

$R_{1}\neq f$ as

$2$ or

$4 \epsilon A$ has no image in B,

$\displaystyle R_{2}=R$ but

$\displaystyle R_{2}\neq f$ as in

$\displaystyle R_{1}$

$\displaystyle R_{3}=\left \{ \left ( 1,7 \right ),\left ( 3,5 \right ),\left ( 5,3 \right ) \right \}$

$\displaystyle x+y=8,x \epsilon A, y\epsilon B$

$\displaystyle \therefore R_{3}=R$ being a subset of

$\displaystyle A\times B$ but

$\displaystyle R_{3}\neq f$ as in

$\displaystyle R_{1}$

$\displaystyle R_{4}=\left \{ \left ( 2,5 \right ),\left ( 3,10 \right )\left ( 4,17 \right ),\left ( 5,26 \right )\right \}$

$\displaystyle \because y=x^{2}+1,x \epsilon A,y\epsilon B$

$\displaystyle R_{4}=R$ but

$\displaystyle R_{4}\neq f$ as in

$\displaystyle R_{1}$

$\displaystyle R_{5}=\left \{ \left ( 1,1 \right ),\left ( 2,3 \right ),\left ( 3,5 \right ),\left ( 4,7 \right ),\left ( 5,9 \right ) \right \}$

$\displaystyle y=2x-1$

$\displaystyle R_{5}=R$ and

$\displaystyle R_{5}=f$ as each element

$\displaystyle \epsilon A$ has a unique image in B.

Find the number of ordered pairs in $\displaystyle R^{-1} o R.$

We first find

$\displaystyle R^{-1}.$ We have

$\displaystyle R^{-1}= \left \{ \left ( 5,4 \right ), \left ( 4,1 \right ), \left ( 6,4 \right ), \left ( 6,7 \right ), \left ( 7,3 \right ) \right \}$
We now obtain the elements of

$\displaystyle R^{-1}$ o R. We first pick the element of R and then of

$\displaystyle R^{-1}$
Since (4,5)

$\displaystyle \epsilon$ R and (5,4)

$\displaystyle \epsilon R^{1},$ we have (4,4)

$\displaystyle \epsilon R^{1},$ o R.Similarly

$\displaystyle \left ( 1,4 \right )\epsilon R, \left ( 4,1 \right )\epsilon R^{-1}\Rightarrow \left ( 1,1 \right )\epsilon R^{-1}\:o\:R,$

$\displaystyle \left ( 4,6 \right )\epsilon R, \left ( 6,4 \right )\epsilon R^{-1}\Rightarrow \left ( 4,4 \right )\epsilon R^{-1}\:o\:R,$

$\displaystyle \left ( 4,6 \right )\epsilon R, \left ( 6,7 \right )\epsilon R^{-1}\Rightarrow \left ( 4,7 \right )\epsilon R^{-1}\:o\:R,$

$\displaystyle \left ( 7,6 \right )\epsilon R, \left ( 6,4 \right )\epsilon R^{-1}\Rightarrow \left ( 7,4 \right )\epsilon R^{-1}\:o\:R,$

$\displaystyle \left ( 7,6 \right )\epsilon R, \left ( 6,7 \right )\epsilon R^{-1}\Rightarrow \left ( 7,7 \right )\epsilon R^{-1}\:o\:R,$

$\displaystyle \left ( 3,7 \right )\epsilon R, \left ( 7,3 \right )\epsilon R^{-1}\Rightarrow \left ( 3,3 \right )\epsilon R^{-1}\:o\:R,$
Hence

$\displaystyle R^{-1}\:o\:R= \left \{ \left ( 1,1 \right ), \left ( 4,4 \right ), \left ( 4,7 \right ), \left ( 7,4 \right ), \left ( 7,7 \right ), \left ( 3,3 \right ) \right \}$

Write the relation, represented by the arrow diagram given below, in roster form. Also, write the domain and range of the relation.

For the required Relation we have to find the ordered pairs where

$x$ co-ordinate is from set A which is mapped to the

$y$ co-ordinate from set B.

So, the relation is

$\{(5,a), (5,b) , (7,b), (7,c), (9,a), (9,b), (9,c)\}$

Domain is the set of all

$x$ co-ordinates

$=\{5,7,9\}$
Range is the set of all

$y$ co-ordinates

$=\{ a,b,c\}$ .

Given ordered pairs : (5, 4), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6), (6, 7), (8, 4), (8, 5), (8, 6), (8,8).
Use these ordered pairs to find the following relation :
$R_2$ = "is equal to"

For the Relation

${ R }_{ 2 }$ we have to find the ordered pairs in the list where

$x$ co-ordinate is equal to

$y$ co-ordinate.

So,

$(5,5), (6,6), (8,8)$ have their

$x$ co-ordinate equal to

$y$ co-ordinate.

Hence,

${ R }_{2 } = \{(5,5), (6,6), (8,8)\}$
Domain is the set of all

$x$ co-ordinates

$=\{ 5,6,8 \}$
Range is the set of all

$y$ co-ordinates

$= \{5,6,8 \}$ .

Given ordered pairs : (5, 4), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6), (6, 7), (8, 4), (8, 5), (8, 6), (8,8).
Use these ordered pairs to find the following relation :
$R_1$ = "is less than"

For the Relation

${ R }_{ 1 }$ we have to find the ordered pairs in the list where

$x$ co-ordinate is less than

$y$ co-ordinate.

So,

$(5,6), (6,7)$ have their

$x$ co-ordinate less than

$y$ co-ordinate.

Hence,

${ R }_{ 1 } = \{(5,6), (6,7) \}$

Let $f$ be a function from the set of positive integers to the set of real number such that
(i) $f(1)=1$
(ii) $\displaystyle \sum_{r=1}^{n} rf(r) =n(n+1)f(n),\forall n\geq 2$
then find the value of $2126 f(1063)$

$f(1)+2f(2)+3f(3)+...+n(f(n))=n(n+1)f(n)$

$\Rightarrow f(1)+2(2)+3(3)+..+(n+1)f(n+1)=(n+1)(n+2)f(n+1)$

$\Rightarrow \displaystyle n(n+1)f(n)=(n+1)^{2}f(n+1)\Rightarrow nf(n)=(n+1)f(n+1)$
ie,

$2f(2)=3f(3)=....=n(f(n))$
ie,

$\displaystyle f(n)=\frac{1}{n}$

$2126\:f(1063)=2$

Let $\displaystyle f\left ( x \right )=\frac{9^{x}}{9^{x}+3}$ . Show $f(x)+f(1-x)=1$ , and hence evaluate $\displaystyle f\left ( \frac{1}{1996} \right )+f\left ( \frac{2}{1996} \right )+f\left ( \frac{3}{1996} \right )+...+f\left ( \frac{1995}{1996} \right )$ .

$\displaystyle f\left ( x \right )=\frac{9^{x}}{9^{x}+3}$ (i)
and

$\displaystyle f\left ( 1-x \right )=\frac{9^{1-x}}{9^{1-x}+3}$

$\Rightarrow$

$\displaystyle f\left ( 1-x \right )=\frac{\frac{9}{9^{x}}}{\frac{9}{9^{x}}+3}=\frac{9}{9+3.9^{x}}$

$\displaystyle f\left ( 1-x \right )=\frac{9}{3\left ( 3+9^{x} \right )}$ (ii)
Adding Eqs. (i) and (ii), we get

$\displaystyle f\left ( x \right )+f\left ( 1-x \right )=\frac{9^{x}}{9^{x}+3}+\frac{9}{3\left ( 3+9^{x} \right )}$

$\displaystyle =\frac{3.9^{x}+9}{3\left ( 9^{x}+3 \right )}=\frac{3\left ( 9^{x}+3 \right )}{3\left ( 9^{x}+3 \right )}$

$\therefore f(x)+f(1-x)=1$ (iii)
Now, putting

$\displaystyle x=\frac{1}{1996}, \frac{2}{1996}, \frac{3}{1996}, ..., \frac{998}{1996}$ in Eq. (iii), we get

$\displaystyle f\left ( \frac{1}{1996} \right )+f\left ( \frac{1995}{1996} \right )=1$

$\Rightarrow$

$\displaystyle f\left ( \frac{2}{1996} \right )+f\left ( \frac{1994}{1996} \right )=1$

$\Rightarrow$

$\displaystyle f\left ( \frac{3}{1996} \right )+f\left ( \frac{1993}{1996} \right )=1$
..........
..........

$\Rightarrow$

$\displaystyle f\left ( \frac{997}{1996} \right )+f\left ( \frac{999}{1996} \right )=1$

$\Rightarrow$

$\displaystyle f\left ( \frac{998}{1996} \right )+f\left ( \frac{998}{1996} \right )=1$
or

$\displaystyle f\left ( \frac{998}{1996} \right )=\frac{1}{2}$
Adding all the above expression, we get

$\displaystyle f\left ( \frac{1}{1996} \right )+f\left ( \frac{2}{1996} \right )+...+f\left ( \frac{1995}{1996} \right )=\left ( 1+1+1+...\text{upto} \, 997\, \text{times} \right )+\frac{1}{2}$

$\displaystyle =997+\frac{1}{2}$

$=997.5$

Which of the following are identity relations of set $A = \{1, 2, 3\}$
1. $\displaystyle R_{1}=\left \{ (1,1),(2,2) \right \}$
2. $R_{2}=\left \{ (1,1),(2,2)(3,3),(1,3) \right \}$
3. $R_{3}=\left \{ (1,1),(2,2),(3,3) \right \}$

The relation

$\displaystyle R_{3}$ is identity relation on set

$A$ .

$\displaystyle R_{1}$ is not identity relation on set

$A$ as

$(3,3)\not\in R_1$ .

$\displaystyle R_{1}$

$\displaystyle R_{2}$ is not identity relation on set

$A$ as

$\displaystyle (1,3)\in R_{2}$

Let $A=\{1, 3, 5, 7\}$ and $B = \{2, 4, 6, 8\}$ be two sets $R$ be a relation from $A$ to $B$ defined by the phrase $\displaystyle “(x,y)\in R\Rightarrow x>y".$ Find relation $R$ and its domain and range.

Under relation

$R$ , we have

$3R2, 5R2, 5R4, 7R4,$ and

$7R6$
i.e.

$R = \{(3,2),(5,2),(5,4),(7,2),(7,6)\}$

$\displaystyle \therefore$ Domain

$(R) = \{3, 5, 7 \}$ and range

$= \{2, 4, 6\}$ .

Define a sequence $\left<f_0(x),\,f_1(x),\,f_2(x),\dots\right>$ of functions by $f_0(x)=1,\;f_1(x)=x,\; \ \ \begin{pmatrix}f_n(x)\end{pmatrix}^2-1=f_{n+1}(x)f_{n-1}(x)$ , for $n\,\ge\,1$ .
Prove that each $f_n(x)$ is a polynomial with integer coefficients.

We have,

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;f_n^2(x)-f_{n-1}(x)f_{n+1}(x)=1=f_{n-1}^2(x)-f_{n-2}(x)f_n(x)$ .

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\therefore f_n(x)\begin{pmatrix}f_n(x)+f_{n-2}(x)\end{pmatrix}=f_{n-1}\begin{pmatrix}f_{n-1}(x)+f_{n+1}(x)\end{pmatrix}$ .
We write this as

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\displaystyle\frac{f_{n-1}(x)+f_{n+1}(x)}{f_n(x)}=\displaystyle\frac{f_{n-2}(x)+f_n(x)}{f_{n-1}(x)}$ .
Using induction, we get

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\displaystyle\frac{f_{n-1}(x)+f_{n+1}(x)}{f_n(x)}=\displaystyle\frac{f_0(x)+f_2(x)}{f_1(x)}$ .
Observe that

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;f_2(x)=\displaystyle\frac{f_1^2(x)-1}{f_0(x)}=x^2-1$ .
Hence

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\displaystyle\frac{f_{n-1}(x)+f_{n+1}(x)}{f_n(x)}=\displaystyle\frac{1+(x^2-1)}{x}=x$ .
Thus we obtain

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;f_{n+1}(x)=xf_n(x)-f_{n-1}(x)$ .
Since

$f_0(x),\,f_1(x)\;and\;f_2(x)$ are polynomials with integer coefficients, induction again shows that

$f_n(x)$ is a polynomial with integer coefficients.
Note: We can get

$f_n(x)$ explicitly:

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;f_n(x)=x^n-\begin{pmatrix}n-1 \\ 1\end{pmatrix}x^{n-2}+\begin{pmatrix}n-2 \\ 2\end{pmatrix}x^{n-4}-\begin{pmatrix}n-3 \\ 3\end{pmatrix}x^{n-6}+\dots$

Solve the
$2\dfrac{1}{5}-\dfrac{-1}{3}$

$R = \{(a, b) : a, b\displaystyle \in N; a = b^{2}\}$
(i) It can be seen that

$\displaystyle 2\in N$ ; however

$\displaystyle 2 \neq 2^{2} = 4$
Therefore the statement

$(a, a) \displaystyle \in R$ for all

$\displaystyle a\in N$ is not true

(ii) It can be seen that

$(9, 3) \displaystyle \in N$ because

$9, 3\displaystyle \in N$ and

$9 = \displaystyle 3^{2}$
Now

$\displaystyle 3\neq 9^{2}=81$

$\therefore (3, 9) \displaystyle \notin N$
Therefore the statement

$\displaystyle (a, b) \in R$ implies

$\displaystyle (b, a) \in R$ is not true.

(iii) It can be seen that

$(9, 3) \displaystyle \in R , (16, 4) \displaystyle \in R$ because

$9, 3, 16, 4 \displaystyle \in N$ and

$9 = \displaystyle 3^{2}$ and

$16 = \displaystyle 4^{2}$
Now

$\displaystyle 9\neq 4^{2}=16$

$\therefore \displaystyle (9,4)\notin N$
Therefore the statement

$\displaystyle \left ( a,b \right )\in R,\left ( b,c \right )\in R$ implies

$\displaystyle \left ( a,c \right )\in R$ is not true

The cartesian product $\displaystyle A\times A$ has $9$ elements among which are found $(-1, 0)$ and $(0, 1)$ . Find the set $A$ and the remaining elements of $\displaystyle A\times A$

We know that if

$n(A) = p$ and

$n(B) = q$ , then n(

$\displaystyle A\times B) = n(A) \times n(B)=pq$

$\displaystyle \therefore n\left ( A\times A \right )=n\left ( A \right )\times n\left ( A \right )$
It is given that

$\displaystyle n\left ( A\times A \right )=9$

$\displaystyle \therefore n\left ( A \right )\times n\left ( A \right )=9$

$\displaystyle \Rightarrow n\left ( A \right )=3$
The ordered pairs

$(-1, 0)$ and

$(0, 1)$ are two of the nine elements of

$\displaystyle A\times A$

Now,

$\displaystyle A\times A = \{(a, a): a\in A\}$
Therefore

$-1, 0$ and

$1$ are elements of

$A$
Since

$n(A) = 3$ , so set

$A = \{-1, 0, 1\}$
The remaining elements of set

$\displaystyle A\times A$ are

$(-1, -1), (-1, 1), (0, -1), (0, 0), (1, -1), (1, 0)$ and

$(1, 1)$

Let $A = \{1, 2, 3, .... , 14\}$ . Define a relation $R$ from $A$ to $A$ by $R = \{(x, y): 3x - y = 0$ where $x, y \displaystyle \in A\}$ . Write down its domain, co-domain and range.

The relation

$R$ from

$A$ to

$A$ is given as

$R = \{(x, y): 3x - y = 0 ; x, y\in A\}$

i.e.,

$R = \{(x, y): 3x = y ; x, y\in A\}$

$\displaystyle \therefore R = \{(1, 3), (2, 6), (3, 9), (4, 12)\}$

The domain of

$R$ is the set of all first elements of the ordered pairs in the relation

$\displaystyle \therefore$ Domain of

$R = \{1, 2, 3, 4\}$

The whole set

$A$ is the co-domain of the relation

$R$

$\displaystyle \therefore$ Codomain of

$R = A = \{1, 2, 3,....., 14\}$

The range of

$R$ is the set of all second elements of the ordered pairs in the relation.

$\displaystyle \therefore$ Range of

$R = \{3, 6, 9, 12\}$

If $G = \{7, 8\}$ and $H = \{5, 4, 2\}$ , find $\displaystyle G\times H$ and $\displaystyle H\times G$ .

$G = \{7, 8\}$ and

$H = \{5, 4, 2\}$
We know that the Cartesian product of

$\displaystyle P\times Q$ of two non-empty sets

$P$ and

$Q$ is defined as

$\displaystyle P\times Q = \{(p, q): \displaystyle p\in P,q\in Q \}$

$\displaystyle \therefore G\times H = \{(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)\}$
and

$\displaystyle H\times G = \{(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)\}$

Define a relation $R$ on the set $N$ of natural numbers by $R = \{(x, y): y = x + 5, x$ is a natural number less than $4; x, \displaystyle y\in N\}$ . Depict this relationship using roster form. Write down the domain and the range.

Given definition of

$R$ is

$R = \{(x, y): y = x + 5, x \text{ is a natural number less than 4}, x, y\in N \}$
The natural numbers less than

$4$ are

$1, 2$ and

$3$

$\displaystyle \therefore R =\{(1, 6), (2, 7), (3, 8)\}$
The domain of

$R$ is the set of all first elements of the ordered pairs in the relation

$\displaystyle \therefore$ Domain of

$R = \{1, 2, 3\}$
The range of

$R$ is the set of all second elements of the ordered pairs in the relation

$\displaystyle \therefore$ Range of

$R = \{6, 7, 8\}$

Let $f, g$ : $\displaystyle R\rightarrow R$ be defined respectively by $f(x) = x + 1, g(x) = 2x - 3$ . Find $f + g, f - g$ and $\displaystyle \frac{f}{g}$

$f, g: \displaystyle R\rightarrow R$ is defined as

$f(x) = x + 1$

$g(x) = 2x - 3$

Now,

$(f + g) (x) = f(x) + g(x) = (x + 1) + (2x - 3) = 3x - 2$

$\displaystyle \therefore (f + g) (x) = 3x - 2$

Now,

$(f - g) (x) = f(x) - g(x) = (x + 1) - (2x - 3)= x + 1 - 2x + 3 = -x + 4$

$\displaystyle \therefore (f - g) (x) = -x + 4$

$\displaystyle \left ( \frac{f}{g} \right )\left ( x \right )=\frac{f\left ( x \right )}{g\left ( x \right )},g\left ( x \right )\neq 0,x\in R$

$\displaystyle \therefore \left ( \frac{f}{g} \right )\left ( x \right )=\frac{x+1}{2x-3},2x-3\neq 0\ or\ 2x\neq 3$

$\displaystyle \therefore \left ( \frac{f}{g} \right )\left ( x \right )=\frac{x+1}{2x-3},x\neq \frac{3}{2}$

Let $R$ be the relation on $Z$ defined by $R =$ $\{(a, b): a, b \in Z, a -b$ is an integer $\}$ . Find the domain and range of $R.$

$\textbf{Step-1: Apply the property of Domain & Range.}$

$\text{Given,}$

$R = \{(a, b): a, b ∈ Z, a -b$

$\text{is an integer}$

$\}$

$a∈ Z$

$\text{So, Domain of}$

$R$

$\text{is}$

$Z$ .

$\text{R}$

$\text{is such that}$

$a-b∈Z$

$\text{and we have}$

$a ∈Z$

$\text{so}$

$b∈Z$

$\text{So, Range of}$

$R$

$\text{is}$

$Z$ .

$\therefore$

$\text{Domain of}$

$R=Z$

$\text{and Range of}$

$R=Z.$

$\textbf{Hence, Domain and Range of R is Z.}$

The figure shows a relationship between the sets $P$ and $Q$ . Write this relation in
(i) in set-builder form (ii) roster form

The relation mentioned in the figure shows,

$P$ a domain and

$Q$ as range.

Let the relation be

$R$

In roster form

$R=\{(5,3),(6,4),(7,5)\}$

In set builder form

$R=\{(x,y):x\in P,y \in Q,y=x-2\}$

If $A = \{2, 4, 6, \}$ , $B = \{5, 7, 1, 9\}$ .
Let R be the relation ‘is less than’ from A to B. Find Domain (R) and Range (R).

Under this relation $(R)$ , we have

$R = {(4, 5); (4, 7); (4, 9); (6, 7); (6, 9), (8, 9) (2, 5) (2, 7) (2, 9)}$

Therefore, Domain $(R)$ = $\{2, 4, 6, 8\}$ and Range $(R)$ = $\{1, 5, 7, 9\}$

If $R = \left \{(x, y) : x + 2y = 8\right \}$ is a relation on $N,$ write the range of $R.$

$x+2y=8$ , Given

$x,y$ are natural numbers , which implies

$x,y >0$

$y=1$ , we get

$x=6$

$y=2$ , we get

$x=4$

$y=3$ , we get

$x=2$

For

$y\ge 4$ ,

$x$ wont be a natural number

So the range of

$R$ is

$\{1,2,3\}$

If $f(x)=4-3x$ , find $f(-5)$ and $f(13)$

Given,

$f(x)=4-3x$

$\Rightarrow f(-5)=4-3(-5)$

$=4+15$

$= 19$

$\Rightarrow f(13) = 4-3(13)$

$= 4-39$

$= -35$

In the problem below, $f(x)={x}^{2}$ and $g(x)=4x-2$
Find the following function:
$(f\times g)(x)$

Given,

$f(x)={x}^{2}$ and

$g(x)=4x-2$

We need to find

$(f\times g)(x)$

$\Rightarrow (f \times g)(x) = f(x) \times g(x)$

$= {x}^{2}(4x-2)$

$=4{x}^{3}-2{x}^{2}$

A function $f : (-3, 7)\rightarrow R$ is defined as follows:
$f(x) = \left\{\begin{matrix} 4x^{2} -1;& -3 \leq x < 2\\ 3x - 2; & 2\leq x \leq 4\\ 2x - 3; & 4 < x \leq 6\end{matrix}\right.$
Find:
$f(1) - f(-3)$

Given,

$f : (-3, 7)\rightarrow R$ is defined as follows:

$f(x) = \left\{\begin{matrix} 4x^{2} -1;& -3 \leq x < 2\\ 3x - 2; & 2\leq x \leq 4\\ 2x - 3; & 4 < x \leq 6\end{matrix}\right.$

Then,

$f(1) = 4(1)^{2} - 1 = 4 (1) - 1 = 4 - 1 = 3$
and

$f(-3) = 4(-3)^{2} - 1 = 4(9) - 1 = 36 - 1 = 35$
Therefore,

$f(1) - f(-3) = 3 - 35 = -32$

A function $f : (-3, 7)\rightarrow R$ is defined as follows:
$f(x) = \left\{\begin{matrix} 4x^{2} -1;& -3 \leq x < 2\\ 3x - 2; & 2\leq x \leq 4\\ 2x - 3; & 4 < x \leq 6\end{matrix}\right.$
Find:
$\dfrac {f(3) + f(-1)}{2f(6) - f(1)}$

Given,

$f(x) = \left\{\begin{matrix} 4x^{2} -1;& -3 \leq x < 2\\ 3x - 2; & 2\leq x \leq 4\\ 2x - 3; & 4 < x \leq 6\end{matrix}\right.$

For

$f(3)$ ,

$f(x)=3x-2$

Then,

$f(3) = 3(3) - 2 = 9 - 1 = 7$
For

$f(-1), f(x)=4x^2-1$

Then,

$f(-1)= 4(-1)^{2} - 1 = 4 - 1 = 3$
For,

$f(6), f(x)=2x-3$

Then,

$f(6)= 2(6) - 3 = 12 - 3 = 9$
Therefore,

$\dfrac {f(3) - f(-1)}{2f(6) - f(1)} = \dfrac {7 + 3}{2(9) - 3} = \dfrac {10}{18 - 3} = \dfrac {10}{15} = \dfrac {2}{3}$

A function $f : [-7, 6) \rightarrow R$ is defined as follows
$f(x)=\left\{\begin{matrix}x^2+2x+1; &-7 \leq x < -5 \\ x+5;& -5\leq x\leq 2 \\ x-1;& 2 < x < 6 \end{matrix}\right.$
(iii) $2f(-4)+3f(2)$
(ii) $f(-7)-f(-3)$
(iv) $\dfrac{4f(-3)+2f(4)}{f(-6)-3f(1)}$

$f(-4)=-4+5=1\\f(2)=2+5=7\\\therefore 2f(-4)+3f(2)=2+21=23$

$f(-7)={(-7)}^2+2(-7)+1=36\\f(-3)=-3+5=2\\\therefore f(-7)-f(-3)=36-2=34$

$f(4)=4-1=3\\f(-6)={(-6)}^2+2(-6)+1=25\\f(1)=1+5=6\\\therefore \dfrac {4f(-3)+2f(4)}{f(-6)-3f(1)}=\dfrac {8+6}{25-18}=2$

A function $f : (-3, 7)\rightarrow R$ is defined as follows:
$f(x) = \left\{\begin{matrix} 4x^{2} -1;& -3 \leq x < 2\\ 3x - 2; & 2\leq x \leq 4\\ 2x - 3; & 4 < x \leq 6\end{matrix}\right.$
Find:
$f(-2) - f(4)$

$f(x) = \left\{\begin{matrix} 4x^{2} -1;& -3 \leq x < 2\\ 3x - 2; & 2\leq x \leq 4\\ 2x - 3; & 4 < x \leq 6\end{matrix}\right.$

Then,

$f(-2) = 4(-2)^{2} - 1 = 4(4) - 1 = 16 - 1 = 15$
and

$f(4) = 3(4) - 2 = 12 - 2 = 10$
Therefore,

$f(-2) - f(4) = 15 - 10 = 5$

Let $f(x)=\sqrt{x+\sqrt{0+\sqrt{x+\sqrt{0+\sqrt{x+......}}}}}$ .If $f(a)=4$ and $f(a)=\dfrac{p}{q}$ where p and q are relatively prime natural numbers then $(a > 0)(p+q)$ is equal to

We have

$f\left( x \right) = \sqrt {x + \sqrt {0 + \sqrt {x + \sqrt {0 + \sqrt {x + .......} } } } }$

$f\left( a \right) = \sqrt {a + \sqrt {a + \sqrt {a + \sqrt {a + .....} } } }$

$\Rightarrow {4^2} = a + \sqrt {a + \sqrt {a + \sqrt {a + .......} } }$

$\Rightarrow 16 = a + 4$

$\Rightarrow a = 16 - 4$

$\therefore a = 12$

Now,

$f\left( a \right) = \frac{p}{q}$

$\Rightarrow 4 = \frac{p}{q}$

$\Rightarrow p = 4q$

$\therefore p + q = 12$

Hence , the answer of given question is

$12$ .

Let $\quad f(x)=\sqrt { \cfrac { { x }^{ 2 }+ax+4 }{ { x }^{ 2 }+bx+16 } }$ is defined for all real $x$ , then find the number of possible ordered pairs $(a,b)$ (where $a,b\in I$ )

$f(x) = \sqrt{\dfrac{x^{2}+ax+4}{x^{2}+bx+16}}$ to be defined for all real a,

$x^{2}+ax+4\geqslant 0$ &

$x^{2}+bx+16> 0...(1)$

$x^{2}+ax+4< 0$ &

$x^{2}+bx+16< 0$ (both simultaneously)...(2)

$(2)\rightarrow$ [Not Possible as Positive coefficient]

for (1) to be true Discriminant

$\Rightarrow B^{2}-4AC = D$

for

$x^{2}+ax+4\geq 0\rightarrow D\leqslant 0$

$x^{2}+bx+16> 0 \rightarrow D< 0$

$a^{2}-4\times 4\times 1 \leqslant 0$

$b^{2}-4\times 16\times 1< 0$

$a^{2}\leqslant 16$

$b^{2}-64< 0\rightarrow b^{2}< 64$

$a\epsilon [-4,4]$

$b\epsilon (-8,8)$

Ordered Pair,

$(-4,-7)(-4,-6)(-4,-5)(-4,-4)(-4,-3)....(-4,7)$

$(-3,-7)(-3-6).......(-3,7)$

$(3,-7)(3,-6)....(3,7)$

$(4,-7)(4,-6)(4,-5)...(4,0).....(4,7)$

1456100_779861_ans_44c5d7d0c35a46a58579f14323c69639.jpeg

Which term of the A.P. $3,10,17,......$ wil be $84$ more than its $13$ th term?

Given A.P

$3,10,17,...…..$

first term of A.P is

$a_{1}=3$

second term of A.p is

$a_{2}=10$

common difference

$d=a_{2}-a_{1}=10-3=7$

nth term of A.P is given by

$a_{n}=a_{1}+(n-1)d$

$a_{n}=3+(n-1)7$

$a_{n}=3+7n-7$

$a_{n}=-4+7n...eq(1)$

13th term of A.P is

$a_{13}=3+(13-1)7$

$a_{13}=3+(12)7$

$a_{13}=3+84$

$a_{13}=87....eq(2)$

we have to find term which is 84 more than 13th term

$\implies a_{n}=a_{13}+84$

put value of

$a_{n}$ from eq(1) and

$a_{13}=87$ from eq(2) in above equation

$\implies -4+7n=87+84$

$\implies 7n=87+84+4$

$\implies 7n=175$

$\implies n=\dfrac{175}{7}$

$\implies n=25$

hence

$25$ th term of this A.P will be

$84$ more than its

$13$ th term

Find the intervals of monotocity of the function $f(x)=2x^2-\log |x|, x\neq 0$ .

We have

$f(x)=2x^2-\log |x| ,x\ne0$

Now,

$f'(x)=4x-\dfrac{1}{x}=\dfrac{4x^2-1}{x}$

For

$f(x)$ being monotone increasing we must have

$f'(x)\ge 0$

or,

$4x^2-1\ge 0$

or,

$x\in\left(-\infty,-\dfrac{1}{2}\right]\cup\left[\dfrac{1}{2},\infty\right)$

For

$f(x)$ being monotone decreasing we must have

$f'(x)\le 0$

or,

$4x^2-1\le 0$

or,

$x\in\left[-\dfrac{1}{2},\dfrac{1}{2}\right]-\left\{0\right\}$

A function $f(x)$ is defined as $f(x) = x^{2} + 3$ . Find $f(0), f(1), f(x^{2}), f(x + 1)$ and $f(f(1))$ .

$f(0) = o^{2} + 3 = 3$

$f(1) = 1^{2} + 3 = 4; f(x^{2}) = (x^{2})^{2} + 3 = x^{4} + 3$

$f(x + 1) = (x + 1)^{2} + 3 = x^{2} + 2x + 4$

$f(f(1)) = f(4) = 4^{2} + 3 = 19$ .

If A = {a, b, c, d}, B={1, 2, 3}, find whether or not the following sets of ordered pairs are relations from A to B or not
(a) $R_1$ ={(a, 1), (a, 3)}
(b) $R_2$ = { (b, 1), (c, 2), (d, 1)}
(c) $R_3$ = {(a, 1), (b, 2),( 3, c)}

$R_1$ and

$R_2$ are relations as both are subsets of

$A\times B$ but

$R_3$ is not a relation from A to B as the ordered pair (3, c)

$\in B\times$ A and

$\notin A\times B$

N is the set of positive integers and ~ be a relation on $N\times N$ defined (a, b) ~ (c, d) iff ad = bc.Is it equivalence relation.

Consider any

$(a, b), (c, d), (e, f ) \in N \times N$ .

(i) ab = ba for all

$a, b \in N$ (commutative law of multiplication in N)

$\Rightarrow (a, b) R (a, b) \Rightarrow R$ is reflexive.

(ii) Let

$(a, b) R (c, d ) \Rightarrow ad = bc \Rightarrow bc = ad$

$\Rightarrow cb = da \Rightarrow (c, d ) R (a, b)$ .

Thus,

$(a, b) R (c, d ) \Rightarrow (c, d) R (a, b) \Rightarrow R$ is symmetric.

(iii) Let (a, b) R (c, d) and (c, d) R (e, f )

$\Rightarrow ad = bc and cf = de \Rightarrow ad cf = bc de$

$\Rightarrow af = be \Rightarrow (a, b) R (e, f )$ .

Thus, (a, b) R (c, d ) and

$(c, d) R (e, f ) \Rightarrow (a, b) R (e, f ) \Rightarrow R$ is transitive.

Therefore, the relation R is reflexive, symmetric and transitive, and hence it is an equivalence relation

A class has $175$ students. The following table shows the number of students studying one or more of the following subjects in this class :
Subject Number of students
Mathematics $100$
Physics $70$
Chemistry $46$
Mathematics and Physics $30$
Mathematics and Chemistry $28$
Physics and Chemistry $23$
Mathematics, Physics and Chemistry $18$
How many students are enrolled in Mathematics alone, Physics alone and Chemistry alone ? Are there students who have not offered any of these three subjects ?

Ans

$60M$ ,

$35P$ ,

$13C$ ,

$22\ None$ .
Here

$n(U)=175$ ,

$n(M)=100$ ,

$n(P)=70$

$n(C)=46$

$n\left( M\cap P \right) =30$ ,

$n\left( M\cap C \right) =28$

$n\left( P\cap C \right) =28$
and

$n\left( M\cap P\cap C \right) =18$
To find

$n\left( M\cap P'\cap C' \right)$ ,

$n\left( P\cap M'\cap C' \right)$

$n\left( C\cap M'\cap P' \right)$
and

$n\left( P'\cap M'\cap C' \right)$
Now
similarly

$n\left( P'\cap M'\cap C' \right) =n(P)-n(P\cap M)-n(P\cap C)+n(P\cap M\cap C)\}$

$=70-30-23+18=35$

$n\left( C\cap M'\cap P' \right) =n(C)-n(C\cap M)-n(C\cap P)+n(C\cap M\cap P)\}$

$=46-28-23+18=13$
and

$n\left( M'\cap P'\cap C' \right) =n(M\cup P\cup C)'$

$\bar { { z }_{ 2 } } -\bar { { z }_{ 1 } } =\left( \overline { { z }_{ 2 }-{ z }_{ 1 } } \right)$

$=n(U)-\{ { S }_{ 1 }\cap { S }_{ 2 }\cap { S }_{ 3 }\}$

$=n(U)-\{ n(M)+n(P)+n(C)$

$-[(M\cap P)-n(P\cap C)$

$-(M\cap C)+n(M\cap P\cap C)\}$

$=175-{100+70+46-30-23-28+18}$

$175-{216-81+18}=22$

Suppose $f$ is the collection of all ordered pairs of real numbers and x = 6 is the first element of some ordered pair in $f$ . Suppose the vertical line through x = 6 intersects the graph of $f$ twice. Is $f$ a function? Why or why not?

f is not a function.

a function f is the same thing as a subset

$G_f$ of

$X \times Y$ with the following property:

for all

$x \in X$ , there is a unique

$y \in Y$ such that

$(x, y) \in G_f$ and we set y = f(x).

To say that there is a unique

$y \in Y$ says that f(x) is uniquely determined by x, and to say that for every

$x \in X$ there exists an

$(x, y) \in G_f$ says that in fact f(x) is defined for all

$x \in X$ .

This is the so-called vertical line test: for each

$x \in X$ , we have the subset

${x} \times Y of X \times Y$ . (In case

$X = Y = R$ , such subsets are exactly the vertical lines.)

Then G is the graph of a function f if and only if, for every

$x \in X$ , ({x} × Y ) ∩ G consists of exactly one point, necessarily of the form

$(x, y)$ for some

$y \in Y$ .

The unique such y is then

$f(x).$

If $X= \{1, 2, 3, 4, 5\}$ and $Y = \{ 1, 3, 5, 7,9\}$ determine which of the following sets are (i) mappings (ii) relations (ii) neither of X to Y.
(a) $R_1 = { (x, y): y = x + 2 , x \in X, y \in Y}$
(b) $R_2$ $= \{ (1, 1), ( 2, 1), (3, 3), (4, 3), ( 5, 5)\}$
(c) $R_3$ $= \{ (1, 1), (1, 3), (3, 5), (3, 7), (5, 7)\}$
(d) $R_4$ $= \{ (1, 3), (2, 5), (4, 7), (5, 9), (3, 1)\}$

$i)\$ mappings

$(b),\ (d)$

$ii)\$ relations

$(c)$

$iii)\$ neither

$(a)$

Let a relation R be defined by
R = {(4, 5), ( 1, 4), ( 4, 6), (7, 6), (3, 7)}.
Find (i) R o R (ii) $R^{-1}$ o R.

(i) To obtain the elements of R oR, we proceed as follows. Since (1, 4)

$\in$ R and ( 4, 5)

$\in$ R we have (1, 5)

$\in$ R o R.
Again since ( 1, 4)

$\in$ R and ( 4, 5)

$\in$ R we have (1, 6)

$\in$ R o R.
Similarly (3, 6)

$\in$ RoR since (3, 7)

$\in$ R and (7, 6)

$\in$ R
Hence R o R = (1, 5), ( 1, 6), (3, 6)
(ii) We first find

$R^{-1}$ .
We have

$R^{-1}$ = {(5,4), ( 4, 1), (6, 4), (6, 7), (7, 3)}
We now obtain the elements of

$R^{-1}$ o R. We first pick the element of R and then

$R^{-1}$
Since (4, 5)

$\in$ R and (5, 4)

$\in R^{-1}$ We have (4, 4)

$\in$

$R^{-1} o R$
Similarlly

$(1, 4) \in R, (4, 1) \in R^{-1} \Rightarrow (1, 1) \in R^{-1}o R.$

$(4, 6) \in R, (6,4) \in R^{-1} \Rightarrow (4, 4) \in R^{-1}o R.$

$(4, 6) \in R, (6, 7) \in R^{-1} \Rightarrow (4, 7) \in R^{-1}o R.$

$(7, 6) \in R, (6, 4) \in R^{-1} \Rightarrow (7, 4) \in R^{-1}o R.$

$(7, 6) \in R, (6, 7) \in R^{-1} \Rightarrow (7, 7) \in R^{-1}o R.$

$(3, 7) \in R, (7, 3) \in R^{-1} \Rightarrow (3, 3) \in R^{-1}o R.$
Hence

$R^{-1} o R$ = { (1, 1), (4, 4), (4, 7), (7, 4), (7, 7), (3, 3)}

Let $A= {1, 2, 3, 4, 5}$ , $B = { 1, 3, 5, 7, 9, 10, 17, 26, 30}.$ Which of the following sets of ordered pairs are (i) relations (ii) functions (iii) neither, from A to B
(a) $R_1$ = {(1, 3), (3, 3), (5, 17), (1, 2)}
(b) $R_2$ = {(3, 7), (4, 5), ( 5, 30)}
(c) $R_3 = {(x, y): x \in A, y \in B, x+y =8}$
(d) $R_4 = {(x, y): x \in A, y \in B, y=x^2+1}$
(e) $R_5 = {(x, y): x \in A, y \in B, y=2x-1}$

$R_1 \ neq R$ as

$(1, 2) \notin A\times B, R_1\neq f$ as 2 or 4

$\in$ A has no image in B,

$R_2 = R$ but

$R_2 \neq f$ as in

$R_1$

$R_3 = { (1, 7), (3, 5), (5, 3)}$

$x + y = 8, x\in A, y\in B$

$\therefore R_3$ = R being a subset of

$A\times B$ but

$R_3 \neq f$ as in

$R_1$

$R_4= {(2, 5), (3, 10), (4, 17), (5, 26)}$

$\because y = x^2+1, x\in A, y \in B$

$R_4= R$ but

$R_4 \neq f$ as in

$R_1$

$R_5 = {(1, 1), (2, 3), (3, 5), (4, 7), (5, 9)}$

$y = 2x - 1$

$R_5$ = R and

$R_5 = f$ as each element

$\in A$ has a unique image in B.

The following relation is defined on the set of real numbers. $aRb$ if $|a - b| > 0$ .

R is not reflexive since |a -a| = 0 and
So | a- a | > a.
Thus a (-R) a for any real numbers a
R is symmetric since if | a - b| > 0, then
|b - a | = | a - b| > 0.
Thus aRb

$\Rightarrow$ bRa
R is not transitive. For example.
Consider the numbers 3, 7, 3 then we have 3R7 since
|3 - 7| = 4 > 0 and 7R3 since |7-3| = 4>0.
But 3(-R) 3 since |3 -3| = 0
so that |3 - 3| > |0

Let, $f(x+1)=2x^2-3x-1$ , then find the value of $f(0)$ and $f(x+2)$ .

Given,

$f(x+1)=2x^2-3x-1$ ......(1).

Putting

$x=-1$ in the above expression we get,

$f(0)=2.(-1)^2-3.(-1)-1$

or,

$f(0)=4$ .

Now putting

$x+1$ in expression (1) in stead of

$x$ we get,

$f(x+2)=2(x+1)^2-3(x+1)-1=2x^2+x-2$

If $f(x) = x+7$ and $g(x) = x-7$ , $x \in R$ , find $(fog)(7)$ . function $f:R \rightarrow R$ defined by $f(x) = \frac{3x+5}{2}$ is an invertible function, find $f^{-1}$

$f(x)=x+7$

$g(x)=x-7,x\in R$

$fog(x)=f(x-7)(x)$

$=(x-7+7)(x)$

$=x$

$fog(7)=7$

$f(x)=\cfrac { 3x+5 }{ 2 }$

$y=\cfrac { 3x+5 }{ 2 }$

Swapping

$x=\cfrac { 3y+5 }{ 2 }$

$2x=3y+5$

$2x-5=3y$

$y=\cfrac { 2x-5 }{ 3 }$

${ f }^{ -1 }(x)=\cfrac { 2x-5 }{ 3 }$

$f:[0,2] \rightarrow R$ is defined as $f(x) = \sqrt{x^3 +2 - 2\sqrt{x^3 + 1}} + \sqrt{x^3 +10 - 6\sqrt{x^3 + 1}}$ and $g(x) = \int f(x) dx.$ if $g(1) = 3$ then $g(\frac{1}{3}) =$

1313280_1013588_ans_ab04a60cfdc44ff98fb7957c58391497.jpg

Let $f(x)=\cos(\log x)$ the prove that $f(x).f(y)=\dfrac{1}{2}\left[f\left(\dfrac{y}{x}\right)+f(xy)\right]$ .

Given,

$f(x)=\log(\cos x)$ .

Now,

$\dfrac{1}{2}\left[f\left(\dfrac{y}{x}\right)+f(xy)\right]$

$=\dfrac{1}{2}\left[\cos(\log x-\log y )+\cos (\log x+\log y )\right]$

$=\cos(\log x).\cos(\log y)$ [Using formula of

$\cos(A+B)+\cos(A-B)=2\cos A.\cos B$ ]

$=f(x).f(y)$

If $(2.3)^{x} = (0.23)^{y} =1000$ then find the value of $\dfrac{1}{x} -\dfrac{1}{y}$

$(2.3)^x = (0.23)^y = 1000$

$(2.3)^x = 1000 \implies 2.3 = (1000)^{\dfrac{1}{x}}$ ……………(i)

$(0.23)^y = 1000 \implies 0.23 = (1000)^{\dfrac{1}{y}}$ ………..(ii)

$\dfrac{(1000)^{\dfrac{1}{x}}}{(1000)^{\dfrac{1}{y}}} = \dfrac{2.3}{0.23}$

$(1000)^{\dfrac{1}{x} - \dfrac{1}{y}} = 10$

$(10)^{3(\dfrac{1}{x}- \dfrac{1}{y})} = 10^1$

$3(\dfrac{1}{x} - \dfrac{1}{y}) = 1$

$\therefore \dfrac{1}{x} - \dfrac{1}{y} =\dfrac 13$

Let $R$ be a relation $<$ from $A={1,2,3,4}$ to $B={1,3,5},i.e.,(a,b) \in R \Leftrightarrow a < b$ , then $R o R^{-1}$ is:
a) { $(1,3),(1,5),(2,3),(2,5),(3,5),(4,5)$ }
b) { $(3,1),(5,1),(3,2),(5,2),(5,3),(5,4)$ }

1160121_1051712_ans_6a6506954ee6445c83b14ba2a68c3f5f.jpg

Let $A = \{1, 2\}$ and $B = \{3, 4\}$ . Find the number of relations from $A$ to $B$ .

$A=\{1,2\}$

$B=\{3,4\}$

Relation from

$A$ to

$B=\left\{ \left( 1,3 \right) ,\left( 1,4 \right) ,\left( 2,3 \right) ,\left( 2,4 \right) \right\}$ .

Verify $A=\{1,2\}$
$B=\{1,2,3,4\}$ $C=\{5,6\}$ $D=\{5,6,7,8\}$ . Prove $A\times (B \cap C)=(A\times B) \cap (A\times C)$

given

$A=\{1,2\} ,B=\{1,2,3\} ,C=\{5,6\} ,D=\{5,6,7,8\}$

therefore

$(A\times B)={(1,1),(1,2)(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)}$ and

$(A \times C)={(1,5),(1,6),(2,5),(2,6)}$

$\Rightarrow (A\times B)\cap (A \times C)$ is a null set

$B=\{1,2,3,4\} ,C=\{5,6\}$

$(B\cap C)$ is also a null set

therefore

$A\times (B \cap C)$ is a null set

$\Rightarrow A\times (B \cap C)=(A \times B)\cap(A \times C)$

A relation R is defined on the set $z$ of integers as follows. $(x, y)\in R\Leftrightarrow x^2+y^2=25$ . Express $R$ and ${R}^{-1}$ as the set of ordered pairs and hence find their respective domains.

We have

$\left(x,y\right)\in R\Leftrightarrow {x}^{2}+{y}^{2}=25$

$\Leftrightarrow y=\pm\sqrt{25-{x}^{2}}$

We observe that

$x=0\Rightarrow y=\pm 5$

$\therefore \left(0,5\right)\in R$ and

$\left(0,-5\right)\in R$

$x=\pm 3\Rightarrow y=\pm\sqrt{25-9}=\pm 4$

$\therefore \left(3,4\right)\in R,\, \left(-3,4\right)\in R,\, \left(3,-4\right)\in R$ and

$\left(-3,-4\right) \in R$

$x = \pm 4 \Rightarrow y =\pm\sqrt{25-16}=\pm 3$

$\therefore \left(4,3\right)\in R,\, \left(-4,3\right) \in R,\, \left(4,-3\right)\in R$ and

$\left(-4,-3\right)\in R$

$x = \pm 5 \Rightarrow y = \pm\sqrt{25-25}= 0$

$\therefore \left(5,0\right)\in R$ and

$\left(-5,0\right)\in R$

We also notice that for any other integral value of

$x,y$ is not an integer.

$\therefore R =\left\{\left(0,5\right), \left(0,-5\right),\left(3,4\right),\left(-3,4\right),\left(3,-4\right),\left(-3,-4\right),\left(4,3\right),\left(-4,3\right),\left(4,-3\right),\left(-4,-3\right),\left(5,0\right),\left(-5,0\right)\right\}$

and

${R}^{-1}=\left\{\left(5,0\right),\left(-5,0\right),\left(4,3\right),\left(4,-3\right),\left(-4,3\right),\left(-4,-3\right),\left(3,4\right),\left(3,-4\right),\left(-3,4\right),\left(-3,-4\right),\left(0,5\right),\left(0,-5\right)\right\}$

Clearly domain

$\left[R\right]=\left\{0,3,-3,4,-4,5,-5\right\} =$ domain

$\left({R}^{-1}\right)$

Let $A=\left\{x,y,z\right\}$ and $B=\left\{1,2\right\}$ . Find the number of relations from $A$ to $B$ .

Given

$n(A)=3$ and

$n(B)=2$

$\therefore n (A \times B)=3\times 2=6$
Number of relations from

$A$ to

$B=2^{n(A \times B)}=2^6=64$

If $A =\left\{2, 4, 6, 9\right\}, B = \left\{4, 6, 18, 27, 54\right\}$ and a relation $R$ from $A$ to $B$ is defined by $R = \{(a, b) : a \, \in \, A, \, b \, \in \, B, \,\,\, a \, \,\text{is a factor of b and }\ a < b\}$

First let us find

$R=A\times B$

$R=\left\{\left(a,\,b\right):a\in A,\,\,b\in B\right\}$

Given

$a$ is a factor

$b$

$R=\left\{\left(a,\,b\right):a\in A,\,\,b\in B\,\,\,and \,\,\,a<b,\,\,\,a \,\,is \,\,a\,\,factor\,\,of\,\,b\right\}$

$\therefore R=\left\{\left(2,\,4\right),\,\left(2,\,6\right),\,\left(2,\,18\right),\,\left(2,\,54\right),\,\left(6,\,18\right),\,\left(6,\,54\right),\,\left(9,\,18\right),\,\left(9,\,27\right),\,\left(9,\,54\right)\right\}$

If $f(x)=\dfrac{1}{1-x}$ , show that $f[f\{f(x)\}]=x$ .

here

$f(x)=\dfrac {1}{1-x}$

now,

$f\left\{f(x)\right\}=f\left (\dfrac {1}{1-x}\right)$

$\left\{so,\ put\ x\to \dfrac {1}{1-x}\right\}$

$f\left (\dfrac {1}{1-x}\right)=\dfrac {1}{1-\left (\dfrac {1}{1-x}\right)}=\dfrac {1-x}{(1-x)-1}$

$f\left\{f(x)\right\}=\dfrac {1-x}{-x}=\dfrac {x-1}{x}$

now,

$f[f\left\{f(x)\right\}]=f\left [\dfrac {x-1}{x}\right]$

$\left\{so,\ put\ x\to \dfrac {x-1}{x}\right\}$

$f\left [\dfrac {x-1}{x}\right]=\dfrac {1}{1-\left (\dfrac {x-1}{x}\right)} =\dfrac {x}{x-(x-1)}$

$f[f\left\{f(x)\right\}]= \dfrac {x}{x-x+1}=\dfrac {x}{1}=x$

so,

$\boxed {f[f\left\{f(x)\right\}]=x}$

Let $R$ be a relation in $N$ defined as $R = \{(x, y) \in \, N \times N : \, x + 2y = 39\}$ Find the domain, co-domain and range of $R$ . Also write $R^{-1}$ in roster form.

$x+2y=39$

$\Rightarrow y=\dfrac{39-x}{2}$

Since

$y\in N,y> 0$

$\dfrac{39-x}{2}>0$

$39-x>0$

$x<39$

Since

$x\in N,x$ can take values

$=\left \{ 1,2,....38 \right \}$

Domain of R

$=\left \{ 1,2,....38 \right \}$

Range of R is

$y=\dfrac{39-x}{2}, for x=\left \{ 1,2,....38 \right \}$

Range of R

$= \left \{ 1,2,....,37 \right \}$

when

$x=1, y=19$

when

$x=2, y=18.5$

when

$x=37, y=1$

Domain of f

$=\left \{ 1,2,....38 \right \}$

Range of f

$=\left \{ 1,2,....37 \right \}$

Let $A=\{1,2,3\}$ and $B=\{3,4\}$ then find $A\cap B$ and then find $A\times (A\cap B)$ .

Given,

$A=\{1,2,3\}$ and

$B=\{3,4\}$ .

Now,

$A\cap B$

$=\{3\}$ .

Now

$A\times (A \cap B)$

$=\{(1,3),(2,3),(3,3)\}$ .

If $f(x)=2x-5$ , then what is the value of $f(2)+f(5)$ ?

We have

$f(x)=2x-5$

To find

$f(2)+f(5)$

Put

$x=2$ in

$f(x)=2x-5$ , we get

$f(2)=2(2)-5=4-5=-1$

Put

$x=5$ in

$f(x)=2x-5$ , we get

$f(5)=2(5)-5=10-5=5$

Thus

$f(2)+f(5)=-1+5=4$

Hence

$f(2)+f(5)=4$

If $f(x) = x^2- 3x +4$ , then find the value of x satisfying $f(x) = f( 2x+1).$

$f(2x+1)=(2x+1)^2-3(2x+1)+4=4x^2-2x+2$

For,

$f(x)=f(2x+1)\Rightarrow x^2-3x+4=4x^2-2x+2$

$\Rightarrow 3x^2+x-2=0$

$\Rightarrow 3x^2+3x-2x-2=0$

$\Rightarrow (3x-2)(x+1)=0$

$\Rightarrow x=-1,\dfrac23$

If $A = {2,3,4,5,6,7,8,9}$ given by $R = \left\{ \left( x,y \right) :x\in A,y\in A\ and \ divides\ y \right\}$ then find the domain and range of $R$ .

1356747_1128316_ans_be7ab80440674ffbbf88cc5ce4ee239d.jpg

If $f\ {(x)}= \dfrac { \dfrac { 1 }{ x } +1 }{ \dfrac { 1 }{ x } -1 }$ then the value of $f\ {(x)}+f\ {(-x)}$ is

1361002_1129383_ans_19ea227781a04eb696a28248fbea2320.jpg

If $R=\{ (1,-1),(2,-2),(3,-1)\}$ is relation, then find the range of R.

Range :- Set of y-coordinates

From given

$R = \{(1, -1), (2, -2), (3, -1)\}$

$\therefore$ Range

$= \{-1, -2, -1\}$ .

Domain :- Set of x-coordinates

$\therefore$ Domain

$=(1, 2, 3)$

If $f(x) = \dfrac{(x - 1)}{(x + 1)}$ then prove that $f(2x) = \dfrac{3f(x) + 1}{f(x) + 3}$

$f(x)=\cfrac{x-1}{x+1}$

$f(2x)=\cfrac{3f(x)+1}{f(x)+3}$

$L.H..S$

$f(2x)=\cfrac{2x-1}{2x+1}\left[as, f(x)=\cfrac{x-1}{x+1}\right]$

$R.H.S$

$\cfrac{2f(x)+1}{f(x)+3}=\cfrac{3\left(\cfrac{x-1}{x+1}\right)+1}{\cfrac{x-1}{x+1}+3}$

$\cfrac { 2f(x)+1 }{ f(x)+3 } =\cfrac { \cfrac { 3x-3+x+1 }{ x+1 } }{ \cfrac { x-1+3x+3 }{ x+1 } }$

$\Rightarrow \cfrac{4x-2}{4x+2}\Rightarrow \cfrac{2x-1}{2x+1}$

Find the value $\left| 7 \right|\times\left| -4\right|\times \left|4-9 \right|\times \left|7 \right|\times \left| -4 \right|\times \left|4-9 \right|$

$=7\times 4 \times 5\times 7\times 4 \times 5$

$=49\times 16\times 25$

$=19,600‬$

Let $''$ be a binary operation on set Q of rational number defined as $ab=\dfrac{ab}{5}$ . Write the identity for $'*'$ , if any.

Given, binary operation is

$a*b=\dfrac{ab}{5}$

Now, let e be the identity element of

$'*'$ on Q.

Then,

$a*e=a=e*a$ for every

$a\in Q$

$\dfrac{ae}{5}=a=\dfrac{ea}{5}$

$ae=5a \implies e=5, a\neq 0$

Also, for

$0\in Q$ , we have,

$0*5=0=5*0$

Hence,

$e=5$ is the identity element.

Find range & domain of following $\sqrt {4 - {x^2}}$

1302506_1136365_ans_18b41053818644f0b13e6dde06b71e25.JPG

If $f\left( x \right) = x + \dfrac{1}{x}$ , prove that ${\left( {f\left( x \right)} \right)^3} = f\left( {{x^3}} \right) + 3f\left( {\dfrac{1}{x}} \right)$

$\\LHS=(f(x))^3=(x+(\dfrac{1}{x}))^3$

$\\=x^3+(\dfrac{1}{x^3})+3\cdot x\cdot(\dfrac{1}{x})(x+(\dfrac{1}{x}))$

$\\=[x^3+(\dfrac{1}{x^3})]+3(x+(\dfrac{1}{x}))$

$\\=f(x^3)+3f(\dfrac{1}{x})=RHS$

Hence Proved!

If ${e^{f\left( x \right)}} = \dfrac{{10 + x}}{{10 - x}},x \in \left( { - 10,0} \right)$ and $f\left( x \right) = k.f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right)$ then $k$ =

$\displaystyle e^{f(x)} = \frac{10+x}{10-x}, x\epsilon (-10,0)$

$\displaystyle \Rightarrow f(x) = log (\frac{10+x}{10-x})...(1)$

and

$\displaystyle f(x) = k.f (\frac{200x}{100+x^{2}})$

$\displaystyle f\left(\frac{200x}{100+x^{2}}\right) = log\left(\tfrac{10+\tfrac{200x}{100+x^{2}}}{10-\tfrac{200x}{100+x^{2}}}\right)$

$\displaystyle = log \left(\frac{1000+10x^{2}+200x}{1000+10x^{2}-200x}\right)$

$\displaystyle = log \left(\frac{100+20x+x^{2}}{100-20x+x^{2}}\right)$

$\displaystyle = log \left \{ \left(\frac{10+x}{10-x}\right)^{2} \right \}$

$\displaystyle \therefore f(x) = k.log\left \{ \left(\frac{10+x}{10-x}\right)^{2} \right \} = log \left \{ \left(\frac{10+x}{10-x}\right)^{2} \right \}^{k}$

$\displaystyle \Rightarrow log\left(\frac{10+x}{10-x}\right) = log \left(\frac{10+x}{10-x}\right)^{2k}$

$\displaystyle \Rightarrow 2k = 1$

$\displaystyle \Rightarrow k = \frac{1}{2}$

Let $A=\{-1,1\}$ then find $A\times A$ .

Given,

$A=\{-1,1\}$ .

Now

$A\times A=\{(-1,-1),(-1,1),(1,-1),(1,1)\}$ .

If $A=\left\{1,2,3\right\}$ and $B=\left\{3,4\right\}$ and $C=\left\{1,3,5\right\}$ .Find $\left(A\times\,B\right)\cap\left(A\times\,C\right)$

$\textbf{Hint: Cartesian product is set of all possible ordered pairs where first element belongs}$ $\textbf{to first set and second element belongs to second set. }$

Solution:

$\textbf{Step 1: Use definition of cartesian prdoucts of to sets}$

We have given,

$A=\left\{1,2,3\right\}$ , $B=\left\{3,4\right\}$ and $C=\left\{1,3,5\right\}$

$\because$ The Cartesian product of two non-empty sets P and Q is given by

${\bf{P}}{\rm{ }} \times {\bf{Q}}{\rm{ }} = {\rm{ }}\{ \left( {{\bf{a}},{\rm{ }}{\bf{b}}} \right):{\rm{ }}{\bf{a}} \in {\rm{ }}{\bf{P}},{\rm{ }}{\bf{b}} \in {\bf{Q}}\}$

$\therefore$ $A\times\,B=\left\{\left(1,3\right),\left(1,4\right),\left(2,3\right),\left(2,4\right),\left(3,3\right),\left(3,4\right)\right\}$

$A\times\,C=\left\{\left(1,1\right),\left(1,3\right),\left(1,5\right),\left(2,1\right),\left(2,3\right),\left(2,5\right),\left(3,1\right),\left(3,3\right),\left(3,5\right)\right\}$

$\textbf{Step 2: Use the definition of intersection of two sets}$

Now,

$\left(A\times\,B\right)\cap\left(A\times\,C\right)$

$=\left\{\left(1,3\right),\left(1,4\right),\left(2,3\right),\left(2,4\right),\left(3,3\right),\left(3,4\right)\right\}\cap\left\{\left(1,1\right),\left(1,3\right),\left(1,5\right),\left(2,1\right),\left(2,3\right),\left(2,5\right),\left(3,1\right),\left(3,3\right),\left(3,5\right)\right\}$

$\because \ {\bf{A}} \cap \;{\bf{B}}{\rm{ }} = {\rm{ }}\{ {\bf{x}}:{\rm{ }}{\bf{x}}\; \in \;{\bf{A}}{\rm{ }}{\bf{\ and\ }}{\rm{ }}{\bf{x}} \in \;{\bf{B}}\}$

$\therefore \ \left(A\times\,B\right)\cap\left(A\times\,C\right)$ $=\left\{\left(1,3\right),\left(2,3\right),\left(3,3\right)\right\}$

$\textbf{Hence, the value of}$ $\boldsymbol{ \ \left(A\times\,B\right)\cap\left(A\times\,C\right)}$ $\textbf{is}$ $\boldsymbol{\left\{\left(1,3\right),\left(2,3\right),\left(3,3\right)\right\}}$

The simplified form of $\left ({\dfrac {x(x - 5y)} {{x^2} - 6xy + 5{y^2}} } \times {\dfrac {(x + y)(x - y)} {{x^2} + xy}}\right )$ is

$\dfrac{x(x-5y)}{x^2-6xy+5y^2}\times \dfrac{(x+y)(x-y)}{x^2+xy}$

$\Rightarrow \dfrac{x(x-5y)}{x^2-xy-5xy+5y^2}\times \dfrac{(x+y)(x-y)}{x(x+y)}$

$\Rightarrow \dfrac{x(x-5y)}{x(x-y)-5y(x-y)}\times \dfrac{(x+y)(x-y)}{x(x+y)}$

$\Rightarrow \dfrac{x(x-5y)}{(x-5y)(x-y)}\times \dfrac{(x+y)(x-y)}{x(x+y)}$

$=1$ .

The number of relations from $A = \{ 2,6 \}$ to $B = \{ 1,3,5,6,7 \}$ that are not functions from A to $B$ is

According to the problem,

$|A|=2$ and

$|B|=5$ .

Now number of relations from the set

$A$ to

$B$ is

$2^{2\times 5}=2^{10}$ .

Also number of mappings from the set

$A$ to

$B$ is

$5^2$ .

So number of relations from

$A$ to

$B$ that are not functions is

$2^{10}-5^2=1024-25=999$ .

$y=\dfrac{x-2}{2x+3}$
Express in the form $f(x,y) = 0$

$y=\dfrac{x-2}{2x+3}$

$(2x+3)y=x+2$

$2xy+3y=x+2$

$2xy+3y-x-2=0$

Let $A=\left\{1, 2, 3, 4, 5\right\}$ and $B=\left\{1, 4, 5\right\}$ . Let $R$ be a relation 'is less than' from $A$ to $B$ . Find the domain, co domain and range of $R$ .

$A=\{ 1,2,3,4,5\} \quad \quad B=\{ 1,4,5\}$

$R\rightarrow A$ is less than

$B$

Domain is

$\{ 1,2,3,4\}$

Co-domain is

$\{ 1,4,5\}$

range is

$\{ 4,5\}$

1207476_1243485_ans_10aba777edd147ef8bf8352fde0f9827.png

A function $f\::\:R\rightarrow R$ satisfy the equation $f\left(x\right)f\left(y\right)-f\left(xy\right)=x+y$ for all $x,y\in R$ and $f\left(y\right)>0$ , then ?

Let

$x, y=1$

$\Rightarrow y^2-y-2=0$

$(f(1)=y)$

$\Rightarrow y^2-2y+y-2$

$\Rightarrow y(y-2)+1(y-2)=0$

$\Rightarrow y=-1$ (y > 0)

y=2$$

Let

$y=1$ now

$f(x).2-f(x)=x+1$

$\Rightarrow f(x)=x+1$ .

1134150_1242083_ans_9a81effe3dea403d9831b6ef35394258.jpg

If the function $f(x)$ given by $f(x)=\begin{cases}3ax+b & if \, x > 1 \\ 11 & if\, x=1 \\5ax-2b & if \, x < 1\end{cases}$ is continuous at x=1, then find the values of a and b.

Here,

$f(x)=\begin{cases}3ax+b & if \, x > 1 \\ 11 & if\, x=1 \\5ax-2b & if \, x < 1\end{cases}$

Now RHL,

$\underset{x \rightarrow 1^{+}}{\lim} 3ax + b$

$\underset{h \rightarrow 0}{\lim} 3a (1 + h) + b$

$= 3a + b$

Again, LHL ,

$\underset{x \rightarrow 1^{-}}{\lim} 5ax - 2b$

$\underset{h \rightarrow 0^{-}}{\lim} 5a (1 - h ) - 2b$

$= 5a - 2b$

Now, RHL = LHL

$= 11$

$3a + b = 11$ ........(i)

$\times 2$

$5a - 2b = 11$ .......(ii)

$\times 1$

$6a + 2b = 22$

$5a - 2b = 11$

_____________

$11a = 33$

$a = 3$

$\therefore 3 \times 3 + b = 11$

$b = 11 - 9 = 2$

$\boxed{a = 3,\ \ b = 2}$

For x $\in$ R, the range of the given expression $\dfrac{x^2 + 34x - 71}{x^2 + 2x - 7}$ is

R.E.F image

let

$y = \frac{x^{2}+34x-71}{x^{2}+2x-7} \Rightarrow x^{2}(y-1)+x(2y-34)-7y+71=0$

$x \epsilon R, \Rightarrow D \geqslant 0$

$\Rightarrow (2y - 34)^{2} + 4(y-1)(7y-71) \geqslant 0$

$\Rightarrow (y-17)^{2}+(7y^{2}-78y+71)\geqslant 0$

$\Rightarrow 8y^{2}-112y+306\geqslant 0$

$\Rightarrow 4y^{2}-56y+153\geqslant 0$

$D=(56)^{2}-16(153)=688 > 0 \Rightarrow Roots \Rightarrow y = \frac{56\pm \sqrt{688}}{8}$

$\Rightarrow$ Range

$= y \,\epsilon (-\infty , \alpha ]\cup [\beta ,\infty )$

1167302_1246926_ans_6309b0aa5db14319bff569884311364f.jpg

Number of solutions satisfying $| x | + | x - 2 | \leq 2$ for $x \in R$ is

$\left| x \right| +\left| x-2 \right| \le 2$

Case

$(1)$

$x<0$

$\Rightarrow \left| x \right| +\left| x-2 \right| \le 2$

$=-x+2-x\le 2$

$\Rightarrow -2x\le 0\quad =x\ge 0\quad x\in \left[ 0,\infty \right]$

Case

$(2)$

$0\le x\le 2$

$x+2-x\le 2$

$2\le 2\quad =x\in \left[ 0,2 \right]$

Case

$(3)$

$x\ge 2$

$x+x-2\le 2$

$\Rightarrow 2x\le 4$

$\Rightarrow x\le 2\quad \quad =x\in \left[ -\infty ,2 \right]$

$\therefore$

$x\in \left[ 0,2 \right]$

$\therefore$ No. of solutions

$=$ infinite.

Let $f:$ $R \rightarrow R$ be given by $f ( x ) = x ^ { 2 } + 3 .$ Find ( a ) $\{ x : f ( x ) = 28 \}$ ( b ) the pre-images of $39$ and $2$ under $f.$

We have function R

$\rightarrow$ R defined by f(x)

$= x^{2}+3$

i.e its domain & range are real no.

1) f(x) = 28 =

$x^{2}+3$

$x^{2} = 25 \Rightarrow x = \pm 5$

Therefore , for range 28 Pre - image is 5 & -5

2) f(x) = 39.

$x^{2}+3 = 39$

$x^{2} = 36 \Rightarrow x = \pm 6$

Now f(x) = 2

$x^{2} + 3 = 2$

$x^{2} = -1$

$x = \pm i$ But the value of x is not are

value, both i & -i are complex no.

therefore, 2 is not an image under f.

1196915_1292136_ans_0f70a05873bc4518981ce4d6dd7e40f2.jpg

Find the range
$f(x)=\dfrac{1}{\sqrt{x-5}}$

Given,

$f\left(x\right)=\dfrac{1}{\sqrt{x-5}}$

Domain of

$\dfrac{1}{\sqrt{x-5}}:x>5$

Range of

$\dfrac{1}{\sqrt{x-5}}: \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)>0\:\\ \:\mathrm{Interval\:Notation:}&\:\left(0,\:\infty \:\right)\end{bmatrix}$

Find the domain of $\cos^{-1}[X]$ where $[X]$ is the greatest integer functions.

Domain of

$\cos^{-1}t$ is

$t\in [-1, 1]$

$\Rightarrow [x]\in [-1, 1]$

$\Rightarrow$ From graph of

$[x]$ ,

$x\in [-1, 2)$ .

1189688_1344412_ans_215af2009ff14a3eb49e9f2cf1665570.jpg

Find the domain and range of the following real functions :
$f ( x ) = - | x |$

value of $x$	value of $[x]$	$f(x)=-\|x\|$	whether real number
2	\|2 \|=2	-2	yes
1	\|1\|=0	-1	yes
0	\|0\|=0	0	yes
-1	\|-1\|=1	-1	yes
-2	\|-2\|=2	-2	yes

Here we are given a real function

Hence, both domain and range should be real numbers

From the above table

$X$ can be any real number

and

$f(X)$ will always be negative or zero

All these are real values.

Here, value of domain

$(Xx$ $ can be negative number

Hence, Domain =

$R$ (all real number)

we note that the range

$f(x)$ is

$0$ or negative numbers

So, range cannot be positive

Hence, Range =Non positive real number.

If $A=\left\{1,2,3\right\}$ and $B=\left\{3,4\right\}$ and $C=\left\{1,3,5\right\}$ .Find $A\times\left(B\cup\,C\right)$

We have

$B=\left\{3,4\right\}$ and

$C=\left\{1,3,5\right\}$

$B\cup\,C=\left\{3,4\right\}\cup\left\{1,3,5\right\}$

$=\left\{1,3,4,5\right\}$

$A\times\left(B\cup\,C\right)=\left\{1,2,3\right\}\times\left\{1,3,4,5\right\}$

$=\left\{\left(1,1\right),\left(1,3\right),\left(1,4\right),\left(1,5\right),\left(2,1\right),\left(2,3\right),\left(2,4\right),\left(2,5\right),\left(3,1\right),\left(3,3\right),\left(3,4\right),\left(3,5\right)\right\}$

$n ( A ) + n ( B ) - n ( A \cap B ) =$

$\textbf{Step 1: Find the required value.}$

$\text{Let,}$

$n(A)=a+c \quad \quad \ldots (i)$

$n(B)=b+c \quad \quad \ldots (ii)$

$n(A\cup B)=a+b+c \quad \quad \ldots (iii)$

$n(A\cap B)=c \quad \quad \ldots (iv)$

$\text{Adding equations (i) and (ii)}$

$\Rightarrow n(A)+n(B)=a+c+b+c$

$\Rightarrow n(A)+n(B)=a+b+2c$

$\Rightarrow n(A)+n(B)=(a+b+c)+c$

$\Rightarrow n(A)+n(B)=n(A\cup B)+n(A\cap B)$

$\Rightarrow n(A)+n(B)-n(A\cap B)=n(A\cup B)$

$\textbf{Hence the required answer is } \mathbf{n(A\cup B)}.$

Is $a\leq a^2$ , $a\in R$ true? Why or why not?

$a\le a ^2$ is not true always.

when

$a\ \epsilon\ (0,1)$ then

$a^2$ approaches to zero.

example

Let

$a=1/2\implies a^2=1/4$

$\implies a>a^2$

So,

$a\le a^2, a\ \epsilon\ R-(0,1)$

If $a,b,c\in{R}^{+}$ such that $a+b+c=18$ , then the maximum value of ${a}^{2}{b}^{3}{c}^{4}$ is

1302346_1358654_ans_d35c015df3944ebda386bc5f8d0f6164.jpg

If $f ( x ) = \frac { a ^ { x } + a ^ { - x } } { 2 }$ then $f ( x + y ) f ( x - y )$ is equal to:

$\textbf{Step-1: Finding }$

$\mathbf{f(y)}$

$\text{From the given value of }f(x)\text{ we get,}$

${f(y) = \dfrac{a^x+a^{-x}}{2}}$

$\textbf{Step-2: Finding the value of }$

$\boldsymbol{f(x+y)}$

$\textbf{and}$

$\boldsymbol{f(x-y)}$

$\text{From the above equation, we get}$

$f(x+y)=\dfrac{a^{x+y}+a^{-(x+y)}}{2}$

$f(x-y)=\dfrac{a^{x-y}+a^{-(x-y)}}{2}$

${f(x+y)f(x-y)= \dfrac{a^{x+y}+a^{-(x+y)}}{2} \times \dfrac{a^{x-y}+a^{-(x-y)}}{2}}$

$\text{By separating the exponents, we get}$

${f(x+y)f(x-y)= \dfrac{a^{x}a^{y}+a^{-x}a^{-y}}{2}\times\dfrac{a^{x}a^{-y}+a^{-x}a^{y}}{2}}$

$\text{ By rearranging the terms, we get}$

${f(x+y)f(x-y)=\dfrac{a^xa^ya^xa^{-y}+a^xa^ya^{-x}a^y+a^{-x}a^{-y}a^{x}a^{-y}+a^{-x}a^{-y}a^{-x}a^{y}}{4}}$

$=\dfrac{a^{2x}+a^{2y}+a^{-2y}+a^{-2x}}{4}$

$=2 f(x)+2f(y)$

$\textbf{Hence , the value of}$

$\mathbf{f(x+y)f(x-y)}$

$\textbf{is}$

$\mathbf{2f(x)+2f(y)}$

A relation $R$ is defined from a set $A=[2,3,4,5]$ to a set $B=[3,6,7,10]$ as follows:
$(x,y)\in R\Leftrightarrow x$ is relatively prime to $y$ .Express R as a set of ordered pairs and determine its domain and range.

A relation

$R$ from the set

$A =\left\{2, 3, 4, 5\right\}$ to

$B =\left\{3, 6, 7, 10\right\}$ is defined as,

$\left(x, y\right)\in R \Rightarrow x$ is relatively prime to

$y$ .

Now, it is seen that

$2$ is relatively prime to

$3$ , [HCF of

$2$ and

$3$ is

$1$ ]

So,

$\left(2, 3\right)\in R$

Similarly,

$2$ is relatively prime to

$7$ so that

$\left(2, 7\right)\in R$ etc.

So, we get

$R =\left\{\left(2, 3\right), \left(2, 7\right), \left(3, 7\right), \left(3, 10\right), \left(4, 3\right), \left(4, 7\right), \left(5, 3\right), \left(5, 6\right), \left(5, 7\right)\right\}$

Thus, domain

$=\left\{2, 3, 4, 5\right\}$ and range

$=\left\{3, 6, 7, 10\right\}$ .

If $A=\left\{1, 2, 4,5\right\}$ and $B=\left\{a,b\right\}$ Find the total number of relations from $A$ to $B$ .

Given:

$A=\left\{1,2,4,5\right\}$ and

$B=\left\{a,b\right\}$

Number of relations from

$A$ to

$B={2}^{Number\,\, of\,\, elements\,\, in\,\, A\times B}$

$={2}^{n\left(A\right)\times n\left(B\right)}$

$={2}^{4\times 2}={2}^{8}=256$

Write the relations as sets of ordered pairs $\left\{\left(x,y\right) : y = 3x, x\in \left\{1,2,3\right\}, y\in \left\{3,6,9,12\right\}\right\}$

$\left\{\left(x,y\right):y=3x,x\in\left\{1,2,3\right\},y\in\left\{3,6,9,12\right\}\right\}$

When

$x=1$ , we have

$y=3\left(1\right)=3$

When

$x=2$ , we have

$y=3\left(2\right)=6$

When

$x=3$ we have

$y=3\left(3\right)=9$

Thus,

$R=\left\{\left(1,3\right),\left(2,6\right),\left(3,9\right)\right\}$

If $A={1, 2, 4}$ and $B={a, b}$ Find the total number of relations from A to B.

Given:

$A=\left\{1,2,4\right\}$ and

$B=\left\{a,b\right\}$

Number of relations from

$A$ to

$B={2}^{number\,\, of \,\, elements\,\, in \,\, A\times B}$

$={2}^{number\,\, of \,\, elements\,\, in \,\, A\times number\,\, of \,\, elements\,\, in \,\, B}$

$={2}^{n\left(A\right)\times n\left(B\right)}$

Number of elements in set

$A=3$

Number of elements in set

$B=2$

Number of relations from

$A$ to

$B={2}^{n\left(A\right)\times n\left(B\right)}$

$={2}^{3\times 2}={2}^{6}=64$

If f = {(4, 5), (5, 6), (6, -4) and g = {(4, -4), (6, 5),(8, 5)} then find
$A)f+g\ \ \ \ B)f-g\ \ \ \ C)2f+4g\ \ \ \ D)f+4\ \ \ \ E)fg\ \ \ \ F)f/g\ \ \ \ G)|f| \ \ \ \ H)\sqrt{f}\ \ \ \ I)f^2\ \ \ \ G)f^3$

$f =\{(4, 5), (5, 6), (6, -4)\}$ and

$g = \{(4, -4), (6, 5),(8, 5)\}$

Binary operations are done on common domain.

$A)f+g=\{(4,1)\}\\$

$B)f-g=\{(4,9)\}\\$

$C)2f+4g=\{(4,-6)\}\\$

$D)f+4=\{(4,9),(5,10),(6,0)\}\\$

$E)fg=\{(4,-20)\}\\$

$F)f/g=\left\{\left(4,\dfrac 5{-4}\right)\right\}\\$

$G)|f|=\{(4,5),(5,6)(6,4)\}\\$

$H) \sqrt f=\{(4,\sqrt 5),(5,\sqrt 6)\} \\$

$I)f^2=\{(4,25),(5,36),(6,16)\}\\$

$J)f^3=\{(4,125),(5,216),(6,-64)\}$

If $f:A\rightarrow A$ , then range of $f$ where $A=\left\{1,2,3,4\right\}$ and $f\left(x\right)={x}^{2}+x+2$

Given:

$f:A\rightarrow A$

$f\left(1\right)={1}^{2}+1+2=1+1+2=4$

$f\left(2\right)={2}^{2}+2+2=4+2+2=8$

$f\left(3\right)={3}^{2}+3+2=9+3+2=14$

$f\left(4\right)={4}^{2}+4+2=16+4+2=22$

$\therefore$ Range of

$A=\left\{4,8,14,22\right\}$

If $f\left(x\right)={x}^{2}+2x+3,$ then find $\dfrac{f\left(x+h\right)-f\left(x-h\right)}{h},\,\,h\neq 0$

$f\left(x\right)={x}^{2}+2x+3$

$f\left(x+h\right)={\left(x+h\right)}^{2}+2\left(x+h\right)+3$

$={x}^{2}+2xh+{h}^{2}+2x+2h+3$

$f\left(x-h\right)={\left(x-h\right)}^{2}+2\left(x-h\right)+3$

$={x}^{2}-2xh+{h}^{2}+2x-2h+3$

Now,

$f\left(x+h\right)-f\left(x-h\right)$

$={x}^{2}+2xh+{h}^{2}+2x+2h+3-{x}^{2}+2xh-{h}^{2}-2x+2h-3$

$f\left(x+h\right)-f\left(x-h\right)=4xh+4h$

Now,

$\dfrac{f\left(x+h\right)-f\left(x-h\right)}{h}=\dfrac{4xh+4h}{h}=4x+4$

Let $X=\left\{1,2,3,4,5\right\}$ .If the relation $g=\left\{\left(1,2\right),\left(2,3\right),\left(3,4\right),\left(4,5\right),\left(5,1\right)\right\}$ on $X$ is a function from $X$ to $X$ .then find $g\left(g\left(g\left(2\right)\right)\right)$

Now,

$g\left(2\right)=3$

$g\left(g\left(2\right)\right)=g\left(3\right)=4$

$g\left(g\left(g\left(2\right)\right)\right)=g\left(4\right)=5$

$\therefore\,g\left(g\left(g\left(2\right)\right)\right)=5$

If $f\left(x\right)+f\left(\dfrac{1}{x}\right)=f\left(x\right)f\left(\dfrac{1}{x}\right)$ and $f\left(3\right)=28$ find $f\left(\dfrac{1}{3}\right)$

$f\left(x\right)+f\left(\dfrac{1}{x}\right)=f\left(x\right)f\left(\dfrac{1}{x}\right)$

$\Rightarrow f\left(3\right)+f\left(\dfrac{1}{3}\right)=f\left(3\right)f\left(\dfrac{1}{3}\right)$

$\Rightarrow 28+f\left(\dfrac{1}{3}\right)=28f\left(\dfrac{1}{3}\right)$

$\Rightarrow f\left(\dfrac{1}{3}\right)=28f\left(\dfrac{1}{3}\right)-28$

$\Rightarrow f\left(\dfrac{1}{3}\right)=28\left(f\left(\dfrac{1}{3}\right)-1\right)$

Let

$t=f\left(\dfrac{1}{3}\right)$

$\Rightarrow t=28\left(t-1\right)$

$\Rightarrow t=28t-28$

$\Rightarrow 28t-t=28$

$\Rightarrow 27t=28$

$\Rightarrow t=\dfrac{28}{27}$

$\therefore \,f\left(\dfrac{1}{3}\right)=\dfrac{28}{27}$

If $A \times B = \left\{ {\left( {p,q} \right),\left( {p,r} \right),\left( {m,q} \right),\left( {m,r} \right)} \right\}$
Find A and B

$A=$ set of first elements

$=\left\{ p, m \right\}$

$B=$ set of second elements

$=\left\{ q, r\right\}$

If $f\left(x\right)=\dfrac{x+1}{x-1}$ the show that $f\left(x\right)+f\left(\dfrac{1}{x}\right)=0$

$f\left(x\right)=\dfrac{x+1}{x-1}$

$f\left(\dfrac{1}{x}\right)=\dfrac{\dfrac{1}{x}+1}{\dfrac{1}{x}-1}$

$=\dfrac{\dfrac{1+x}{x}}{\dfrac{1-x}{x}}$

$=\dfrac{1+x}{1-x}=\dfrac{x+1}{-\left(x-1\right)}=-\dfrac{x+1}{x-1}$

$f\left(x\right)+f\left(\dfrac{1}{x}\right)=\dfrac{x+1}{x-1}-\dfrac{x+1}{x-1}=0$

Hence proved.

Find whether $f\left(x\right)=\dfrac{{a}^{x}+1}{{a}^{x}-1}$ is even, odd or neither odd nor even.

$f\left(x\right)=\dfrac{{a}^{x}+1}{{a}^{x}-1}$

$f\left(-x\right)=\dfrac{{a}^{-x}+1}{{a}^{-x}-1}$

$f\left(-x\right)=\dfrac{\dfrac{1}{{a}^{x}}+1}{\dfrac{1}{{a}^{x}}-1}$

$f\left(-x\right)=-\dfrac{{a}^{x}+1}{{a}^{x}-1}=-f\left(x\right)$

Thus

$f\left(x\right)$ is an odd function

If $\dfrac { 4 }{ x } <\dfrac { 1 }{ 3 },$ what is the possible range of value for $x$ ?

$\dfrac{4}{x} < \dfrac{1}{3}$

$\Rightarrow x > 4\times 3$

$\Rightarrow x > 12$

So Range

$=(12, \infty)$ .

1199985_1446014_ans_20824cf2aaa24bceab5045fe22ed248a.jpg

Let $X=\left\{1,2,3,4,5\right\}$ .If the relation $g=\left\{\left(1,2\right),\left(2,3\right),\left(3,4\right),\left(4,5\right),\left(5,1\right),\right\}$ on $X$ is a function from $X$ to $X$ .then find $g\left(g\left(g\left(4\right)\right)\right)$

Now,

$g\left(4\right)=5$

$g\left(g\left(4\right)\right)=g\left(5\right)=1$

$g\left(g\left(g\left(4\right)\right)\right)=g\left(1\right)=2$

$\therefore\,g\left(g\left(g\left(4\right)\right)\right)=2$

Let $f(x)=\left\{\begin{matrix} x^3+x^2+10x, & x < 0\\ x^2e^x, & x\geq 0\end{matrix}\right.$ . Investigate the behaviour of the function for $x\in R$ .

1976676_1423356_ans_3743428b3bce44ea805b93f653c2a57b.jpg

The domain of $f(x)=\sqrt { x-2-2\sqrt { x-3 } } -\sqrt { x-2+2\sqrt { x-3 } }$

$\sqrt{x-2-2\sqrt{x-3}}-\sqrt{x-2+2\sqrt{x-3}}$

domain

$\Rightarrow \sqrt{x-2-2\sqrt{x-3}}\geq 0$

$\sqrt{x-2+2\sqrt{x-3}}\geq 0$

$x-2 \geq 2\sqrt{x-3}$

$x-2 \geq -2\sqrt{x-3}$

$x^{2}-4x+x\geq 4x-12$

$(x-2)^{2}\geq 4(x-3)$

$x^{2}-Bx+16\geq 0$

$(x-4)^{2}\geq 0$

$(x-y)^{2}\geq 0$

$x\epsilon [4,\infty )$

1213165_1419597_ans_661bcc5ad3b942c292cb1312e22c8f3d.jpg

Find the range of $\ln{\left[{\left\{x\right\}}^{2}+3\left\{x\right\}+2\right]}$ where $\left\{x\right\}$ is a fractional function.

$\ln{x}$ is strictly increasing function it is sufficient to find range of

$\ln{\left[{\left\{x\right\}}^{2}+3\left\{x\right\}+2\right]}$

$\ln{\left[{\left\{x\right\}}^{2}+3\left\{x\right\}+2\right]}={\left(\left\{x\right\}+\dfrac{3}{2}\right)}^{2}+2-\dfrac{9}{4}$

$={\left(\left\{x\right\}+\dfrac{3}{2}\right)}^{2}+\dfrac{8-9}{4}$

$={\left(\left\{x\right\}+\dfrac{3}{2}\right)}^{2}-\dfrac{1}{4}$

Since

$0\le\left\{x\right\}<1$

$\dfrac{3}{2}\le\left(\left\{x\right\}+\dfrac{3}{2}\right)<\dfrac{5}{2}$

$\dfrac{9}{4}\le {\left(\left\{x\right\}+\dfrac{3}{2}\right)}^{2}<\dfrac{25}{4}$

$\Rightarrow \,\dfrac{9}{4}-\dfrac{1}{4}\le {\left(\left\{x\right\}+\dfrac{3}{2}\right)}^{2}-\dfrac{1}{4}<\dfrac{25}{4}-\dfrac{1}{4}$

$\Rightarrow \,\dfrac{9-1}{4}\le {\left(\left\{x\right\}+\dfrac{3}{2}\right)}^{2}-\dfrac{1}{4}<\dfrac{25-1}{4}$

$\Rightarrow \,\dfrac{1}{4}\le {\left(\left\{x\right\}+\dfrac{3}{2}\right)}^{2}-\dfrac{1}{4}<\dfrac{24}{4}$

$\Rightarrow \,2 \le {\left(\left\{x\right\}+\dfrac{3}{2}\right)}^{2}-\dfrac{1}{4}<6$

$\Rightarrow \ln{2}\le\ln{\left({\left\{x\right\}}^{2}+3\left\{x\right\}+2\right)}<\ln{6}$

Hence Range

$\in\left[\ln{2},\ln{6}\right)$

Find the range of $f\left(x\right)=\dfrac{\left\{x\right\}}{\left\{x\right\}+1}$

Let

$y=\dfrac{\left\{x\right\}}{\left\{x\right\}+1}$

$\Rightarrow\,y\left\{x\right\}+y=\left\{x\right\}$ as

$0\le \left\{x\right\}<1$

Thus,

$0\le\dfrac{y}{1-y}<1$

$\Rightarrow \dfrac{y}{1-y}\ge 0\Rightarrow \dfrac{y}{y-1}\le 0$

$\Rightarrow y\in\left[0,1\right)$

$\dfrac{y}{y-1}-1<0\Rightarrow \dfrac{2y-1}{y-1}>0$

$\Rightarrow \,t\in\left(-\infty,\dfrac{1}{2}\right)\cup\left(1,\infty\right)$

Hence range is

$\left[0,\dfrac{1}{2}\right)$

Let $f:R\rightarrow R$ be defined by $f\left(x\right)=\dfrac{x+1}{{x}^{2}+2},\,x\in \,R$ .Find $f\left(4\right)$

We have

$f\left(x\right)=\dfrac{x+1}{{x}^{2}+2},\,x\in \,R$ .

$f\left(4\right)=\dfrac{4+1}{{4}^{2}+2}=\dfrac{5}{16+2}=\dfrac{5}{18}$

Let $f:R\rightarrow R$ be defined by $f\left(x\right)=\dfrac{x+1}{{x}^{2}+2},\,x\in \,R$ .Find $f\left(0\right)$

We have

$f\left(x\right)=\dfrac{x+1}{{x}^{2}+2},\,x\in \,R$ .

$f\left(0\right)=\dfrac{0+1}{{0}^{2}+2}=\dfrac{1}{2}$

If $n(A)=15, n(A\cup B)=29, n(A\cap B)=7$ then $n(B)=$ ?

1293592_1462756_ans_84baabc851bd4c6d9bdcbb2ef82d45e9.jpg

Let $A$ and $B$ be two sets such that $n\left(A\right)=5$ and $n\left(B\right)=2$ .If $a,b,c,d,e$ are distinct and $\left(a,2\right),\left(b,3\right),\left(c,2\right),\left(d,3\right),\left(e,2\right)$ are in $A\times B$ ,find $A$ and $B$

Since

$\left(a,2\right),\left(b,3\right),\left(c,2\right),\left(d,3\right),\left(e,2\right)$ are elements of

$A\times B$

$\therefore\,a,b,c,d,e\in A$ and

$2,3\in B$

It is given that

$n\left(A\right)=5$ and

$n\left(B\right)=2$

$\therefore\,A=\left\{a,b,c,d,e\right\}$ and

$B=\left\{2,3\right\}$

Express $A=\left\{\left(a,b\right):2a+b=5,a,b\in\,W\right\}$ as the set of ordered pairs.

Here ,

$W$ denotes the set of whole numbers

We have

$2a+b=5$ where

$a,b\in\,W$

$\therefore\,a=0\Rightarrow\,b=5$

$\Rightarrow\,a=1\Rightarrow\,b=5-2=3$

and

$a=2\Rightarrow\,b=1$

For

$a>3,$ the values of

$b$ given by the above relation are not whole numbers.

$\therefore\,A=\left\{\left(0,5\right),\left(1,3\right),\left(2,1\right)\right\}$

Suppose that f(x) is a function of the form f(x)= $\frac{ax^{8}+bx^{6}+cx^{4}+dx^{2}+15x+1}{x}(x\neq 0)$ If f(5)=-28 then the value of f(-5) /14 is equal to

1777060_1499522_ans_fa9cb2bbe52e4125a289028e3a1417ea.jpg

$f=\{(4,3),(6,7)\}\ \ \ g=\{(4,9),(6,5)\}$ find $f+g$

$f=\{(4,3),(6,7)\}$

$g=\{(4,9),(6,5))\}$

$f+g=\{(4,12),(6,12)\}$

If $f(x+y)=f(x).f(y)$ then prove that $f\left( x \right) ={ \left[ f\left( 1 \right) \right] }^{ x }$

$f(x+y)=f(x).f(y)\\f(2)=f(1).f(1)=[f(1)]^2\\f(3)=f(2).f(1)=[f(1)]^3\\ \Rightarrow f(x)=[f(1)]^x$

If $f(x+y)=f(x)+f(y)$ . Then prove that $f(x)=x.f(1)$

$\textbf{Hint: Use the definition of derivative of a function.}$

Solution:

$\textbf{Step 1: Use the definition of derivative of a function.}$

We have given

$f(x + y) = f(x) + f(y)$ … (1)

$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$

$\therefore$ $f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( x \right) + f\left( h \right) - f\left( x \right)}}{h}$ $\boldsymbol{[using(1)]}$

$\Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( h \right)}}{h}$ …(2)

$\textbf{Step 2: Use L’Hospital rule}$

Using L’Hospital rule

$\mathop {\lim }\limits_{h \to 0} \frac{{f\left( h \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{d}{{dh}}f\left( h \right)}}{{\frac{{dh}}{{dh}}}}$

$\Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{f\left( h \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{f'\left( h \right)}}{1}$

$\Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{f\left( h \right)}}{h} = f'\left( 0 \right)$ …..(3)

Using (2) and (3), we get

$f'\left( x \right) = f'\left( 0 \right)$

Integrating both sides with respect to $x$ .

$\Rightarrow \int {f'\left( x \right)\,dx} = \int {f'\left( 0 \right)} \,dx$

$\Rightarrow f\left( x \right)\, = xf'\left( 0 \right) + c$ …(4)

$\textbf{Step 3: Find the value of the arbitrary constant}$

Put $x = 0$ in eq(4), we get

$f\left( 0 \right)\, = c$

From eq(1), for $x = 0,y = 0$

$f(0 + 0) = f(0) + f(0)$

$\Rightarrow f(0) = 2f(0)$

$\Rightarrow f(0) = 0$

$\therefore \,\,c = 0$

$\Rightarrow f\left( x \right)\, = xf'\left( 0 \right)$ ...(5)

$\textbf{Step 4: Find the value of}$ $\boldsymbol{f'\left( 0 \right)}$

For $x = 1$ in eq (5), we get

$f\left( 1 \right) = f'\left( 0 \right)$

Substituting the value of $f'\left( 0 \right)$ in eq(5), we get

$f\left( x \right)\, = x \cdot f\left( 1 \right)$

$\textbf{Hence, proved.}$

Let $|X|$ denote the number of elements in a set $X$ .let $S=\{1,2,3,4,5,6\}$ be a sample space ,where each elements is equally likely to occur.If A and B are independent events associated with $S$ , then the number of ordered pairs $(A,B)$ such that $1\le |B|<|A|$ equals.

A and B are Independent events

$P(A)P(B)=P(A\cap B) \rightarrow \dfrac{a}{6}\times \dfrac{b}{6}=\dfrac{c}{6}\rightarrow ab=6c$

$|A| =a,|B|=b,|A\cap B|=c$

$(a,b,c)=(3,2,1)$ so

$^6C_1^5C_2^3C_1=180$

$=(4,3,2)$ so

$^6C_2^4C_2^2C_1=180$

$=(6,1,1)$ so

$^6C_1=6$

$=(6,2,2)$ so

$^6C_2=15$

$=(6,3,3)$ so

$^6C_3=20$

$=(6,4,4)$ so

$^6C_4=15$

$=(6,5,5)$ so

$^6C_5=6$
Total=360+62=422$$

Let $f:R^+ \to R$ , where $R^+$ is the set of all positive real numbers, be such that $f(x)=\log_e x$ .
Determine
whether $f(xy)=f(x)+f(y)$ holds.

1416778_1664512_ans_e4cbba6dd94f4348b78696b39b0a9369.jpg

Find the domain and range of each of the following real value functions:
$f(x)= -|x|$

We have,

$f(x)=-|x|$

We observe that

$f(x)$ is defined for all

$x\in R$ . So domain

$(f)=R.$

Also,

$|x| \ge 0$ for all

$x\in R\Rightarrow f(x)\le 0$ for all

$x\in R$

So, range

$(f)=(-\infty , 0)$

Which one of the given relations on $A=\left\{ 1,2,3 \right\}$ is a function?
$f=\left\{ \left( 1,3 \right) ,\left( 2,3 \right) ,\left( 3,2 \right) \right\}$ , $g=\left\{ \left( 1,2 \right) ,\left( 1,3 \right) ,\left( 3,1 \right) \right\}$

$f=\{(1,3),(2,3),(3,2)\}$

$\Rightarrow$ As we can see, in

$f$ , each element of the domain set has unique image.

$\therefore$

$f$ is a function.

$g=\{(1,2),(1,3),(3,1)\}$

$\Rightarrow$ In

$g$ , the element

$1$ of the domain set has two images

$2$ and

$3$ of the range set.

$\therefore$

$g$ is not a function.

1400740_1659053_ans_0fcf309ed4a14da99d1387503c8add7c.png

what is the fundamental difference between a relation and a function? Is every relation a function?

1416767_1664479_ans_f42da2c96f374a318331f4fdae2275a0.jpg

Find the domain and range of each of the following real value functions:
$f(x)=\sqrt {x-1}$

$f(x)=\sqrt {x-1}$ is defined for all

$x$ satisfying

$x-1 \ge 0$ i.e.,

$x\ge 1$ .

So, domain

$(f)=[1, \infty]$ .

Let

$y=\sqrt {x-1}$ . Clearly,

$y\ge 0$ for all

$x\in [1, \infty]$ .

So, range

$(f)=[1, \infty]$ .

Find the domain of each of the following real valued functions of real variable:
$f(x)=\dfrac {x^2+2x+1}{x^2-8x+12}$

$f(x)=\dfrac {x^2 +2x+1}{x^2 -8x+12}=\dfrac {(x+1)^2}{(x-6)(x-2)}$ is defined for all satisfying

$(x-6)(x-2)\neq 0$ i.e.,

$x\neq 2, 6$ .

$\therefore \$ Domain

$(f)=R-\left\{2, 6\right\}$

Determine the domain and range of the relation R defined by $R=\{(x, x+5):x\in \{0, 1, 2, 3, 4, 5\}\}$ .

Domain

$R=\{0, 1, 2, 3, 4, 5\}$ , Range

$R=\{5, 6, 7, 8, 9, 10\}$ .

We have,

$R=\{(x, x+5):x \in \{0, 1, 2, 3, 4, 5\}\}=\{(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)\}$

$\therefore$ Domain

$(R)=\{0, 1, 2, 3, 4, 5\}$ and Range

$(R)=\{5, 6, 7, 8, 9, 10\}$ .

Let $f:(0, \infty)\to R$ and $g:R\to R$ be defined by $f(x)=\sqrt x$ and $g(x)=x$ . Find $f+g, f-g, fg$ and $\dfrac {f}{g}$ .

It is given that

$f:(0, \infty)\to R$ and

$g:R\to R$ are such that

$f(x)=\sqrt {x}$ a nd

$g(x)=x$ .

$D(f+g)=[0, \infty]\cap R=[0, \infty]$

So,

$f+g[0, \infty]\to R$ is given by

$(f+g)(x)=f(x)+g(x)=\sqrt x+x$

$D(f-g)=D(f)\cap D(g)=[0, \infty] \cap R=[0, \infty]$

So,

$f-g:[0, \infty]\to R$ is given by

$(f-g)(x)-g(x)=\sqrt x -x$

$D(fg)=D(f)\cap D(g)=[0, \infty]\cap R=[0, \infty]$

So,

$f(g):(0, \infty)\to R$ is diven by

$(fg)(x)=f(x)g(x)=\sqrt x=x^{3/2}$

$D\left (\frac {f}{g}\right)=D(f)\cap D(g)- \left\{x:g(x)=0\right\}=(0,\infty)$

So,

$\dfrac {f}{g}: (0, \infty)\to R$ is given by

$\left(\dfrac {f}{g}\right)(x)=\dfrac {f(x)}{g(x)}=\dfrac {\sqrt x}{x}=\dfrac {1}{\sqrt x}$

Determine the domain and range of the relation R defined by $R=\{(x, x^3):x$ is a prime number less than $10\}$ .

Domain

$R=\{2, 3, 5, 7\}$ , Range

$R=\{8, 27, 125, 343\}$ .

We have,

$R=\{(x, x^3) : x$ is a prime number less than

$10\}=\{(2, 8), (3, 27), (5, 125), (7, 343)\}$

$\therefore$ Domain

$(R)=\{2, 3, 5, 7\}$ , and Range

$(R)=\{8, 27, 125, 343\}$ .

Let $A=\{1, 2\}$ and $B=\{3, 4\}$ . Find the total number of relations from A into B.

We have,

$n(A)=2, n(B)=2$

$\therefore n(A\times B)=2\times 2=4$

So, there are

$2^4=16$ relations from A to B.

A relation R is defined from a set $A=\{2, 3, 4, 5\}$ to a set $B=\{3, 6, 7, 10\}$ as follows:
$(x, y)\in R\Leftrightarrow x$ is relatively prime to y.
Express R as a set of ordered pairs

Relatively prime number :- When two numbers have no common factors other than 1.

Hence, We have to select pairs of x and y such that it does not have common factors other than 1.

So,

$R=\{(2, 3), (2, 7), (3, 7), (3, 10), (4, 3), (4, 7), (5, 3), (5, 7)\}$ .

Write the following relation as the sets of ordered pairs.
A relation R on the set $\{1, 2, 3, 4, 5, 6, 7\}$ defined by $(x, y)\in R\Leftrightarrow x$ is $\{1, 2, 3, 4, 5, 6, 7\}$ defined by $(x, y)\in R\Leftrightarrow x$ is relatively prime to y.

$R=\left\{\begin{matrix}(2, 3), (2, 5), (2, 7), (3, 2), (3, 4), (3, 5), (3, 7), (4, 3), (4, 5), (4, 7), (5, 2), (5, 3), (5, 4), (5, 6), (5, 7), (6, 5), \end{matrix}\right.$

$\left.\begin{matrix}(6, 7), (7, 2), (7, 3), (7, 4), (7, 5), (7, 6), (7, 7)\end{matrix}\right\}$ .

1641083_1666332_ans_0e92436eef1640718c621ddd7b6f6306.jpg

Let $A=\{1, 2\}$ and $B=\{3, 4\}$ . Find the total number of relations from A into B.

We have,

$n(A)=2, n(B)=2$

$\therefore n(A\times B)=2\times 2=4$

So, there are

$2^4=16$ relations from A to B.

Determine the domain and range of the following relations.

$S=\{(a, b): b=|a-1|, a\in Z$ and $|a|\leq 3\}$ .

We have,

$S=\{(-3, 4), (-2, 3), (-1, 2), (0, 1), (1, 0), (2, 1), (3, 2)\}$

$\therefore$ Domain

$(S)=\{-3, -2, -1, 0, 1, 2, 3\}$

Range

$(S)=\{0, 1, 2, 3, 4\}$ .

Determine the domain and range of the relation R defined by $R=\{(x, x+5):x\in \{0, 1, 2, 3, 4, 5\}\}$ .

Domain

$R=\{0, 1, 2, 3, 4, 5\}$ , Range

$R=\{5, 6, 7, 8, 9, 10\}$ .

We have,

$R=\{(x, x+5):x \in \{0, 1, 2, 3, 4, 5\}\}=\{(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)\}$

$\therefore$ Domain

$(R)=\{0, 1, 2, 3, 4, 5\}$ and Range

$(R)=\{5, 6, 7, 8, 9, 10\}$ .

Determine the domain and range of the following relation.
$R=\{(a, b):a\in N, a < 5, b=4\}$ .

We have,

$R=\{(1, 4), (2, 4), (3, 4), (4, 4)\}$

$\therefore$ Domain

$(R)=\{1, 2, 3, 4\}$ , Range

$(R)=\{4\}$ .

Determine the domain and range of the following relation.

$R=\{(a, b):a\in N, a < 5, b=4\}$ .

Domain

$R=\{1, 2, 3, 4\}$ ,

Range

$R=\{4\}$ .

Let $A=\{a, b\}$ . List all relations on A and find their number.

1499273_1666344_ans_2aca1d132fd146da8b56920287ad98b1.jpg

Determine the domain and range of the following relations.
$S=\{(a, b): b=|a-1|, a\in Z$ and $|a|\leq 3\}$ .

We have,

$S=\{(-3, 4), (-2, 3), (-1, 2), (0, 1), (1, 0), (2, 1), (3, 2)\}$

$\therefore$ Domain

$(S)=\{-3, -2, -1, 0, 1, 2, 3\}$

Range

$(S)=\{0, 1, 2, 3, 4\}$ .

Let $A=\{x, y, z\}$ and $B=\{a, b\}$ . Find the total number of relations from A into B.

1499162_1666346_ans_3cc125805e384282b94d8f9e7a1e0744.jpg

Let R be the relation on Z defined by $R=\{(a, b):a, b\in Z, a-b$ is an integer $\}$ . Find the domain and range of R.

The relation R on Z is defined by

$R=\{(a, b):a, b\in Z, a-b$ is an integer

$\}$

Since

$a-b$ is an integer for all a, b

$\in$ Z.

So, Domain

$(R)=Z$ , Range

$(R)=Z$ .

Let $A=\{1, 2, 3, 4, 5, 6\}$ . Let R be a relation on A defined by $R=\{(a, b):a, b\in A, b$ is exactly divisible by a $\}$ Find the range of R.

We have,

$R=\{(a, b): a, b\in A, b$ is exactly divisible by a

$\}$ ,

where

$A=\{1, 2, 3, 4, 5, 6\}$ .

Range

$(R)=\{1, 2, 3, 4, 5, 6\}$ .

Let $A=\{1, 2, 3,....., 14\}$ . Define a relation on a set A by $R=\{(x, y): 3x-y=0$ , where $x, y\in A\}$ .
Depict this relationship using an arrow diagram. Write down its domain, co-domain and range.

$A=\{1,2,3,....,14\}$

$R=\{(x,y):3x-y=0,$ where

$x,y\in A\}$

Or,

$R=\{(x,y):2x=y,\,$ where

$x,y\in A\}$

$3\times 1=3$

$3\times 2=6$

$3\times 3=9$

$3\times 4=12$

$R=\{(1,3),(2,6),(3,9),(4,12)\}$

Domain

$(R)=\{1,2,3,4\}$

Range

$(R)=\{3,6,9,12\}$

Co-domain

$(R)=A$

1635371_1666352_ans_72172ba47b0444d6ae40d68f38f30295.JPG

Answer the following question in one word or one sentence or as per the exact requirement of the question. If $R=\{(x, y):x, y\in Z, x^2+y^2\leq 4\}$ is a relation defined on the set Z of integers, then write domain of R.

1445017_1666459_ans_366a330d6ffb479fa052474f746788ed.jpg

Let $A=\{1, 2, 3, 4, 5, 6\}$ . Let R be a relation on A defined by $R=\{(a, b):a, b\in A$ , b is exactly divisible by a $\}$ . Find the domain of R.

We have,

$R=\{(a, b): a, b\in A, b$ is exactly divisible by a

$\}$ ,

where

$A=\{1, 2, 3, 4, 5, 6\}$ .

Domain

$(R)=\{1, 2, 3, 4, 5, 6\}$ .

$A=\{1, 2, 3, 5\}$ and $B=\{4, 6, 9\}$ . Define a relation R from A to B by $R=\{(x, y):$ the difference between x and y is odd, $x\in A, y\in B\}$ . Write R in Roster form.

We have,

$A=\{1, 2, 3, 5\}, B=\{4, 6, 9\}$

$R=\{(x, y):$ the difference between x and y is odd,

$x\in A, y\in B\}$ .

$R=\{(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)\}$ .

Define a relation R on the set N of natural numbers by $R=\{(x, y):y=x+5$ , x is a natural number less than $4, x, y\in N\}$ .
Depict this relationship using (i) roster form (ii) an arrow diagram. Write down the domain and range or R.

(i)We have,

$R=\{(x, y) :y=x+5$ ,

x is a natural number less than

$4, x, y\in N$

$R=\{(1, 6), (2, 7), (3, 8)\}$

(ii) Domain

$(R)=[1, 2, 3]$ ,

Range

$(R)=\{6, 7, 8\}$ .

1422108_1666354_ans_3fbbc09d55f8430090a25d5d21d69e03.png

Answer the following question in one word or one sentence or as per exact requirement of the question.
If $R=\{(x, y): x, y\in W, 2x+y=8\}$ , then write the domain and range of R.

1445026_1666467_ans_0730fda62eed4c8e820ab1a08ab0513b.jpg

Determine the domain and range of the relation R defined by $R=\{(x, x^3):x$ is a prime number less than $10\}$ .

We have,

$R=\{(x, x^3) : x$ is a prime number less than

$10\}=\{(2, 8), (3, 27), (5, 125), (7, 343)\}$

$\therefore$ Domain

$(R)=\{2, 3, 5, 7\}$ , and Range

$(R)=\{8, 27, 125, 343\}$ .

Determine the domain and range of the following relations.
$S=\{(a, b): b=|a-1|, a\in Z$ and $|a|\leq 3\}$ .

$S=\{(0, 1), (-1, 2), (-2, 3), (-3, 4), (1, 0), (2, 1), (3, 2)\}$ .

We have,

$S=\{(-3, 4), (-2, 3), (-1, 2), (0, 1), (1, 0), (2, 1), (3, 2)\}$

$\therefore$ Domain

$(S)=\{-3, -2, -1, 0, 1, 2, 3\}$

Range

$(S)=\{0, 1, 2, 3, 4\}$ .

If $f(x)+f(x+4)=f(x+2)+f(x+6)$ then find the period of $f(x)$ .

$\Rightarrow$

$f(x)+f(x+4)=f(x+2)+f(x+6)$ [ Given ] ---- ( 1 )

Lets replace

$x$ by

$x+2$

$\Rightarrow$

$f(x+2)+f(x+6)=f(x+4)+f(x+8)$ ---- ( 2 )

From equation ( 1 ) and ( 2 ) we get,

$\Rightarrow$

$f(x)+f(x+4)=f(x+4)+f(x+8)$

$\Rightarrow$

$f(x)=f(x+8)$

$\therefore$ Period of

$f(x)$ is

$8.$

Answer the following question in one word or one sentence or as per exact requirement of the question.
If $A=\{1, 3, 5\}$ and $B=\{2, 4\}$ , list the elements of R, if $R=\{(x, y):x, y\in A\times B$ and $x > y\}$ .

1445025_1666465_ans_ce7fed2030624ba285c0c47510917d19.jpg

Let $R=\{(x, y):x+2y=8\}$ be a relation on N. Write the range of R.

Given,

$R=\left \{ (x,y):x+2y=8 \right \}$

when,

$x=2,y=3$

$2+2\times3= 2+6=8$

when,

$x=4,y=2$

$4+2\times2= 4+4=8$

when,

$x=6,y=1$

$6+2\times1= 6+2=8$

Domain

$\left \{ 3,2,1 \right \}$

Range

$R=\left \{ (2,3),(4,2),(6,1) \right \}$

Find the domain and range of the relation
$R=\{(-1, 1), (1, 1), (-2, 4), (2, 4)\}$ .

Given,

R = {

$(-1,1), (1,1), (-2,4), (2,4)$ }

Therefore, from given we have,

R = {

$a, a^2 : a$ is a natural number}

$dom(R)=\left \{ −1,1,−2,2 \right \}$ and

$range (R)=\left \{ 1,4 \right \}$

Let $R=\{(a, a^3):a$ is a prime number less than $10\}$ . Find dom(R).

Given,

R={

$(a, a^3):$ a is prime number less than 10}

2, 3, 5 and 7 are the prime numbers less than 10.

Therefore, we have,

when

$a = 2, a^3 = 2^3=8$

when

$a = 3, a^3 = 3^3=27$

when

$a = 5, a^3 = 5^3=125$

when

$a = 7, a^3 = 7^3=343$

$\therefore R =$ {

$(2,8), (3,27), (5,125), (7,343)$ }

And the domain of R is {

$2,3,5,7$ }

Let $R=\{(a, a^3):a$ is a prime number less than $10\}$ . Find R.

Given,

R={

$(a, a^3):$ a is prime number less than 10}

2, 3, 5 and 7 are the prime numbers less than 10.

Therefore, we have,

when

$a = 2, a^3 = 2^3=8$

when

$a = 3, a^3 = 3^3=27$

when

$a = 5, a^3 = 5^3=125$

when

$a = 7, a^3 = 7^3=343$

$\therefore R =$ {

$(2,8), (3,27), (5,125), (7,343)$ }

Let $R=\{(a, a^3):a$ is a prime number less than $10\}$ . Find range (R).

Given,

R={

$(a, a^3):$ a is prime number less than 10}

2, 3, 5 and 7 are the prime numbers less than 10.

Therefore, we have,

when

$a = 2, a^3 = 2^3=8$

when

$a = 3, a^3 = 3^3=27$

when

$a = 5, a^3 = 5^3=125$

when

$a = 7, a^3 = 7^3=343$

$\therefore R =$ {

$(2,8), (3,27), (5,125), (7,343)$ }

The range of R is {

$8,27,125,343$ }

Let $R=\{(a, a^3):a$ is a prime number less than $5\}$ . Find the range of R.

Given,

R={

$(a, a^3):$ a is prime number less than 5}

2 and 3 are the prime numbers less than 5.

Therefore, we have,

when

$a = 2, a^3 = 2^3=8$

when

$a = 3, a^3 = 3^3=27$

$\therefore R =$ {

$(2,8), (3,27)$ }

The range of R is {

$8,27$ }

Let $f: R\rightarrow R: f(x)=3x+2$ , find $f\{f(x)\}$ .

$f\{f(x)\}=f(3x+2)=3(3x+2)+2=(9x+8)$

$\therefore f\{f(x)\}=(9x+8)$ .

Let $f: R\rightarrow R$ be defined by
$f(x)=\left\{\begin{matrix} 2x+3, & when & x < -2\\ x^2-2, & when & -2\leq x\leq 3\\ 3x-1, & when & x > 3\end{matrix}\right.$
Find $f(2)$ .

1544137_1705491_ans_547bcb9083984644a7595b0494c52005.png

Let $R=\{\left(a, \dfrac{1}{a}\right):a\in N$ and $1 < a < 5\}$ . Find the doman and range of R.

Dom

$(R)=\{2, 3, 4\}$
and Range

$(R)=\left\{\dfrac{1}{2}, \dfrac{1}{3}, \dfrac{1}{4}\right\}$ .

Let $f: R\rightarrow R$ be defined by
$f(x)=\left\{\begin{matrix} 2x+3, & when & x < -2\\ x^2-2, & when & -2\leq x \leq 3\\ 3x-1, & when & x >3\end{matrix}\right.$
Find $f(-3)$ .

1544144_1705497_ans_86c938c8a8a74d0cb5da90b42ea99f28.png

Let $R=\{(a, b):b=|a-1|, a\in Z$ and $|a| < 3\}$ . Find the domain and range of R.

1546001_1705171_ans_4e0cbb80b04947a9bde5340fbb994a73.jpeg

Let $f: R\rightarrow R$ be defined by
$f(x)=\left\{\begin{matrix} 2x+3, & when & x < -2\\ x^2-2, & when & -2\leq x \leq 3\\ 3x-1, & when & x > 3\end{matrix}\right.$
Find $f(4)$ .

1544140_1705494_ans_6089fa4041ae4442bb8ae7685bf4ae68.png

Let $R=\{(a, b):a, b\in N$ and $a+3b=12\}$ . Find the domain and range of R.

The relation which can be formed is

$R=\{(3, 3), (6, 2), (9, 1)\}$ .

$dom(R)=\{3, 6, 9\}$ and range

$(R)=\{3, 2, 1\}$ .

Let $R=\{(a, b):a, b\in N$ and $b=a+5, a < 4\}$ . Find the domain and range of R.

1546002_1705176_ans_e8d2d0dea73449dab8093b0afff8a68f.jpeg

For any sets $\displaystyle A, B$ and $\displaystyle C$ , prove that:
$\displaystyle A \times (B \cup C) = (A \times B)\cup (A \times C)$

Let

$A=\left\{ -3,-1 \right\}$ ,

$B=\left\{ 1,3 \right\}$ and

$C=\left\{ 3,5 \right\}$

$B\cup C=\left\{ 1,3 \right\} \cup \left\{ 3,5 \right\}$

$\therefore B\cup C=\left\{ 1,3,5 \right\}$

$A\times \left( B\cup C \right) =\left\{ -3,-1 \right\}\times \left\{ 1,3,5 \right\}$

$\therefore A\times \left( B\cup C \right) =\left\{ \left( -3,1 \right) ,\left( -3,3 \right) ,\left( -3,5 \right) ,\left( -1,1 \right) ,\left( -1,3 \right) ,\left( -1,5 \right) \right\}$ Equation (1)

$A\times B=\left\{ -3,-1 \right\} \times \left\{ 1,3 \right\}$

$\therefore A\times B=\left\{ \left( -3,1 \right) ,\left( -3,3 \right) ,\left( -1,1 \right) ,\left( -1,3 \right) \right\}$

$A\times C=\left\{ -3,-1 \right\}\times \left\{ 3,5 \right\}$

$\therefore A\times C=\left\{ \left( -3,3 \right) ,\left( -3,5 \right) ,\left( -1,3 \right) ,\left( -1,5 \right) \right\}$

$\therefore \left( A\times B \right) \cup \left( A\times C \right) =\left\{ \left( -3,1 \right) ,\left( -3,3 \right) ,\left( -1,1 \right) ,\left( -1,3 \right) \right\}\cup \left\{ \left( -3,3 \right) ,\left( -3,5 \right) ,\left( -1,3 \right) ,\left( -1,5 \right) \right\}$

$\therefore \left( A\times B \right) \cup \left( A\times C \right) =\left\{ \left( -3,1 \right) ,\left( -3,3 \right) ,\left( -1,1 \right) ,\left( -1,3 \right) ,\left( -3,5 \right) ,\left( -1,5 \right) \right\}$ Equation (2)

From equation (1) and (2),

$A\times \left( B\cup C \right)=\left( A\times B \right) \cup \left( A\times C \right)$

Let $f:R\rightarrow R$ be defined by
$f(x)=\left\{\begin{matrix} 2x+3, & when & x < -2\\ x^2-2, & when & -2\leq x \leq 3\\ 3x-1, & when & x > 3\end{matrix}\right.$
Find $f(-1)$ .

1544143_1705495_ans_e4cc1613fce94dcbb4f6b71457d285b6.png

If $\displaystyle A = \{1, 3, 5\}, B = \{3, 4\}$ and $\displaystyle C = \{2, 3\}$ , verify that
$\displaystyle A \times (B \cap C) = ( A \times B) \cap (A \times C)$

1848268_1708135_ans_b565d07004db42c89951eaa92c393902.jpeg

If $\displaystyle A = \{1, 3, 5\}, B = \{3, 4\}$ and $\displaystyle C = \{2, 3\}$ , verify that
$\displaystyle A \times (B \cup C) = ( A \times B) \cup (A \times C)$

1848246_1708134_ans_b6ab954a6db34fb78d625d57e5f87e78.jpeg

Let $\displaystyle A = \{ x \in W : x < 2 \}, B = x \in N : 1 < x \leq 4\}$ and $\displaystyle C = \{3, 5\}$ . Verify that :
$\displaystyle A \times (B \cap C) = (A \times B) \cap (A \times C)$

1972419_1708137_ans_03b9155035e843b7b94d84ba3a81535d.jpeg

If $\displaystyle A = \{2, 3, 5\}$ and $\displaystyle B = \{5, 7\}$ , find $A \times A$ and enter the number of elements of it.

The number of elements are 9
1847570_1708085_ans_17b5124c53f74bd9a17df19a41667fad.jpeg

Let $\displaystyle A \times B = \{ (a, b) : b = 3a - 2\}.$ If $\displaystyle (x, -5)$ and $\displaystyle (2, y)$ belong to $\displaystyle A \times B$ , find the values of $\displaystyle x$ and $\displaystyle y$ .

If the answer is $\displaystyle x = -1, y = 4$ then $1,$ else $0$

1848718_1708178_ans_6886513cfa974373a2df4a2f9550dd35.jpeg

Let $\displaystyle A = \{2, 3 \}$ and $\displaystyle b = \{ 4, 5\}$ . Find $\displaystyle (A \times B)$ . How many subsets will $\displaystyle (A \times B)$ have?

1847926_1708166_ans_a99400d382aa454a9c4111223b7a91ca.jpeg

Enter 1 if it is true else enter 0.
If $\displaystyle A = \{2, 3, 5\}$ and $\displaystyle B = \{5, 7\}$ , then $\displaystyle A \times B = \{(2, 5), (2, 7), (3, 5), (3, 7), (5, 5), (5, 7)\}$

Given

$A=\{2,3,5\},B=\{5,7\}$

$A\times B=\{\left( 2,5 \right),\left( 2,7 \right),\left( 3,5 \right),\left( 3,7 \right),\left( 5,5 \right),\left( 5,7 \right)\}$

The correct answer is 1.

If $\displaystyle A = \{2, 3, 5\}$ and $\displaystyle B = \{5, 7\}$ , find $B \times B$ and enter the value of no of pair which contain the same element.

1847537_1708091_ans_b3863bc676244ab98ea722d059a8a3cd.jpeg

If $\displaystyle A = \{2, 3, 5\}$ and $\displaystyle B = \{5, 7\}$ , find:
$B \times A$

Given:

$\displaystyle A = \{2, 3, 5\}$ and

$\displaystyle B = \{5, 7\}$

therefore

$B \times A$ can be written as,

$\displaystyle B \times A = \{(5, 2), (5, 3), (5, 5), (7, 2), (7, 3), (7, 5)\}$

Let $\displaystyle A = \{ x \in W : x < 2 \}, B = x \in N : 1 < x \leq 4\}$ and $\displaystyle C = \{3, 5\}$ . Verify that :
$\displaystyle A \times (B \cup C) = (A \times B) \cup (A \times C)$

$A=\{0,1\},\,B=\{2,3,4\}$ and

$C=\{3,5\}$

$\cup$ is combination of two sets.

$B\cup C=\{2,3,4,5\}$

Now,

$L.H.S=A\times(B\cup C)$

$=\{0,1\}\times (2,3,4,5\}$

$=\{(0,2),(0,3),(0,4),(0,5),(1,2),(1,3),(1,4),(1,5)\}$ ---- ( 1 )

$R.H.S=(A\times B)\cup (A\times C)$

$=\{0,1\}\times \{2,3,4\} \cup \{0,1\}\times \{3,5\}$

$=\{(0,2),(0,3),(0,4),(1,2),(1,3),(1,4)\}\cup\{(0,3),(0,5),(1,3),(1,5)\}$

$=\{(0,2),(0,3),(0,4),(0,5),(1,2),(1,3),(1,4),(1,5)\}$ ---- ( 2 )

From ( 1 ) and ( 2 )

$\Rightarrow$

$L.H.S=R.H.S$

$\therefore$

$A\times (B\cup C)=(A\times B)\cup (A\times C)$

Let $\displaystyle A = \{-3, -1\}, B = \{1, 3\}$ and $\displaystyle C = \{3, 5\}$ . Find $\displaystyle A \times B$ , and enter the value of number of elements of it.

The Cartesian product of two sets

$A$ and

$B$ , denoted

$A \times B$ , is the set of all possible ordered pairs where the elements of

$A$ are first and the elements of

$B$ are second.

$A=\{-3,-1\},\,B=\{1,3\}$ and

$C=\{3,5\}$

$\therefore$

$A\times B=\{(-3,1),(-3,3),(-1,1),(-1,3)\}$

If $\displaystyle A$ and $\displaystyle B$ are nonempty sets, prove that
$\displaystyle A \times B = B \times A \iff A = B$ .

$A=\phi ,B=\phi \,\,\,\,\,\left[ given \right]$

$Let$

$a\in A,b\in B\,\,\,\,\,\,..............\left( i \right)$

$i)a\in A\,\left( fix \right)$

$\left( a,b \right)\in A\times B$

$\left( a,b \right)\in B\times A$

$\Rightarrow a\in B,b\in A\,\,\,\,\forall b\in B$

$\Rightarrow B\subseteq A\,\,\,\,\,\,\,.........\left( ii \right)$

$ii)b\in B\,\left( fix \right)$

$\left( a,b \right)\in A\times B$

$\left( a,b \right)\in B\times A$

$\Rightarrow b\in A,a\in B\,\,\,\,\forall a\in A$

$\Rightarrow A\subseteq B\,\,\,\,\,\,\,.........\left( ii \right)$

$from\,\left( i \right)and\,\left( ii \right)$

$A=B$

If $\displaystyle A = \{5, 7\}$ , find $\displaystyle A \times A$ and enter the value of number of elements of it.

The Cartesian product of two sets

$A$ and

$A$ , denoted

$A \times A$ , is the set of all possible ordered pairs where the elements of

$A$ are first and the elements of

$A$ are second.

$A=\{5,7\}$

$A\times A=\{5,7\}\times\{5,7\}$

$A\times A=\{(5,5),(5,7),(7,5),(7,7)\}$

If $\displaystyle A \subseteq B$ , prove that $\displaystyle A \times C \subseteq B \times C$ for any set $\displaystyle C$ .

1947793_1708313_ans_67647f05f7ec4e1ea410dea9ba07b819.jpg

For any sets $\displaystyle A, B$ and $\displaystyle C$ , prove that:
$\displaystyle A \times (B - C) = (A \times B)- (A \times C)$

Let

$A=\left\{ a,b,c,d,e \right\}$ ,

$B=\left\{ a,c,e,g \right\}$ and

$C=\left\{ b,c,f,g \right\}$

$B-C=\left\{ a,c,e,g \right\}-\left\{ b,c,f,g \right\}$

$\therefore B-C=\left\{ a,e \right\}$

$A\times \left( B-C \right) =\left\{ a,b,c,d,e \right\}\times \left\{ a,e \right\}$

$\therefore A\times \left( B-C \right) =\left( \left( a,a \right) ,\left( a,e \right) ,\left( b,a \right) ,\left( b,e \right) ,\left( c,a \right) ,\left( c,e \right) ,\left( d,a \right) ,\left( d,e \right) ,\left( e,a \right) ,\left( e,e \right) \right)$ Equation (1)

$A\times B=\left\{ a,b,c,d,e \right\} \times \left\{ a,c,e,g \right\}$

$\therefore A\times B=\left\{ \left( a,a \right) ,\left( a,c \right) ,\left( a,e \right) ,\left( a,g \right) ,\left( b,a \right) ,\left( b,c \right) ,\left( b,e \right) ,\left( b,g \right) ,\left( c,a \right) ,\left( c,c \right) ,\left( c,e \right) ,\left( c,g \right) ,\left( d,a \right) ,\left( d,c \right) ,\left( d,e \right) ,\left( d,g \right) ,\left( e,a \right) ,\left( e,c \right) ,\left( e,e \right) ,\left( e,g \right) \right\}$

$\therefore A\times C=\left\{ a,b,c,d,e \right\}\times \left\{ b,c,f,g \right\}$

$\therefore A\times C=\left\{ \left( a,b \right) ,\left( a,c \right) ,\left( a,f \right) ,\left( a,g \right) ,\left( b,b \right) ,\left( b,c \right) ,\left( b,f \right) ,\left( b,g \right) ,\left( c,b \right) ,\left( c,c \right) ,\left( c,f \right) ,\left( c,g \right) ,\left( d,b \right) ,\left( d,c \right) ,\left( d,f \right) ,\left( d,g \right) ,\left( e,b \right) ,\left( e,c \right) ,\left( e,f \right) ,\left( e,g \right) \right\}$

$\left( A\times B \right) -\left( A\times C \right) =\left\{ \left( a,a \right) ,\left( a,c \right) ,\left( a,e \right) ,\left( a,g \right) ,\left( b,a \right) ,\left( b,c \right) ,\left( b,e \right) ,\left( b,g \right) ,\left( c,a \right) ,\left( c,c \right) ,\left( c,e \right) ,\left( c,g \right) ,\left( d,a \right) ,\left( d,c \right) ,\left( d,e \right) ,\left( d,g \right) ,\left( e,a \right) ,\left( e,c \right) ,\left( e,e \right) ,\left( e,g \right) \right\}- \left\{ \left( a,b \right) ,\left( a,c \right) ,\left( a,f \right) ,\left( a,g \right) ,\left( b,b \right) ,\left( b,c \right) ,\left( b,f \right) ,\left( b,g \right) ,\left( c,b \right) ,\left( c,c \right) ,\left( c,f \right) ,\left( c,g \right) ,\left( d,b \right) ,\left( d,c \right) ,\left( d,f \right) ,\left( d,g \right) ,\left( e,b \right) ,\left( e,c \right) ,\left( e,f \right) ,\left( e,g \right) \right\}$

$\left( A\times B \right) -\left( A\times C \right) =\left\{ \left( a,a \right) ,\left( a,e \right) ,\left( b,a \right) ,\left( b,e \right) ,\left( c,a \right) ,\left( c,e \right) ,\left( d,a \right) ,\left( d,e \right) ,\left( e,a \right) ,\left( e,e \right) \right\}$ Equation (2)

From equation (1) and (2),

$A\times \left( B-C \right)=\left( A\times B \right) -\left( A\times C \right)$

If $\displaystyle A \subseteq B$ and $\displaystyle C \subseteq D$ then prove that $\displaystyle A \times C \subseteq B \times D$ .

1947795_1708314_ans_9c16f1fb65cf4a76a2fe49e096d08e55.jpg

Let $\displaystyle A = \{-3, -1\}, B = \{1, 3\}$ and $\displaystyle C = \{3, 5\}$ . Find $\displaystyle A \times (B \times C)$ and enter the value of number of elements of it.

Given

$A=\left\{ -3,-1 \right\}$ ,

$B=\left\{ 1,3 \right\}$ and

$C=\left\{ 3,5 \right\}$

Now,

$B\times C=\left\{ 1,3 \right\}\times \left\{ 3,5 \right\}$

$\therefore B\times C=\left\{ \left( 1,3 \right) ,\left( 1,5 \right) ,\left( 3,3 \right) ,\left( 3,5 \right) \right\}$

Now,

$\therefore A\times \left( B\times C \right) =\left\{ -3,-1 \right\} \times \left\{ \left( 1,3 \right) ,\left( 1,5 \right) ,\left( 3,3 \right) ,\left( 3,5 \right) \right\}$

$\therefore A\times \left( B\times C \right) =\left\{ \left( -3,1,3 \right) ,\left( -3,1,5 \right) ,\left( -3,3,3 \right) ,\left( -3,3,5 \right) ,\left( -1,1,3 \right) ,\left( -1,1,5 \right) ,\left( -1,3,3 \right) ,\left( -1,3,5 \right) \right\}$

For any sets $\displaystyle A, B$ and $\displaystyle C$ , prove that:
$\displaystyle A \times (B \cap C) = (A \times B)\cap (A \times C)$

Let

$A=\left\{ a,b,c,d,e \right\}$ ,

$B=\left\{ a,c,e,g \right\}$ and

$C=\left\{ b,c,f,g \right\}$

$B\cap C=\left\{ a,c,e,g \right\}\cap \left\{ b,c,f,g \right\}$

$\therefore B\cap C=\left\{ c,g \right\}$

Now,

$A\times \left( B\cap C \right) =\left\{ a,b,c,d,e \right\} \times \left\{ c,g \right\}$

$\therefore A\times \left( B\cap C \right) =\left\{ \left( a,c \right) ,\left( a,g \right) ,\left( b,c \right) ,\left( b,g \right) ,\left( c,c \right) ,\left( c,g \right) ,\left( d,c \right) ,\left( d,g \right) ,\left( e,c \right) ,\left( e,g \right) \right\}$ Equation (1)

$A\times B=\left\{ a,b,c,d,e \right\}\times \left\{ a,c,e,g \right\}$

$\therefore A\times B=\left\{ \left( a,a \right) ,\left( a,c \right) ,\left( a,e \right) ,\left( a,g \right) ,\left( b,a \right) ,\left( b,c \right) ,\left( b,e \right) ,\left( b,g \right) ,\left( c,a \right) ,\left( c,c \right) ,\\\qquad\qquad\quad\left( c,e \right) ,\left( c,g \right) ,\left( d,a \right) ,\left( d,c \right) ,\left( d,e \right) ,\left( d,g \right) ,\left( e,a \right) ,\left( e,c \right) ,\left( e,e \right) ,\left( e,g \right) \right\}$

$A\times C=\left\{ a,b,c,d,e \right\}\times \left\{ b,c,f,g \right\}$

$\therefore A\times C=\left\{ \left( a,b \right) ,\left( a,c \right) ,\left( a,f \right) ,\left( a,g \right) ,\left( b,b \right) ,\left( b,c \right) ,\left( b,f \right) ,\left( b,g \right) ,\left( c,b \right) ,\left( c,c \right) ,\left( c,f \right) ,\\\qquad\qquad\quad\left( c,g \right) ,\left( d,b \right) ,\left( d,c \right) ,\left( d,f \right) ,\left( d,g \right) ,\left( e,b \right) ,\left( e,c \right) ,\left( e,f \right) ,\left( e,g \right) \right\}$

$\therefore \left( A\times B \right) \cap \left( A\times C \right) =\left\{ \left( a,a \right) ,\left( a,c \right) ,\left( a,e \right) ,\left( a,g \right) ,\left( b,a \right) ,\left( b,c \right) ,\left( b,e \right) ,\left( b,g \right) ,\left( c,a \right) ,\left( c,c \right) ,\left( c,e \right) ,\\\qquad\qquad\quad\qquad\qquad\left( c,g \right) ,\left( d,a \right) ,\left( d,c \right) ,\left( d,e \right) ,\left( d,g \right) ,\left( e,a \right) ,\left( e,c \right) ,\left( e,e \right) ,\left( e,g \right) \right\}\times \left\{ \left( a,b \right) ,\left( a,c \right) ,\left( a,f \right) ,\\\qquad\qquad\qquad\qquad\quad\left( a,g \right) ,\left( b,b \right) ,\left( b,c \right) ,\left( b,f \right) ,\left( b,g \right) ,\left( c,b \right) ,\left( c,c \right) ,\left( c,f \right) ,\left( c,g \right) ,\left( d,b \right) ,\\\qquad\qquad\qquad\qquad\quad\left( d,c \right) ,\left( d,f \right) ,\left( d,g \right) ,\left( e,b \right) ,\left( e,c \right) ,\left( e,f \right) ,\left( e,g \right) \right\}$

$\therefore \left( A\times B \right) \cap \left( A\times C \right) =\left\{ \left( a,c \right) ,\left( a,g \right) ,\left( b,c \right) ,\left( b,g \right) ,\left( c,c \right) ,\left( c,g \right) ,\left( d,c \right) ,\left( d,g \right) ,\left( e,c \right) ,\left( e,g \right) \right\}$ Equation (2)

From equation (1) and (2),

$A\times \left( B\cap C \right)=\left( A\times B \right) \cap \left( A\times C \right)$

Let $\displaystyle A = \{-3, -1\}, B = \{1, 3\}$ and $\displaystyle C = \{3, 5\}$ . Find $\displaystyle B \times C$ and enter the value of number of elements of it.

The Cartesian product of two sets

$B$ and

$C$ , denoted

$B \times C$ , is the set of all possible ordered pairs where the elements of

$B$ are first and the elements of

$C$ are second.

$A=\{-3,-1\},\,B=\{1,3\}$ and

$C=\{3,5\}$

$\therefore$

$B\times C=\{(1,3),(1,5),(3,3),(3,5)\}$

Lets $\displaystyle A = \{1, 2\}$ and $\displaystyle B = \{2, 3\}$ . Then, write down the no. of all possible subsets of $\displaystyle A\times B$ .

$\displaystyle A\times B = \{(1, 2), (1, 3), (2, 2), (2, 3)\}$ . So, it will have

$2^4 = 16$ subsets.

$\phi, \{(1, 2)\}, \{(1, 3)\},\{(2, 2)\}, \{(2, 3)\}, \{(1, 2), (1, 3)\}, \{(1, 2), (2, 2)\}, \{(1, 2), (2, 3)\},$

$\{(1, 3), (2, 2)\}, \{(1, 3), (2, 3)\}, \{(2, 2), (2, 3)\}, \{(1, 2), (1, 3), (2, 2)\},\{(1, 2), (1, 3), (2, 3)\},$

$\{(1, 2), (2, 2), (2, 3)\}, \{(1, 2), (2, 2), (2, 3)\}, \{(1, 2), (1, 3), (2, 2), (2, 3)\}$

Let $\displaystyle A = \{a, b, c, d\}, B = \{c, d, e\}$ and $\displaystyle C = \{d, e, f, g \}$ . Then verify each of the following identities:
$\displaystyle A \times (B \cap C) = ( A \times B ) \cap (A \times C)$

1851896_1708397_ans_bf41ab33dbdd4701bad926e9538a5c26.jpg

Let $R^+$ be the set all positive real numbers.
Let $f : R^+ \rightarrow : f(x) = log_ex$
Find $\{x:x \epsilon R^+ and f(x) = -2 \}$

1858787_1708522_ans_cfa197b17ce3403d95ed0043879c5288.jpeg

Let $R^+$ be the set all positive real numbers.
Let $f : R^+ \rightarrow : f(x) = log_ex$
Find out whether $f(xy) = f(x) +f (y)$ for all $x, y \epsilon R^+$

1858788_1708524_ans_80e8bb6c7f3d42aba671a88f19cf1d21.jpeg

Let $\displaystyle A = \{2, 3, 5\}$ and $\displaystyle R = \{(2, 3), (2, 5), (3, 3), (3, 5)\}$ .
Show that $\displaystyle R$ is a binary relation on $\displaystyle A$ . Which element is common between domain and range of $R$ .

The correct answer is 3
1858775_1708502_ans_50961d3c10674eba85ff8f16669a5f2f.jpeg

Let $\displaystyle A = \{a, b, c, d\}, B = \{c, d, e\}$ and $\displaystyle C = \{d, e, f, g \}$ . Then verify each of the following identities:
$\displaystyle (A \times B) \cap ( B \times A) = ( A \cap B) \times ( A \cap B)$

1851886_1708405_ans_05399f2fc76a46929288b77356b91143.jpg

Let $\displaystyle A = \{0, 1, 2, 3, 4, 5, 6, 7, 8 \}$ and let $\displaystyle R = \{(a, b): a, b \in A$ and $2a + 3b = 12\}$ .
Express $\displaystyle R$ as a set of ordered pairs. Show that $\displaystyle R$ is a binary relation on $\displaystyle A$ .
Which element is common between domain and range.

1858789_1708531_ans_55537e93a8f648a39c2c54937f8a8791.jpeg

Let $\displaystyle A = \{a, b, c, d\}, B = \{c, d, e\}$ and $\displaystyle C = \{d, e, f, g \}$ . Then verify each of the following identities:
$\displaystyle A \times (B - C) = (A \times B ) - ( A \times C )$

1851356_1708402_ans_e91f5d24ce1e4a159a64e92f62f50ec4.jpg

If $f(x) = x^3$ , find the value of $\frac{ {f(5) - f(1) } }{ (5-1)} .$

1858785_1708515_ans_90fa054310f448559a2fb4658d8ad89b.jpeg

If $f(x) = \dfrac {x+1}{x-1}$ then show that $f[\left\{ f(x) \right\} ]=x$

Given

$f\left( x \right) =\dfrac { x+1 }{ x-1 }$

Replace

$x$ by

$f\left( x \right)$ , we get,

$f\left[ f\left( x \right) \right] =\dfrac { f\left( x \right) +1 }{ f\left( x \right) -1 }$

$f\left[ f\left( x \right) \right] =\dfrac { \dfrac { x+1 }{ x-1 } +1 }{ \dfrac { x+1 }{ x-1 } -1 }$

$\therefore f\left[ f\left( x \right) \right] =\dfrac { \dfrac { x+1+x-1 }{ x-1 } }{ \dfrac { x+1-\left( x-1 \right) }{ x-1 } }$

$\therefore f\left[ f\left( x \right) \right] =\dfrac { \dfrac { 2x }{ x-1 } }{ \dfrac { x+1-x+1 }{ x-1 } }$

$\therefore f\left[ f\left( x \right) \right] =\dfrac { \dfrac { 2x }{ x-1 } }{ \dfrac { 2 }{ x-1 } }$

$\therefore f\left[ f\left( x \right) \right] =\dfrac { 2x }{ x-1 } \times \dfrac { x-1 }{ 2 }$

$\therefore f\left[ f\left( x \right) \right] =\dfrac { 2x }{ 2 }$

$\therefore f\left[ f\left( x \right) \right] =x$

Let $\displaystyle R$ be a relation on $\displaystyle Z$ , defined by $\displaystyle (x, y) \in R \iff x^2 + y^2 = 9$ . Then, write $\displaystyle R$ as set of ordered pairs. How many elements are there in its domain?

Let

$R=\left\{ \left( x,y \right) :x,y\in R\quad and\quad { x }^{ 2 }+{ y }^{ 2 }=9 \right\}$

1) Put x=0 in above equation,

$0+{ y }^{ 2 }=9$

$\therefore { y }^{ 2 }=9$

$\therefore { y }=\pm 3$

Thus,

$\left( 0,3 \right)$ and

$\left( 0,-3 \right)$ are ordered pairs

2) Put y=0 in above equation,

${ x }^{ 2 }+0=9$

$\therefore { x }^{ 2 }=9$

$\therefore { x }=\pm 3$

Thus,

$\left( 3,0 \right)$ and

$\left( -3,0 \right)$ are ordered pairs

Thus,

$R=\left\{ \left( 3,0 \right) ,\left( -3,0 \right) ,\left( 0,3 \right) ,\left( 0,-3 \right) \right\}$

Domain of R is,

$dom\left( R \right) =\left\{ -3,0,3 \right\}$

There are 3 elements in domain

If $f(x) = \frac {x-1}{x+ 1}$ then show that $f( \frac {1}{x}) = - f(x)$

Given

$f\left( x \right) =\dfrac { x-1 }{ x+1 }$

Replace

$x$ by

$\frac { 1 }{ x }$ , we get,

$f\left( \frac { 1 }{ x } \right) =\dfrac { \frac { 1 }{ x } -1 }{ \frac { 1 }{ x } +1 }$

$\therefore f\left( \frac { 1 }{ x } \right) =\dfrac { \frac { 1-x }{ x } }{ \frac { 1+x }{ x } }$

$\therefore f\left( \frac { 1 }{ x } \right) =\dfrac { 1-x }{ 1+x }$

$\therefore f\left( \frac { 1 }{ x } \right) =\dfrac { -\left( x-1 \right) }{ x+1 }$

$\therefore f\left( \frac { 1 }{ x } \right) =-\left( \dfrac { x-1 }{ x+1 } \right)$

$\therefore f\left( \frac { 1 }{ x } \right) =-f\left( x \right)$

If $f(x) = x^3 - \frac {1}{x^3}$ then show that $f(x) + f( \frac {1}{x}) = 0$

Given

$f\left( x \right) ={ x }^{ 3 }-\frac { 1 }{ { x }^{ 3 } }$ Equation (1)

Replace

$x$ by

$\frac { 1 }{ x }$ , we get,

$f\left( \frac { 1 }{ x } \right) ={ \left( \frac { 1 }{ x } \right) }^{ 3 }-\frac { 1 }{ { \left( \frac { 1 }{ x } \right) }^{ 3 } }$

$\therefore f\left( \frac { 1 }{ x } \right) =\frac { 1 }{ { x }^{ 3 } } -\frac { 1 }{ \frac { 1 }{ { x }^{ 3 } } }$

$\therefore f\left( \frac { 1 }{ x } \right) =\frac { 1 }{ { x }^{ 3 } } -\frac { { x }^{ 3 } }{ 1 }$

$\therefore f\left( \frac { 1 }{ x } \right) =\frac { 1 }{ { x }^{ 3 } } -{ x }^{ 3 }$ Equation (2)

From equations (1) and (2),

$f\left( x \right) +f\left( \frac { 1 }{ x } \right) ={ x }^{ 3 }-\frac { 1 }{ { x }^{ 3 } } +\frac { 1 }{ { x }^{ 3 } } -{ x }^{ 3 }$

$\therefore f\left( x \right) +f\left( \frac { 1 }{ x } \right) =\left( { x }^{ 3 }-{ x }^{ 3 } \right) +\left( -\frac { 1 }{ { x }^{ 3 } } +\frac { 1 }{ { x }^{ 3 } } \right)$

$\therefore f\left( x \right) +f\left( \frac { 1 }{ x } \right) =0$

Find the domain and range of $f(x) = \dfrac {1}{x}$ .

The given function is,

$f\left( x \right) =\frac { 1 }{ x }$

We know that we cannot divide any number by zero.

$\therefore x\neq 0$

Thus, domain of the function is,

$dom\left[ f\left( x \right) \right] =R-\left\{ 0 \right\}$

Range of function is set of reciprocals of all real numbers except zero.

$\therefore Range=\left( -\infty ,0 \right) \cup \left( 0,\infty \right)$

If $f (x) =x^2 - 3x + 4$ and $f(x) = f(2x+ 1)$ , find the values of $x$ .

Given

$f\left( x \right) ={ x }^{ 2 }-3x+4$

Replace

$x$ by

$2x+1$ , we get,

$f\left( 2x+1 \right) ={ \left( 2x+1 \right) }^{ 2 }-3{ \left( 2x+1 \right) }+4$

$\therefore f\left( 2x+1 \right) ={ 4x }^{ 2 }+4x+1-6x-3+4$

$\therefore f\left( 2x+1 \right) ={ 4x }^{ 2 }-2x+2$

As given in problem,

$f\left( x \right) =f\left( 2x+1 \right)$

$\therefore { x }^{ 2 }-3x+4={ 4x }^{ 2 }-2x+2$

$\\ \therefore { 4x }^{ 2 }-2x+2-{ x }^{ 2 }+3x-4=0$

$\\ \therefore { 3x }^{ 2 }+x-2=0$

$\\ \therefore x=\dfrac { -1\pm \sqrt { { \left( 1 \right) }^{ 2 }-4\times 3\times -2 } }{ 2\times 3 }$

$\\ \therefore x=\dfrac { -1\pm \sqrt { 1+24 } }{ 6 }$

$\\ \therefore x=\dfrac { -1\pm \sqrt { 25 } }{ 6 }$

$\\ \therefore x=\dfrac { -1\pm 5 }{ 6 }$

$\\ \therefore x=\dfrac { -1+5 }{ 6 }$ or

$x=\dfrac { -1-5 }{ 6 }$

$\\ \therefore x=\dfrac { 4 }{ 6 }$ or

$\\ x=-1$

$\\ \therefore x=\dfrac { 2 }{ 3 }$ or

$\\ x=-1$

If $f(x) = \frac {1}{(1-x)}$ then show that $f[f\left\{ f(x) \right\} ]=x$

Given

$f\left( x \right) =\dfrac { 1 }{ 1-x }$ Equation (1)

Replace

$x$ by

$f\left( x \right)$ , we get,

$f\left[ f\left( x \right) \right] =\dfrac { 1 }{ 1-f\left( x \right) }$

$\therefore f\left[ f\left( x \right) \right] =\dfrac { 1 }{ 1-\dfrac { 1 }{ 1-x } }$

$\therefore f\left[ f\left( x \right) \right] =\dfrac { 1 }{ \dfrac { 1-x-1 }{ 1-x } }$

$\therefore f\left[ f\left( x \right) \right] =\dfrac { 1-x }{ 1-x-1 }$

$\therefore f\left[ f\left( x \right) \right] =\dfrac { 1-x }{ -x }$

Now, in equation (1), replace

$x$ by

$f\left[ f\left( x \right) \right]$ , we get,

$f\left\{ f\left[ f\left( x \right) \right] \right\} =\dfrac { 1 }{ 1-\left( \frac { 1-x }{ -x } \right) }$

$\therefore f\left\{ f\left[ f\left( x \right) \right] \right\} =\dfrac { 1 }{ 1+\left( \frac { 1-x }{ x } \right) }$

$\therefore f\left\{ f\left[ f\left( x \right) \right] \right\} =\dfrac { 1 }{ \dfrac { x+1-x }{ x } }$

$\therefore f\left\{ f\left[ f\left( x \right) \right] \right\} =\dfrac { x }{ 1 }$

$\therefore f\left\{ f\left[ f\left( x \right) \right] \right\} =x$

If $f(x) = \dfrac {x-1}{x+ 1}$ then show that $f ( \frac {-1}{x} ) = \frac {-1}{f(x)}$

Given

$f\left( x \right) =\dfrac { x-1 }{ x+1 }$

Replace

$x$ by

$\dfrac { -1 }{ x }$ , we get,

$f\left( \frac { -1 }{ x } \right) =\dfrac { \dfrac { -1 }{ x } -1 }{ \dfrac { -1 }{ x } +1 }$

$f\left( \frac { -1 }{ x } \right) =\dfrac { \dfrac { -1-x }{ x } }{ \dfrac { -1+x }{ x } }$

$\therefore f\left( \frac { -1 }{ x } \right) =\dfrac { -1-x }{ -1+x }$

$\therefore f\left( \frac { -1 }{ x } \right) =\dfrac { -\left( x+1 \right) }{ x-1 }$

$\therefore f\left( \frac { -1 }{ x } \right) =\dfrac { -1 }{ \left( \dfrac { x-1 }{ x+1 } \right) }$

$\therefore f\left( \frac { -1 }{ x } \right) =\dfrac { -1 }{ f\left( x \right) }$

If $f(x) = \frac {1}{(2x+ 1)} and x \neq \frac {-1}{2}$ then prove that $f\left\{ f(x) \right\} = \frac {2x+ 1}{2x+ 3}$ , when it is given that $x \neq \frac {-3}{2}$

Given

$f\left( x \right) =\frac { 1 }{ 2x+1 }$

Replace

$x$ by

$f\left( x \right)$ , we get,

$f\left[ f\left( x \right) \right] =\frac { 1 }{ 2f\left( x \right) +1 }$

$\therefore f\left[ f\left( x \right) \right] =\frac { 1 }{ 2\left( \frac { 1 }{ 2x+1 } \right) +1 }$

$\therefore f\left[ f\left( x \right) \right] =\frac { 1 }{ \left( \frac { 2 }{ 2x+1 } \right) +1 }$

$\therefore f\left[ f\left( x \right) \right] =\frac { 1 }{ \left( \frac { 2+2x+1 }{ 2x+1 } \right) }$

$\therefore f\left[ f\left( x \right) \right] =\frac { 2x+1 }{ 2x+3 }$

$x\neq \frac { -3 }{ 2 }$ , as for

$x=\frac { -3 }{ 2 }$ ,

$f\left( x \right) =\frac { 1 }{ 0 }$ which is undefined

If $\displaystyle A = \{5 \}$ and $\displaystyle B = \{5, 6\}$ , write down all possible subsets of $\displaystyle A \times B$ .

1847824_1708742_ans_e475645a4c6d44629622b67da64fbea0.jpg

Prove that $\displaystyle A \times B = B \times A \implies A = B$ .

1847829_1708739_ans_55a71b0422894a64992d769a7e7437a7.jpg

Let $\displaystyle A = \{ a, b\}$ . List all relations on $\displaystyle A$ and find their number.

$\displaystyle A \times A = \{ (a, a), (a, b), (b, a), (b, b)\}$ and every subset of

$\displaystyle A \times A$ is a relation on

$\displaystyle A$ .So, their number =

$\displaystyle 2^4 = 16$

Find the domain and range of $f(x) = \dfrac {1}{2 -sin x}$

1860148_1708661_ans_a1d4afc5eefc4341bf19adfa856ef0bf.jpg

Let $\displaystyle R = \{( x, x^2) : x$ is a prime number less than $\displaystyle 10 \}$ .
Find dom $\displaystyle (R)$ and range $\displaystyle (R)$ .

1847257_1708748_ans_30c321743aaa4767a9f055ce6b36709d.jpeg

Find the domain and range of $f(x) = \dfrac {x-3}{2-x}$ .

We have,

$f(x)=\frac{x-3}{2-x}$

Domain of

$f$ : Clearly,

$f(x)$ is defined for all

$x$ satisfying

$2-x\ne0$ i.e.,

$x\ne2$ .

Hence, Doamin

$(f)=R-\{2 \}$ .

Range of

$f$ : Let

$y=f(x)$ i.e.,

$y=\frac{x-3}{2-x}$

$\Rightarrow 2y-xy=x-3$

$\Rightarrow x(y+1)=2y+3$

$\Rightarrow x=\frac{2y+3}{y+1}$

Clearly,

$x$ assumes real values for all

$y$ except

$y+1=0$ i.e.,

$y=-1$ .

Hence, Range

$(f)=R-\{-1 \}$

Let $\displaystyle A = \{1, 2, 3\}$ and $\displaystyle B = \{4\}$ .
How many relations can be defined from $\displaystyle A$ to $\displaystyle B$ ?

1845921_1708751_ans_0c87d5c2443a4017b2550afa59da588c.jpg

Find the domain and range of $f(x) = \dfrac {x^2 -16}{x -4}$ .

$f(x)=\dfrac{x^2-16}{x-4}$

Here,

$x\ne 4$

So, The domain of the function is

$R-\{4\}$

Let

$y=\dfrac{x^2-16}{x-4}$

$y=\dfrac{(x+4)(x-4)}{x-4}$

$y=x+4$

For,

$R-\{4\}$

$y\ne 4+4$

$y\ne 8$

So,

$f(x)\ne 8$

$\therefore$ Range

$=R-\{8\}$

Find the domain and range of $f(x) = \dfrac {x^2 -9}{x- 3}$

$f(x)=\dfrac{x^2-9}{x-3}$

Here,

$x\ne 3$

So, the domain of the function is

$R-\{3\}$

Let

$y=\dfrac{x^2-9}{x-3}$

$y=\dfrac{(x+3)(x-3)}{x-3}$

$y=x+3$

For ,

$R-\{3\}$

$y\ne 3+3$

$y\ne 6$

So,

$f(x)\ne 6$

$\therefore$ Range

$=R-\{6\}$

Let $\displaystyle A$ and $\displaystyle B$ be two nonempty sets.
What do you mean by the domain and range of a relation?

If A and B are two non-empty sets, then a relation from A to B is set of ordered pairs (p, q) where

$p \in A$ and

$q \in B$

For example:

Let A = { 0, 1, 3 }

B = { 2, 5, 7, 9 }

$f: A \rightarrow B$ can be like { (0, 5), (1, 9) , (3, 7) }

Hence, the domain is the set of A i.e first element of ordered pairs, whereas the set of second elements of ordered pair constitutes the range.

Here, entire B cannot be range as {2} is not used.

Domain = {0, 1, 3} range = {5, 7, 9}

Number of positive integers $x$ for which $f(x)=x^{3}-8 x^{2}+20 x-13$ is a prime number is.

1609826_1745398_ans_f2441667dc07401db918377808536ff9.png

Let $\displaystyle A = \{ 1, 3, 5, 7\}$ and $\displaystyle B = \{2, 4, 6, 8\}$ .
Let $\displaystyle R = \{(x, y) : x \in A, y \in B \ and \ x > y \}$ .
Find dom $\displaystyle (R)$ and range $\displaystyle (R)$ .

1910036_1709533_ans_56154b39ffe947efb9408fcb4a393681.jpeg

Find the domain and range of each of the relations given below:
$\displaystyle R = \{ (x, y) : x + 2y = 8$ and $\displaystyle x, y \in N\}$ .

1909793_1709468_ans_182fc211417f44b9b6123e4e80ca17bc.jpeg

Find the domain and range of each of the relations given below:
$\displaystyle R = \{ (x, y) : y = |x - 1|, x \in Z$ and $\displaystyle |x|\leq 3\}$ .

1910030_1709474_ans_97bd77857d464204930ce2ba605331f7.jpeg

Define a relation $\displaystyle R$ from $\displaystyle Z$ to $\displaystyle Z$ , given by

$\displaystyle R = \{ (a, b) : a, b \in Z$ and $\displaystyle (a - b)$ is an integer $\displaystyle \}$ .

Find dom $\displaystyle (R)$ and range $\displaystyle (R)$

2016174_1709605_ans_c5fb6d011856417ebb535632292e8551.png

Let $\displaystyle A = \{ 2, 4, 5\}$ and $\displaystyle B = \{1, 2, 3, 4, 6, 8\}$ .
Let $\displaystyle R = \{(x, y) : x \in A, y \in B \ and \ x$ divides $\displaystyle y \}$ .
How many common elements are there between dom $\displaystyle (R)$ and range $\displaystyle (R)$ .

$A=\{2,4,5\},\,B=\{1,2,3,4,6,8\}$

$x\in A$ means

$x\in\{2,4,5\}$

$y\in B$ means

$y\in\{1,2,3,4,6,8\}$

It is given that

$x$ divides

$y$

Then roster form is,

$\Rightarrow$

$R=\{(2,2),(2,4),(2,6),(2,8),(4,4),(4,8)\}$

dom

$(R)$ means elements of

$x$ in

$R$ and range of

$(R)$ means elements of

$y$ in

$R$ .

$\therefore$ Dom

$(R)=\{2,4\}$

$\therefore$ Range

$(R)=\{2,4,6,8\}$

Let $\displaystyle A$ and $\displaystyle B$ be two nonempty sets.
What do you mean by a relation from $\displaystyle A$ to $\displaystyle B$ ?

1909746_1709442_ans_5f771bb0668544d292b1647b8f5fd877.jpeg

Let $f : R \rightarrow R f(x^2 + x + 3) + 2f(x^2 - 3x + 5) = 6x^2 - 10x + 17 \space \forall \space x \epsilon \space R$ , then the value of f(5) is

1771077_1754020_ans_a80bd127b5534562bcfcfeca3076b7ab.jpeg

If $R=\left \{ \left ( x,y \right ):x+2y=8 \right \}$ is a relation on $N$ , write the range of $R$ .

1961187_1782093_ans_ec5fa75631e24384a7b69997488af2af.jpg

Let $A=\left\{a,b,c \right\}$ and the relation $R$ be define on $A$ as follows:
$R=\left\{(a,a),(b,c),(a,b)\right\}$ .
Then, write the minimum number of ordered pairs to be added in $R$ to make $R$ reflexive and transitive.

Given relation

$R=\left\{(a,a),(b,c),(a,b)\right\}$

To make

$R$ reflexive we must add

$(b,b)$ and

$(c,c)$ to

$R$ .

To make

$R$ transitive we must add

$(a,c)$ to

$R$

So, the minimum number of ordered pairs to be added is

$3$ .

If $f:R \rightarrow R$ is defined by $f(x)=x^2-3x+2$ , find $f(f(x))$

Given,

$f(x)=x^2-3x+2$

$\therefore f(f(x))=f(y)$ where

$y=x^2-3x+2$

$f(f(x))=(x^2-3x+2)^2 -3 (x^2-3x+2)+2$

$=x^4+9x^2+4-6x^3+4x^2-12x-3x^2+9x-6+2$

$=x^4-6x^3+10x^2-3x$

Let R be a relation from N to N defined by R = {(a, b) : a, b $\epsilon$ N and a = $b^2$ }. Are the following true?
(ii) (a, b) $\epsilon$ R, implies (b, a) $\epsilon$ R

R = {(a, b) : a, b

$\epsilon$ N and a =

$b^2$ }
(ii) It can be seen that (9, 3)

$\epsilon$ N because 9, 3

$\epsilon$ N and 9 =

$3^2$ . Now,

$3 \neq 9^2 = 81$ ; therefore,
(3, 9) N
Therefore, the statement "(a, b)

$\epsilon$ R, implies "(b, a)

$\epsilon$ R" is not true.

Let A = {1, 2, 3...14}. Define a relation R from A to A by R = {(x, y) : 3x - y = 0}, where x, y $\epsilon$ A. Write down its domain, codomain and range.

The relation R from A to A is given as R = {(x, y) : 3x - y = 0, where x, y

$\epsilon$ A}

i.e., R = {(x, y) : 3x = y, where x, y

$\epsilon$ A}

$\therefore$ R = {(1, 3), (2, 6), (3, 9), (4, 12)}

The domain of R is the set of all first elements of the ordered pairs in the relation.

$\therefore$ Domain of R = {1, 2, 3, 4}

The whole set A is he codomain of the relation R.

$\therefore$ codomain of R = A {1, 2, 3.....14}

The range of R is the set of all second elements of the ordered pairs in the relation.

$\therefore$ Range of R = {3, 6, 9, 12}

Show that the relation $R$ in the set
$A=\left\{ x \in Z :0 \le x \le 12 \right\}$ , given by
$R=\left\{ (a, b): | a-b| \ is\ a\ multiple\ of\ 4 \right\}$

$a\ R\ a \Rightarrow |a-a|=0$ is a multiple of

$4\forall a \in A$ hence

$R$ is reflexive

$a\ R\ b\Rightarrow a-b=$ multiple of

$4$

$\Rightarrow |-(b-a)| =$ multiple of

$4$

$\Rightarrow |b-a|$ is a multiple of

$4$

$\Rightarrow b \ R\ a$ , hence

$R$ is symmetric

$a\ R\ b$ and

$b\ R\ c$

$\Rightarrow |a-b|$ is a multiple of

$4$ and

$| b-c|$ is a multiple of

$4$

$\Rightarrow (a-b)$ and

$(b-c)$ are multiple of

$4$

$\Rightarrow a-b+b-c=a-c$ is a multiple of

$4$

$\Rightarrow | a-c |$ is a multiple of

$4\Rightarrow a\ R\ c$
hence

$R$ is transitive.
Since

$R$ is reflexive, symmetric and transitive

$R$ is an equivalence relation.

$a\ R\ 1 \Rightarrow |a-1|=m(4)$

$\Rightarrow a=1, 5, 9$ as

$| 1-1|=0, |5-1|=4, |9-1|=8$ are multiple of

$4$ .
Hence elements related to

$1$ are

$\left\{ 1, 5, 9\right\}$

Is $g=\left\{(1,1),(2,3),(3,5),(4,7) \right\}$ a function? If $g$ is described by $g(x)=\alpha x +\beta$ , then what value should be assigned to $\alpha$ and $\beta$ .

Yes,

$g$ is a function since every element in the domain has a unique image

Now, given

$g(x)=\alpha x +\beta$

Then,

$g(1)=\alpha+\beta=1$ and

$g(2)=2\alpha+\beta=3$

Subtracting

$g(1)$ from

$g(2)$ gives

$(2\alpha+\beta)-(\alpha+\beta)=2$

$\Rightarrow \alpha=2$

Substituting this value in

$g(1)$

We get

$\beta=-1$ .

Hence,

$\alpha=2 \ \ \&\ \ \beta=-1.$

The given figure shows a relationship between the sets P and Q. Write this relation
(i) in set-builder for.
(ii) in roster form.
What is its domain and range?

According to the given figure, P = {5, 6, 7}, Q = {3, 4, 5}

(i) R = {(x, y) : y = x -2; x

$\epsilon$ P} or R = {(x, y) : y = x - 2 for x = 5, 6, 7}

(ii) R = {(5, 3), (6, 4), (7, 5)}

Domain of R = {5, 6, 7}

Range of R = {3, 4, 5}

If $f: R\to R$ is defined by $f(x)=x^2 -3x+2$ , write $f(f(x))$ .

Given

$f(x)=x^2 -3x+2$

$\therefore f(f(x))=f(x^2 -3x+2)$

$=(x^2 -3x+2)^2 -3(x^2-3x+2)+2$

$=x^4+9x^2+4-6x^3 -12x+4x^2 -3x^2+9x-6+2$

$=x^4+10x^2-6x^3 -3x$

Hence.

$f(f(x))=x^4 -6x^3 +10x^2-3x$

Let R be a relation from N to N defined by R = {(a, b) : a, b $\epsilon$ N and a = $b^2$ }. Are the following true?
(i) (a,a) $\epsilon$ R, for all a $\epsilon$ N

R = {(a, b) : a, b

$\epsilon$ N and a =

$b^2$ }
(i) It can be seen that 2

$\epsilon$ N; however,

$2 \neq$

$2^2 = 4$ .
Therefore, the statement "(a, a)

$\epsilon$ R, for all a

$\epsilon$ N" is not true.

Let R be a relation from N to N defined by R = {(a, b) : a, b $\epsilon$ N and a = $b^2$ }. Are the following true?
(iii) (a, b) $\epsilon$ R, (b, c) $\epsilon$ R implies (a, c) $\epsilon$ R.
Justify your answer in each case.

R = {(a, b) : a, b

$\epsilon$ N and a =

$b^2$ }
(iii) It can be seen that (9, 3)

$\epsilon$ R, (16, 4)

$\epsilon$ R because 9,3,16, 4

$\epsilon$ N and 9 =

$3^2$ and 16 =

$4^2$ .
Now, 9

$\neq$

$4^2$ = 16; therefore, (9, 4)

$\epsilon$ N
Therefore, the statement "(a, b)

$\epsilon$ R, (b, c)

$\epsilon$ R implies (a, c)

$\epsilon$ R" is not true.

If $A=\left\{1,2,3 \right\}$ and $f, g$ are relation corrosponding to the subset of $A\times A$ indicated against them, which of $f, g$ is a function? Why?
$f=\left\{(1,3),(2,3),(3,2)\right\}$
$g=\left\{(1,2),(1,3),(3,1)\right\}$

$f$ is a function since each element of

$A$ in the first place in the ordered to only one element of

$A$ in the second place while

$g$ is not a function because

$1$ is related to more than one element of

$A$ , namely,

$2$ and

$3$ .

If $R$ is the set of all real numbers, what do the Cartesian products $R \times R$ and $R \times R \times R$ represent?

We have

$R\times R =\left\{ (x, y):x, y\in R\right\}$ which represents the coordinates of all the points in two dimensional space and

$R\times R\times R=\left\{ (x, y, z) x, y, z \in R \right\}$ which represents the coordinates of all the points in three-dimensional space.

Let $A=\left\{ 1, 2, 3, 4, 5, 6\right\}$ . Define a relation $R$ from $A$ to $A$ by $R =\left\{(x, Y): y=x+1\right\}$
Write down the domain, condomain and range of $R$

Domain

$=\left\{ 1, 2, 3, 4, 5 \right\}$ ; Co-domain

$=A$
Range

$=\left\{ 2, 3, 4, 5, 6\right\}$ .

If $(x+1, y-2)=(3, 1)$ , find the values of $x$

Given

$(x+1, y-2)=(3, 1)$

$\Rightarrow x+1=3 \therefore x=2$

$y-2=1 \therefore y=3$

Let $A=\left\{ 1, 2, 3,.......14\right\}$ . Define a relation $R$ from $A$ to $A$ by $R =\left\{ (x, y):3x-y=0, x, y \in A\right\}$ . Write down its domain, co-domain and range.

Given

$R=\left\{ (x, y):3x-y=0, x, y \in A \right\}$

$=\left\{ (1, 3), (2, 6), (3, 9), (4, 12) \right\}$
Domain

$=\left\{ 1, 2, 3, 4\right\}$
Co-Domain

$=A$ Range

$=\left\{ 3, 6, 9, 12\right\}$

Define range of a relation.

The set of all second elements in a relation

$R$ from a set

$A$ to a set

$B$ is called the range of the relation

$R$ . The whole set

$B$ is called the co-domain of the relation

$R$ .

If $P=\left\{ a, b, c \right\}$ and $Q=\left\{ r \right\}$ , from $P \times Q$ and $Q \times P$ . Are these two products equal?

$P \times Q =\left\{ (a, r ), (b, r), (c, r) \right\}$

$Q \times P=\left\{ (r, a), (r, b), (r, c) \right\}$
Clearly

$P \times Q \neq Q \times P$

Define ordered pair.

Two numbers

$a$ and

$b$ listed in specific order and enclosed in parentheses from is called an ordered pair

$(a, b)$ .
Keen Eye: Equality of two ordered pairs:
We have

$\left\{\right. a, b)-(c, d) \Leftrightarrow a-c$ and

$b-d$ .

Define domain of a relation.

The set of all first elements of the ordered pairs in a relation

$R$ from a set

$A$ to a set

$B$ is called the domain of the relation

$R$ .

If $\left( \dfrac x3+1, y-\dfrac 23 \right)=\left( \dfrac 53, \dfrac 13 \right)$ find x and y

Given

$\left( \dfrac x3+1, y-\dfrac 23 \right)=\left( \dfrac 53, \dfrac 13 \right)$

$\Rightarrow \dfrac x3+1=\dfrac 53 \therefore \dfrac x3=\dfrac 53-1=\dfrac 23 \therefore x=2$

$y-\dfrac 23 =\dfrac 13 \therefore y=\dfrac 23+\dfrac 13=1$

Let $A=\left\{ 1, 2, 3, 4, 6 \right\}$ . Let $R$ be the relation on $A$ defined by $\left\{(a, b): a, b \in A, b\right.$ is exactly divisible by $a\left. \right\}$ ,
Find the domain of $R$

Given

$A=\left\{ 1, 2, 3, 4, 6\right\}$

$R=\left\{ (a, b), a, b \in A, b\ is\ exactly\ divisible\ by\ a\right\}$
Domain of

$R=\left\{ 1, 2, 3, 4, 6\right\}=A$

Let $R$ be a relation from $Q$ to $Q$ defined by $R=\left\{(a, b)\ a, b\in Q\right.$ and $\left.a-b \in Z \right\}$ . Show that $(a, a)gR$ , for all $a \in Q$

Given

$R=\left\{ \left\{ a, b ) \right. : a , b \in Q\ and\ a-b \in Z )\right.$

$\forall a \in Q, a-a=0 \in Z\Rightarrow (a, a) \in R$

If A = {1 , 2 , 3} B = { 4 , 5 , 6} , then which of the following is relation from A to B ? Justify your answer also.
${( 1 , 6) , (2 , 6) ( 3 , 6) }$

Here, A = A = {1 , 2 , 3} B = { 4 , 5 , 6}
Then

$A \times B = {(1 , 4), (1 , 5), (1 , 6)( 2 , 4)( 2 , 5)( 2 , 6) ( 3 , 4)( 3 , 5),(3 , 6)}$

Let

$R_{2} = { (1 , 6 ) , (2 , 6 ) , ( 3 , 6)}$

$(1 , 6) \in A \times B$

$( 2 , 6) \in A \times B$

$( 3 , 6) \in A \times B$

$R_{2} \subseteq A \times B$

$R_{2} = {( 1 , 6)( 2 , 6)( 3 , 6)}$
Hence

$R_{2}$ is a relation from A to B.

Let $A=\left\{ \right.1, 2, 3, 4), B=\left\{1, 5, 9, 11, 15, 16\right\}$ and $f=\left\{ (1, 5), (2, 9), (3, 1), (4, 5), (2, 11)))\right.$ .
Is the following true?
$f$ is a relation from $A$ to $B$
Justify your answer.

Every element off is an element of

$A\times B$ , so

$f$ is a relation.

If A = {1 , 2 , 3} B = { 4 , 5 , 6} , then which of the following is relation from A to B ? Justify your answer also.
$( 1 , 4), (3 , 5), (3 , 6 )$

Here, A = A = {1 , 2 , 3} B = { 4 , 5 , 6}
Then

$A \times B = {(1 , 4), (1 , 5), (1 , 6)( 2 , 4)( 2 , 5)( 2 , 6) ( 3 , 4)( 3 , 5),(3 , 6)}$
Let ,

$R_{1} = {(1 , 4)(3 , 5)(3 , 6)}$

$(1 , 4)\in A \times B$

$(3 ,5) \in A \times B$

$(3 , 5)\in A \times B$

Let $N$ be the set of natural numbers and the relation $R$ be defined on $N$ such that $R=\left\{ (x, y):y=2x, x, y\in N \right\}$ . What is the domain, co-domain and range of $R$ ?
Is this relation a function?

Given

$R=\left\{ x, y):y=2x; x, y \in N \right\}$
Domain of

$R$ = set of natural numbers Co-domain of

$R=$ set of natural number
Range of

$R=$ set of even natural numbers
Clearly, every natural number is related to unique image, so this relation is a function.

If A = {1 , 2 , 3} B = { 4 , 5 , 6} , then which of the following is relation from A to B ? Justify your answer also.
${ (1 , 5), ( 3 , 4) ( 5 , 1), (3 ,6)}$

Here, A = A = {1 , 2 , 3} B = { 4 , 5 , 6}
Then

$A \times B = {(1 , 4), (1 , 5), (1 , 6)( 2 , 4)( 2 , 5)( 2 , 6) ( 3 , 4)( 3 , 5),(3 , 6)}$

$R_{3} = {(1 , 5), (3 , 4), ( 5 , 1), (3 , 6)}$

$( 1 , 5)\in A \times B$

$( 3 , 4) \in A \times B$

$( 5 , 1) \notin A \times B$

$R_{3}$ is not relation from A to B.

Let $A=\left\{ 1, 2, 3, 4, 6 \right\}$ . Let $R$ be the relation on $A$ defined by $\left\{(a, b): a, b \in A, b\right.$ is exactly divisible by $a\left. \right\}$ ,
Find the range of $R$

Given

$A=\left\{ 1, 2, 3, 4, 6\right\}$

$R=\left\{ (a, b), a, b \in A, b\ is\ exactly\ divisible\ by\ a\right\}$
Range of

$R=\left\{ 1, 2, 3, 4, 6\right\}=A$

Express the following relation in the rules from defined in N:
${(2 , 3) , ( 4 , 2) , (6 , 1) , }$

Relation in N is expressed as :

${(2 , 3) , ( 4 , 2) , (6 , 1) } = {( 6 , 1), (4 , 2), ( 2 , 3)}$
Here 6 , 4 , 2 are in A.P
Its general term

$T_{n} = 6 + ( n -1) \times \times (-2)$

$T_{n} = 6 - 2n + 2$

$T_{n} = 8 - 2n$

If a relation $\phi$ RR from set C of complex numbers to a set R of real numbers is so defined that $x \phi y \Leftrightarrow |x| = y$
$( 1 + i) \phi 3$

Given set

$\left \{ a + ib : a \in R , b \in R , i = \sqrt{-1} \right \}$ = set of complex number Relation

$\phi$ From C to R is defined as:

$x \phi y \Leftrightarrow |x| = y\forall x \in c , y \in R$

$( 1 + i)\phi 3$

$| 1 + i| = \sqrt{( 1^{2} + 1^{2})} = \sqrt{2} \neq 3$
Hence

$( 1 + i) \phi 1$ is

Express the following relation in the rules from defined in N:
${(1 , 3) , ( 2 , 5) , (3 , 7) , (4 , 9).........}$

$N = { 1 , 2 , 3, ...........}$
The relation from N to N is given by :

${(1 , 3),( 2 , 5) , ( 3 , 7),(4 , 9) ,...........}$
when ,

$x = 1$ , then

$y = 3$

$x = 2$ then

$y = 5$

$x = 3$ then

$y = 7$

$x = 4$ then

$y = 9$

$3 , 5 , 7 , 9$ is an

$A .P$
Hence its nth term

$= a + n ( n - 1) , d$ where a is first term and d is common difference.

In in a set of integers Z , a relation R is defined in such a way that $x Ry \Leftrightarrow x^{2} + y^{2} = 25$ , then write R and $R^{-1}$ in the from of a set order pairs and also write their domain .

Given set = 2
Z = {Set of integer} =

${ 0, \pm 1 , \pm 2 , \pm 3.........}$ Relation in i.e a relation R from Z to Z defined as

$xRy \Leftrightarrow x^{2} + y^{2} = 25 \forall x , y \in Z$
When x = 0 then

$y = \pm 5$ because from

$0^{2} + y ^{2} = 25$

If a relation $\phi$ RR from set C of complex numbers to a set R of real numbers is so defined that $x \phi y \Leftrightarrow |x| = y$
$3 \phi (-3)$

$|3| = 3 \neq - 3$
Relation

$\phi$ is false.

If a relation $\phi$ RR from set C of complex numbers to a set R of real numbers is so defined that $x \phi y \Leftrightarrow |x| = y$
$( 2 + 3i) \phi 13$

$| 2 + 3i| = \sqrt{( 2^{2} + 3^{2})} = \sqrt{(4 + 9)} = \sqrt{13} = \neq 13$
Relation is false.

If A = {1 , 2 , 3} B = { 4 , 5 , 6} , then which of the following is relation from A to B ? Justify your answer also.
${( 2 , 4), (2 , 6)( 3 , 6)( 4 , 2)}$

Here, A = A = {1 , 2 , 3} B = { 4 , 5 , 6}
Then

$A \times B = {(1 , 4), (1 , 5), (1 , 6)( 2 , 4)( 2 , 5)( 2 , 6) ( 3 , 4)( 3 , 5),(3 , 6)}$

Let

$R_{4} = {(2 , 4) , (2 , 6 ) , ( 3 , 6 ) , ( 4 , 2 )}$

$( 2 , 4) \in A \times B$ But

$( 4 , 2 )\notin A \times B$
Hence

$R_{4}$ is not a relation from A to B.

A relation R from set A = { 2 , 3 , 4 , 5) to set B = { 3 , 6 , 7 , 10 } is defined in such a way that $xRY \Leftrightarrow x$ is a prime number related to y . Write relation R in the set from order pairs and also fined domain and range of if.

Given A = { 2 , 3 , 4 , 5) to set B = { 3 , 6 , 7 , 10 }
Relation R from A to B is defined as:

$xRY \Leftrightarrow x$ is a prime number related by

$\forall x , y \in R$
(NUmbers a and b are called co-prime if there is no common factor axcept 1 for example 3 is co-prime related to 10 )

$x = 2 \in A$ then 2 is prime number related to 3 and 7.
Then

$( 2 , 3) \in A and (2 , 7 ) \in R$
When

$x = 3 \in A$
Then 3 is a prime number related to 7 and 10

If A = {1 , 2 , 3} B = { 4 , 5 , 6} , then which of the following is relation from A to B ? Justify your answer also.
$A \times B$

$A \times B \subseteq A \times B$
Hence , it is a relation

Express the following relation in the rules from defined in N:
${(2 , 1) , ( 3 , 2) , (4 , 3) ,( 5 , 4)....... }$

Relation in N is expressed as :

${(2 , 1) , ( 3 , 2) , (4 , 3) ,( 5 , 4)....... }$
Here , 2 , 3 , 4 , 5.............are in A .P.
So , nth term is

$T_{n} = 2 + ( n -1) \times 1 = 2 + n - 1 = n + 1$
Here , putting n = x and

$T_{n} = y$

Express the following relations in the from of sets or orders pairs :
$R_{2}$ is relation from set A = {8 ,9 , 10 , 11 } to set B = (5 , 6 , 7 8} such that " y = x - 2

Given relation

$R_{2}$ is defined as " y = x - 2 " and A = { 8 ,9, 10 , 11 } and B = { 5 , 6, 7 , 8}
So , from y = x - 2
when x = 8 then y = 8 - 2 = 6 So ,

$( 8 , 6) \in R_{2}$
when x = 10 then y = 10 - 2 = 8 So, ( 10 , 8)

$\in R_{2}$

$R_{2}= {(8,6) , (9, 7) , ( 10 , 8)}$

If $f : R \rightarrow R , f (x) = x^{2} ,$ then find
$\{ x | f (x) = 4 \}$

f (x) = 4

$\Rightarrow x^{2} = 4$

$\Rightarrow x = \pm 2$

Hence

$, { x : y (x) = 4} = \{ - 2 , 2\}$

If f : R \rightarrow R , f (x) = x^{2} ,$$ then find
{ y | f (y) = - 1 }

f (y) = -1

$\Rightarrow y^{2} = - 1$

$\Rightarrow y = \pm \sqrt{1}$

$\Rightarrow ( y : f (y) ) = \phi$ null set.

Any relation P is defined in set R0 of non zero real numbers by following ways :
$xPy \Leftrightarrow x^{2} + y^{2} = 1$

Given set
R0 = Set of n on -sero real numbers
Relation P is R0 defined as

$xPy = x^{2} + y^{2} = 1 \forall x, y \in R0$
Reflexivity : P is not reflexivie when

$2 \in R0$
Then

$(2)^{2} + (2)^{2} \neq 1 so , ( 2 , 2) \notin R_{0}$

If a relation $\phi$ RR from set C of complex numbers to a set R of real numbers is so defined that $x \phi y \Leftrightarrow |x| = y$
$( 1 + i) \phi 1$

$| 1 + i| = \sqrt{(1^{2} + 1^{2})} = \sqrt{2} \neq 1$
Relation is false.

Express the following relations in the from of sets or orders pairs :
$R_{1}$ is relation from set A = { 1 , 2 , 3 , 4 , 5 , 6 } to set B = (1 , 2 , 3} such that "x = 2y "

Given relation R1 is defined as " x = 2y " and A = {1 , 2 , 3 , 4 , 5 , 6} and in B = {1 . 2 , 3}
x = 2y

Express the following relations in the from of sets or orders pairs :
$R_{3}$ is relation from set A = {0 , 1 , 2......10} defined by 2x + 3y = 12.

Given relation

$R_{3}$ is defined as
2x + 3y = 12"
where A = { 0 , 1 , 2 , 3 ......10}
So , 2x + 3y = 12
3y = 12 - 2x

$y = \dfrac{12 - 2x}{3}$

Any relation P is defined in set R0 of non zero real numbers by following ways :
$xPy \Leftrightarrow + xy = 1$

Given set

$R_{0}$ = Set of real number
Relation P in R0 is defined as

$xPy = x^{2} + y^{2} = 1 \forall x, y \in R0$
Reflexivity : P is not reflexivie when

$2 \in R0$
Then

$(2)^{2} + (2)^{2} \neq 1 so , ( 2 , 2) \notin R_{0}$

Any relation P is defined in set $R_o$ of non zero real numbers by following ways :
$xPy \Leftrightarrow ( x + y)$ is a rational number.

Given : Set :

$R_{0}$ = Set of real number
Relation P in

$R_{0}$ is defined as a rational number

$xPy = \dfrac{x}{y}$ is rational number

$\forall x , y \in R_{0}$
Relexivity : Let

$a \in R_{0}$

$a \in R_{0} \Rightarrow \dfrac{a}{a} = 1$ is a rational number

$\Rightarrow (a ,a) \in P$
Symmetricity : Let a , b

$\in R_{0}$ is in this way

$( a , b ) \in P$

$\Rightarrow \dfrac{a}{b}$ is a rational number

$\Rightarrow \dfrac{b}{a}$ is a rational number

Express the following relations in the from of sets or orders pairs :
$R_{3}$ is relation from set A = {5 , 6 , 7 ,8} to set B = { 10 , 12 , 15 , 16 , 18} is defined such that "x is divisor of y''

Given relation R4 is defined as " x , y "
A = { 5 , 6 , 7 , 8}
and B = { 10 , 12 , 15 , 16 , 18}
So ,

$xR_{4}y \Leftrightarrow x$ is divisor of y

$x \in A , y \in B$
when x = 5 then 5 is a divisor of 10 and 15
So ,

$( 5 , 10 ) , ( 5 , 15 ) \in R_{4}$
when x = 6 then 6 is a divisor of 12 and 18
So ,

$(6 , 12 ) ( 6 , 18 ) \in R_{4}$

If A = { 1 ,2} , then write all non - zero relations defined in A.

All non - zero relation are:

${( 1 ,1)}, {( 2 ,2)}, {( 1 ,2)} , {( 2,1)}$

${(1 , 1)} , {(1 ,2)} , {( 1 ,1) , ( 2 ,1)( 2 ,2)( 1 ,2)( 2 ,1)}$

Find the domain and range of the following relations
$R_{1} = \left \{ ( x ,y) : x , y \in N , x + y = 10\right \}$

1967329_1873499_ans_389ae0a40eb247f1ad2ef5c4f129531e.jpg

In the set of real numbers R , two relations $R_{1} and R_{2}$ can be defines as
$aR_{1}b \Leftrightarrow a - b >0$

Given ser R = Set of real numbers
Reflection

$aR_{1}b \Leftrightarrow a - b >0 \forall a , b \in R$
Reflexivity : Let

$a \in R$

$a - a = 0 > 0$

In the set of real numbers R , two relations $R_{1} and R_{2}$ can be defines as
$aR_{2}b \Leftrightarrow |a| \leq b$

Given ser R = Set of real numbers

Reflection

$aR_{2}b \Leftrightarrow a - b >0 \forall a , b \in R$

Reflexivity : Let

$a \in R$

$2 - 2 = 0 \geq0$

Hence, from above it is clear that

$R_{2}$ is not a reflexive and symmetric relation.

It is only a transitive relation.

In the set of real numbers R two , relations $R_{1} and R_{2}$ are defined as
$aR_{1}b \Leftrightarrow |a| = |b|$

Given set R = set of real numbers
A relation

$R_{1}$ in R defined as

$aR_{1}b = |a| = |b| \forall a , b \in R$
For proving is equivalence relation we have to prove

$R_{1}$ is reflexive,

Find the domain and range of the relaton R
$R = { ( x + 1, x + 5)} : x \in { 0 , 1 , 2 , 3 , 4 , 5}$

Given relation

$R = { ( x + 1, x + 5)} : x \in { 0 , 1 , 2 , 3 , 4 , 5}$

Then , domain of relation

$R = { ( x + 1, x + 5)} : x \in { 0 , 1 , 2 , 3 , 4 , 5}$

Domain of R = { 1 , 2 , 3 , 4 , 5, 6}

and Range of

$R = { ( x + 5) ; x \in ( 0 , 1 , 2 , 3 , 4 , 5)}$

Range of

$R = ( 5,6,7,8,9,10)$

Solve :
$f(x) = 5 - 3x$
Find $f(16)$ .

Given

$f\left( x \right)=5-3x$

Put

$x=16$ , we get,

$f\left( 16 \right) =5-3\left( 16 \right)$

$\therefore f\left( 16 \right) =5-48$

$\therefore f\left( 16 \right) =-43$

In the set of real numbers R two , relations $R_{1} and R_{2}$ are defined as
$aR_{1}b \Leftrightarrow |a| \leq |b|$

Given set R = set of real numbers
A relation

$R_{1}$ in R defined as

$aR_{1}b = |a| = |b| \forall a , b \in R$
For proving is equivalence relation we have to prove

$R_{1}$ is

If statement is $P(n): (n+ 3) < 2^{n+3}$ then write $P(4)$ .

$P(n) : (n+3) < 2^{n+3}$
Putting

$n=4$

$P(4): (4+3) < 2^{4+3}$

$P(4): 7 < 2^7$

$P(4): 7< 128$ is true
Hence,

$P(4): 7< 2^7$

Given $A = \{5, 6, 7\}$ and $B = \{6, 8, 10\}$ , then : $(A \cup B)\times B$
Find the number of elements in the above set obtained.

$A=\left\{ 5,6,7 \right\}$ and

$B=\left\{ 6,8,10 \right\}$

$\therefore A\cup B=\left\{ 5,6,7,8,10 \right\}$

$\left( A\cup B \right) \times B=\left\{ 5,6,7,8,10 \right\} \times \left\{ 6,8,10 \right\}$

$=\left\{ \left( 5,6 \right) ,\left( 5,8 \right) ,\left( 5,10 \right) ,\left( 6,6 \right) ,\left( 6,8 \right) ,\left( 6,10 \right) ,\left( 7,6 \right) ,\left( 7,8 \right) ,\left( 7,10 \right) ,\left( 8,6 \right) ,\left( 8,8 \right) ,\left( 8,10 \right) ,\left( 10,6 \right) ,\left( 10,8 \right) ,\left( 10,10 \right) \right\}$

Thus there are total

$15$ elements obtained in the above set.

$(A \cup B) \times (A \cap B)$ .

$A=\left\{ 5,6,7,8 \right\}$ and

$B=\left\{ 6,8,10 \right\}$

$A\cup B=\left\{ 5,6,7,8,10 \right\}$

$A\cap B=\left\{ 6,8 \right\}$

$\left( A\cup B \right) \times \left( A\cap B \right) =\left\{ 5,6,7,8,10 \right\} \times \left\{ 6,8 \right\}$

$\left\{ \left( 5,6 \right) ,\left( 5,8 \right) ,\left( 6,6 \right) ,\left( 6,8 \right) ,\left( 7,6 \right) ,\left( 7,8 \right) ,\left( 8,6 \right) ,\left( 8,8 \right) ,\left( 10,6 \right) ,\left( 10,8 \right) \right\}$

total elements in

$\left( A\cup B \right) \times \left( A\cap B \right)$ are

$10$ .

If the ordered pairs $(a - 3, a + 2b)$ and $(3a - 1, 3)$ are equal, find the value of $a+b$ .

Ordered pairs

$\left( a-3,a+2b \right)$ and

$\left( 3a-1,3 \right)$ are equal.

It means

$a-3=3a-1$

and

$a+2b=3\longrightarrow 1$

$\therefore a-3a=-1+3$

$-2a=2\Longrightarrow \boxed { a=-1 }$

Put value of a in equation 1, we get

$-1+2b=3$

$2b=4$

$\boxed { b=2 }$

$a+b=-1+2=1$

Given $A = \{2, 3, 5\}$ and $B = \{6, 10, 14, 18\}$ ; find the number of elements in relation $R$ such that $R=\{(x, y) \,\,\epsilon\,\, A \times B,x < y \text{ and }\displaystyle \frac{y}{x} \,\,\epsilon \,\,N \}$

Given,

$A=\{2,3,5\}$ and

$B=\{6,10,14,18\}$

Product of two sets is the set of ordered pairs formed by mapping every element from the first set to every element of the second set.
So,

$A \times B = \{ (2,6), (2,10), (2,14), (2,18), (3,6), (3,10), (3,14), (3,18), (5,6), (5,10), (5,14), (5,18) \}$ .
Now, a Relation on this set of ordered pairs of

$A \times B$ , where

$x < y$ and

$\dfrac {y}{x}$ is a natural number will be

$R = \{(2,6), (2,10), (2,14), (2,18), (3,6), (3,18), (5,10) \}$

Thus, the number of elements in

$R$ is

$7$ .

If $M=\{x : x \,\,\epsilon\,\, N, 1 < x \leq 4\}$ and $N = \{y : y\,\,\epsilon\,\,W, y < 3\}$ ; find : N $\times$ M
Also find the number of such ordered pairs.

$M=\left\{ x:x\in N,1<x\le 4 \right\}$

$\therefore M=\left\{ 2,3,4 \right\}$

$N=\left\{ y:y\in W,y<3 \right\}$

$\therefore N=\left\{ 0,1,2 \right\}$ [ 0 is included

$\because y\in$ Whole Number

$N\times M=$ Every element of set

$N$ will be paired with each element of set

$M$

$N\times M=\left\{ \left( 0,2 \right) ,\left( 0,3 \right) ,\left( 0,4 \right) ,\left( 1,2 \right) ,\left( 1,3 \right) ,\left( 1,4 \right) ,\left( 2,2 \right) ,\left( 2,3 \right) ,\left( 2,4 \right) \right\}$

The number of such pairs is

$9$ .

Let $f:R-\left\{ 1 \right\} \rightarrow R$ be defined $f(x)=\cfrac { x+1 }{ x-1 }$ , show that $f(x)+f\left( \cfrac { 1 }{ x } \right) =0$ for $\left( x\neq 0 \right)$ .

Given:

$f\left(x\right)=\dfrac{x+1}{x-1}$ where,

$f:R-\left\{1\right\}\rightarrow R$

To Show:

$f\left(x\right)+f\left(\dfrac1x\right)=0$ for

$\left(x\neq0\right)$

Solution:

$f\left(x\right)=\dfrac{x+1}{x-1}$

$f\left(\dfrac1x\right)=\dfrac{\dfrac1x +1}{\dfrac1x-1}=\dfrac{1+x}{1-x}=-\dfrac{x+1}{x-1}$

Now,

$f\left(x\right)+f\left(\dfrac1x\right)=\dfrac{x+1}{x-1}-\dfrac{x+1}{x-1}=0$

If $f(x) = 2e^{x} - c\ ln\ x$ monotonically increases for every $x\epsilon (0, \infty)$ , then the true set of values of $c$ is ?

$f(x)=2e^x -c/n \ x$ for every

$x\in (0, \infty)$

$\because f(x)$ will strictly increases if

$f(x)\Rightarrow 0$

then,

$f(x)=2e^x -c.\dfrac {1}{x}$

Now, for

$x\in (0, \infty), f(x) > 0$

i.e.

$2e^x -\dfrac {c}{x} > 0$

$\Rightarrow \ 2e^x > \dfrac {c}{x}$

Now, multipluing by a positive no Let say

$x$

$2e^x . x > \dfrac {c}{x} x$

Now, check for

$C$ is negative, then

$2e^x . x > 0$ will be satisfy

Now, check for

$c=0$ , then

$2e^x , x > 0$ will also satisfy

if we check for

$c$ is positive then

$2e^x , x < c$ , it will not satisfy for increasing function

So,

$C$ is lie in

$[-\infty , 0]$

If $f\left( x \right) = \cos \left( {\log x} \right)$ , than show that $f\left( {\frac{1}{x}} \right).f\left( {\frac{1}{y}} \right) - \frac{1}{2}\left[ {f\left( {\frac{x}{y}} \right) + \left( {xy} \right)} \right]$

$f\left(\dfrac{1}{x}\right)f\left(\dfrac{1}{y}\right)-\dfrac{1}{2}\left[f\left(\dfrac{x}{y}\right)+f\left(xy\right)\right]$

$\cos{\left(\log{\dfrac{1}{x}}\right)}\cos{\left(\log{\dfrac{1}{y}}\right)}-\dfrac{1}{2}\left[\cos{\left(\log{\dfrac{x}{y}}\right)}+\cos{\left(\log{xy}\right)}\right]$

$=\cos{\left(-\log{x}\right)}\cos{\left(-\log{y}\right)}-\dfrac{1}{2}\left[2\cos{\left(\log{x}\right)}\cos{\left(\log{y}\right)}\right]$

$=\cos{\left(\log{x}\right)}\cos{\left(\log{y}\right)}-\cos{\left(\log{x}\right)}\cos{\left(\log{y}\right)}=0$

Hence proved.

Solve : $f(x) = \dfrac{x^2 +2}{x^2 + 4}$

2040759_1286548_ans_35f6a00ef753462aa0df68f656f11b8d.JPG

If $f\left(x\right)=x+2,\,g\left(x\right)={x}^{2}-x-2,$ then find $\dfrac{g\left(1\right)+g\left(2\right)+g\left(3\right)}{f\left(-4\right)+f\left(-2\right)+f\left(2\right)}$

We have

$f\left(x\right)=x+2$

$\Rightarrow \,f\left(-4\right)=-4+2=-2$

$\Rightarrow \,f\left(-2\right)=-2+2=0$

$\Rightarrow \,f\left(2\right)=2+2=4$

$g\left(x\right)={x}^{2}-x-2$

$g\left(1\right)={1}^{2}-1-2=1-1-2=-2$

$g\left(2\right)={2}^{2}-2-2=4-2-2=0$

$g\left(3\right)={3}^{2}-3-2=9-3-2=4$

Now substituting the above values in

$\dfrac{g\left(1\right)+g\left(2\right)+g\left(3\right)}{f\left(-4\right)+f\left(-2\right)+f\left(2\right)}$ we get

$\dfrac{g\left(1\right)+g\left(2\right)+g\left(3\right)}{f\left(-4\right)+f\left(-2\right)+f\left(2\right)}=\dfrac{-2+0-2}{-2+0+4}=\dfrac{-4}{2}=-2$

Find the range of $f\left(x\right)={\sin}^{-1}{\left(\ln{\left[x\right]}\right)}+\ln{\left({\sin}^{-1}{\left[x\right]}\right)},$ where $\left[x\right]$ is the greatest function.

${\sin}^{-1}{\left(\ln{\left[x\right]}\right)}$ is defined if only if

$-1\le \ln{\left[x\right]}\le 1$

$\Rightarrow {e}^{-1}\le\left[x\right]\le {e}^{1}$

$\Rightarrow \left[x\right]=1,2$

$\ln{\left({\sin}^{-1}{\left[x\right]}\right)}$ is defined if only if

${\sin}^{-1}{\left[x\right]}>0$ and

$-1\le \left[x\right]\le 1\Rightarrow \left[x\right]=1$

$f\left(x\right)$ is defined if

$\left[x\right]=1$ only,

Range of

$f\left(x\right)={\sin}^{-1}{\left(\ln{\left[1\right]}\right)}+\ln{\left({\sin}^{-1}{\left[1\right]}\right)}$

$={\sin}^{-1}{\left(\ln{1}\right)}+\ln{\left({\sin}^{-1}{1}\right)}$

$={\sin}^{-1}{\left(0\right)}+\ln{\left(\dfrac{\pi}{2}\right)}$

$=\ln{\left(\dfrac{\pi}{2}\right)}$

Hence range is

$\left\{\ln{\left(\dfrac{\pi}{2}\right)}\right\}$

If f(x) = $\frac{x + 1}{x - 1}$ ; then f( $\frac{1}{2}$ )

2044372_1563298_ans_b945d8fd0cc847a4a4d6f2b7fc7a315d.jpg

If f(x) = $\frac{x + 1}{x - 1}$ ; then f( $\frac{1}{2}$ )=

2044414_1564457_ans_fec75110882440f3a20dfbf440f28008.jpg

Let $f : R-\{2\} \rightarrow R$ be defined by $f(x)=\dfrac{x^{2}-4}{x-2}$ and $g: R \rightarrow R$ be defined by $g(x)=x+2$ . Find whether $f \equiv g$ or not.

2042862_1535941_ans_2b915ab658324ea881e68df8ff837df7.JPG

Let f(x)= $\frac{ax+b}{cx+d}$ .Then fof(x)=x provided that.

2044089_1562112_ans_0b2fe83e0a2d4bdc9bf06b0bc4e5335e.jpg

What do you mean by a binary relation on a set $\displaystyle A$ ?
Define the domain and range of a relation on $\displaystyle A$ .

1853338_1708488_ans_96258a1af492459fb982210b346d586a.jpg

For any sets $\displaystyle A$ and $\displaystyle B$ , prove that:
$\displaystyle (A \times B) \cap ( B \times A) = (A \cap B) \times (B \cap A)$ .

Let

$A=\left\{ 1,2,3 \right\}$ and

$B=\left\{ 2,4,6 \right\}$

$A\times B=\left\{ 1,2,3 \right\} \times \left\{ 2,4,6 \right\}$

$\therefore A\times B=\left\{ \left( 1,2 \right) ,\left( 1,4 \right) ,\left( 1,6 \right) ,\left( 2,2 \right) ,\left( 2,4 \right) ,\left( 2,6 \right) ,\left( 3,2 \right) ,\left( 3,4 \right) ,\left( 3,6 \right) \right\}$

$B\times A=\left\{ 2,4,6 \right\}\times \left\{ 1,2,3 \right\}$

$\therefore B\times A=\left\{ \left( 2,1 \right) ,\left( 2,2 \right) ,\left( 2,3 \right) ,\left( 4,1 \right) ,\left( 4,2 \right) ,\left( 4,3 \right) ,\left( 6,1 \right) ,\left( 6,2 \right) ,\left( 6,3 \right) \right\}$

$\left( A\times B \right) \cap \left( B\times A \right) =\\\left\{ \left( 1,2 \right) ,\left( 1,4 \right) ,\left( 1,6 \right) ,\left( 2,2 \right) ,\left( 2,4 \right) ,\left( 2,6 \right) ,\left( 3,2 \right) ,\left( 3,4 \right) ,\left( 3,6 \right) \right\}\cap \left\{ \left( 2,1 \right) ,\left( 2,2 \right) ,\left( 2,3 \right) ,\left( 4,1 \right) ,\left( 4,2 \right) ,\left( 4,3 \right) ,\left( 6,1 \right) ,\left( 6,2 \right) ,\left( 6,3 \right) \right\}$

$\therefore \left( A\times B \right) \cap \left( B\times A \right) =\left\{ \left( 2,2 \right) \right\}$ Equation (1)

$A\cap B=\left\{ 1,2,3 \right\}\cap \left\{ 2,4,6 \right\}$

$\therefore A\cap B=\left\{ 2 \right\}$

$B\cap A=\left\{ 2,4,6 \right\}\cap \left\{ 1,2,3 \right\}$

$\therefore B\cap A=\left\{ 2 \right\}$

$\therefore \left( A\cap B \right) \times \left( B\cap A \right) =\left\{ 2 \right\}\times \left\{ 2 \right\}$

$\therefore \left( A\cap B \right) \times \left( B\cap A \right) =\left\{ \left( 2,2 \right) \right\}$ Equation (2)

From equation (1) and (2),

$\left( A\times B \right) \cap \left( B\times A \right)=\left( A\cap B \right) \times \left( B\cap A \right)$

Let $\displaystyle A = \{ 2, 3, 4, 5\}$ and $\displaystyle B = \{3, 6, 7, 10\}$ .
Let $\displaystyle R = \{(x, y) : x \in A, y \in B$ and $\displaystyle x$ divides $\displaystyle y \}$ .
Which element is common between the domain and range?

Let

$A=\left\{ 2,3,4,5 \right\}$ and

$B=\left\{ 3,6,7,10 \right\}$

We have given that,

$R=\left\{ \left( x,y \right) :x\in A,y\in B\quad and\quad x\quad divides\quad y \right\}$

Thus, all the corresponding ordered pairs are,

$\left( 2,6 \right) ,\left( 2,10 \right) ,\left( 3,3 \right) ,\left( 3,6 \right) ,\left( 5,10 \right)$

Thus,

$R=\left\{ \left( 2,6 \right) ,\left( 2,10 \right) ,\left( 3,3 \right) ,\left( 3,6 \right) ,\left( 5,10 \right) \right\}$

$\therefore dom\left( R \right) =\left\{ 2,3,5 \right\}$

$range\left( R \right) =\left\{ 3,6,10 \right\}$

If $A = \left( \begin{array} { l l } { 1 } & { 2 } \\ { 3 } & { 4 } \end{array} \right) \& f ( x ) = x ^ { 2 } - 3 x + 2 .$ find f ( A )

2044421_1564552_ans_65ef1d604e55481ab71514563b6fe00c.jpg

Let $\displaystyle A = \{ 1, 2, 3, 4, 6\}$ and let $\displaystyle R = \{(a, b) : a, b \in A$ and $\displaystyle a$ divides $\displaystyle b \}$ .
Enter 1 if both dom $\displaystyle (R)$ and range $\displaystyle (R)$ are same else enter 0

1855642_1709601_ans_ccda681323c64610ba22bc594b8109ad.jpeg

Let $\displaystyle A = \{2, 3 \}$ and $\displaystyle B = \{ 3, 5 \}$ .
How many relations can be defined from $\displaystyle A$ to $\displaystyle B$ ?

1854998_1709614_ans_8a28640751364bf597987ecf9c0bd097.jpeg

Let $\displaystyle A = \{ 1, 2, 3, 5, 6\}$
Define a relation $\displaystyle R$ from $\displaystyle A$ to $\displaystyle A$ by $\displaystyle R = \{ (x, y) : y = x + 1 \}$ .
How many common elements are there in between dom $\displaystyle (R)$ and range $\displaystyle (R)$ .

1848732_1709589_ans_370b85a4cf914996ba3c737b2f82298e.jpeg

Let $\displaystyle R = \{ (x, y): x, y \in Z$ and $\displaystyle x^2 + y^2 \leq 4 \}$ .
Enter 1 if dom $\displaystyle (R)$ and range $\displaystyle (R)$ are same else enter 0

Domain and range are same.
1854987_1709610_ans_b6213398b22f45bc8f7a0272cb1b48fc.jpeg

Let $fx=\begin{cases} x^{ 2 }-4x+3,\quad x<3 \\ x-4,\quad x\geq 3 \end{cases}\quad and\quad g(x)=\begin{cases} x-3,\quad x<4 \\ x^{ 2 }+2x+2,\quad x\geq 4 \end{cases}$
Describe the function $f / g$ and find its domain.

$\quad f(x)=\begin{cases}x^{2}-4 x+3, x<3 \\ x-4, x \geq 3\end{cases}$

$\Rightarrow f(x)=\begin{cases}x^{2}-4 x+3, x<3 \\ x-4, 3 \leq x<4 \\ x-4, x \geq 4\end{cases} \quad\quad\quad(i)$

$g(x)=\begin{cases}x-3, x<4 \\ x^{2}+2 x+2, x \geq 4\end{cases}$

$\Rightarrow g(x)=\begin{cases}x-3, x<3 \\ x-3, 3 \leq x<4 \\ x^{2}+2 x+2, x \geq 4\end{cases} \quad\quad\quad(ii)$

From (1) and (2) we have

$\dfrac{f(x)}{g(x)}=\begin{cases}\dfrac{x^{2}-4 x+3}{x-3}, x<3 \\ \dfrac{x-4}{x-3}, 3<x<4 \\ \dfrac{x-4}{x^{2}+2 x+2}, x \geq 4\end{cases}$

Clearly,

$f(x) / g(x)$ is not defined at

$x=3$ , hence the domain is

$R-\{3\}$

Let f(x) be a function satisfying the condition f(-x)=f(x) for all real x. if f(0)exists ,find its value

1773094_1752523_ans_d8b64a6293694f9daccb823b30d517da.jpg

Let $\displaystyle R = \{(x, x + 5) : x \in \{0,1, 2, 3, 4, 5\} \}$
Find dom $\displaystyle (R)$ and range $\displaystyle (R)$ . And enter the common element between them.

1848736_1709593_ans_63faafde5db04b5a9e22a8e7621aeb42.jpeg

If $g(x) = \left\{\begin{matrix} [f(x)], & x \in (0, \pi/2) \cup (\pi / 2, \pi) \\ 3, & x=\pi/2 \end{matrix}\right.$

and $f(x) = \dfrac{2(\sin x - \sin^n x) + |\sin x - \sin^n x|}{2(\sin x - \sin^n x) - |\sin x - \sin^n x|}, n \in N$

Where $[\cdot]$ denotes the greatest integer function. Prove that $g(x)$ is continuous at $x=\pi /2$ when $n>1$ .

1973529_1751578_ans_a718584997a34b9aa183cfdc33ceaccb.jpeg

Consider the set $A=(1,2,3)$ and $R$ be the smallest equivalence relation on $A$ , then $R=$ ______

Given

$A=\left\{ 1,2,3 \right\}$

For relation R to be equivalence relation on A, It must be reflexive, symmetric and transitive.

1) Relation R is reflexive if,

$\left( a,a \right) \in R$

$\forall a\in A$

$\therefore R=\left\{ \left( 1,1 \right) ,\left( 2,2 \right) ,\left( 3,3 \right) \right\}$

2) Relation R is symmetric if,

$\left( a,b \right) \in R\rightarrow \left( b,a \right) \in R$

$\therefore R=\left\{ \left( 1,2 \right) ,\left( 2,1 \right) ,\left( 2,3 \right) ,\left( 3,2 \right) ,\left( 1,3 \right) ,\left( 3,1 \right) \right\}$

3) Relation R is transitive if,

$\left( a,b \right) \in R\quad \& \quad \left( b,c \right) \in R\rightarrow \left( a,c \right) \in R$

$\therefore R=\left\{ \left( 1,2 \right) ,\left( 2,3 \right) ,\left( 1,3 \right) \right\}$

Thus, smallest equivalent relation on R is,

$R=\left\{ \left( 1,1 \right) ,\left( 2,2 \right) ,\left( 3,3 \right) ,\left( 1,2 \right) ,\left( 2,1 \right) ,\left( 2,3 \right) ,\left( 3,2 \right) ,\left( 1,3 \right) ,\left( 3,1 \right) \right\}$

Let $Z$ be the set of integers and $R$ be the relation defined in $Z$ such that $aRb$ if $a-b$ is divisible by $3$ . Then $R$ partitions the set $Z$ into ______ pairwise disjoint subsets.

1962256_1795388_ans_b187138ede5e4e109fce7a6a7c22b872.jpg

Solve :
$f(x) = 5 - 3x$
Find $ff(x)$ , in its simplest form.

Given

$f\left( x \right) =5-3x$

Replace

$x$ by

$f\left( x \right)$ , we get,

$f\left( f\left( x \right) \right) =5-3f\left( x \right)$

$\therefore f\left( f\left( x \right) \right) =5-3\left( 5-3x \right)$

$\therefore f\left( f\left( x \right) \right) =5-15+9x$

$\therefore f\left( f\left( x \right) \right) =9x-10$

Define a Cartesian product of two sets.

1976305_1861969_ans_7d1c45d0deed4ef8a80bba39ed474f85.jpg

Solve :
$f(x) = 5 - 3x$
Find $f (x + 2)$ .

Given

$f\left( x \right)=5-3x$

Put

$x=x+2$ , we get,

$f\left( x+2 \right) =5-3\left( x+2 \right)$

$\therefore f\left( x+2 \right) =5-3x-6$

$\therefore f\left( x+2 \right) =-1-3x$

If f is a function satisfying
$f(x + y) = f(x) f(y) \forall x, y \epsilon N$ such that $f(1) 3$ and $\sum_{x}^{n} = 1 f(x) = 120$ find the value of n.

1897439_1865512_ans_a0fbff6caaf541dbbcd1a45fb760204e.jpeg

Find the natural number a for which $\sum_{k=1}^{n}f(a + k) = 16(2^n -1)$ , where the function f satisfies the relation f(x+y) = f(x) f(y) for all natural numbers x, y and further f(1) = 2.

1777078_1754131_ans_c8cec581f82b4f61b13ad1e78f644b97.jpg

Subject	Number of students
Mathematics	$100$
Physics	$70$
Chemistry	$46$
Mathematics and Physics	$30$
Mathematics and Chemistry	$28$
Physics and Chemistry	$23$
Mathematics, Physics and Chemistry	$18$

Relations And Functions - Class 11 Engineering Maths - Extra Questions

f(x)=1\displaystyle f\left ( x \right )=1 and ϕ(x)=sin2x+cos2x. \phi \left ( x \right )=\sin ^{2}x+\cos ^{2}x.Show that both functions are equal?

State whether the given statement is true or false. If true enter 11 or else enter 00.(x,y)(x, y) is an ordered pair, whose first component is xx and the second component is yy.

The cartesian product A×AA\times A has 99 elements among which are found (−1,0)(-1, 0) and (0,1)(0, 1). Find the set AA and the remaining elements of A×AA\times A.

If xx is real, then find the solution set of √x+1+√x−1=1\sqrt{x+1}+\sqrt{x-1}=1.

Use the elements of set A ={x, y, z} to form all possible ordered pairs. Find the number of elements in the ordered pair.

Given below are some points, with their co-ordinates. P(4,3),Q(−3,2),R(5,1),S(0,−4)P(4, 3), Q(-3, 2), R(5, 1), S(0, -4) What is the y−y-coordinate of point RR ?

If (x3+1,y−23)=(53,13)\displaystyle \left ( \frac{x}{3}+1,y-\frac{2}{3} \right )=\left ( \frac{5}{3},\frac{1}{3} \right ) find the values of x and y

Find the number of ordered pairs in RoRR o R ?

A={1,2,3,5}A = \{1, 2, 3, 5\} and B={4,6,9}B = \{4, 6, 9\}. Define a relation RR from AA to BB by R={(x,y):R = \{(x, y): the difference between xx and yy is odd x∈A,y∈B}\displaystyle x\in A, y\in B\}. Write RR in roster form

Let A={1,2,3,4,6}A = \{1, 2, 3, 4, 6\} and RR be the relation on AA defined by {(a,b):a,b∈A ,b\{(a, b): a, \displaystyle b\in A\ , b is exactly divisible by a}a\}(i) Write RR in roster form(ii) Find the domain of RR(iii) Find the range of RR

State whether the given statement is true or false. If true enter 11 or else enter 00. (x,y)(x, y) is an ordered pair whose components are xx and yy.

Let AA and BB be two sets such that n(A)=3n(A) = 3 and n(B)=2n(B) = 2. If (x,1),(y,2),(z,1)(x, 1), (y, 2), (z, 1) are in A×B\displaystyle A\times B find AA and BB where x,yx, y and zz are distinct elements

Suppose 33 bulbs are selected at random from a lot. Each bulb is tested and classified as defective (D)(D) or non-defective (N)(N). Write the sample space of this experiment?

Let A={x,y,z}A =\{x, y, z\} and B={1,2}B = \{1, 2\}. Find the number of relations from AA to BB.

Determine the domain and range of the relation RR defined by R={(x,x+5):x∈{0,1,2,3,4,5}}R = \{(x, x + 5) : \displaystyle x\in \{0, 1, 2, 3, 4, 5\}\}

Write the relation R={(x,x3):x is a prime number less than 10}R = \{(x,\displaystyle x^{3}): x\mbox{ is a prime number less than 10}\} in roster form

For the given function F= { (1, 3), (2, 5), (4, 7), (5, 9), (3, 1) }, write the domain and range.

If kk is given a real constant f(x)f(x) is a real quadratic in 'xx such that f(x+k)=f(−x)f(x + k) = f(-x). The coefficient of x2x^2 is 11. Find f(x)f(x).

Let R be a relation from A = {1,2,3,4} to B = {1,3,5} i.e. (a,b) ϵ\epsilon R if a < b then find R o R−1R^{-1}

If G={7,8}G=\left\{ 7,8 \right\} and H={5,4,2}H=\left\{ 5,4,2 \right\} , find H×GH\times G.

Let S={1,2,3}S=\{1,2,3\} and a relation RR is defined by aRbaRb iff a=ba=b then find RR.

If f(x)=logesecxf(x)=\log_e\sec x and ϕ(x)=logetanx\phi(x)=\log_e\tan x then prove that e2f(x)−e2ϕ(x)=1e^{2f(x)}-e^{2\phi(x)}=1.

Find the domain of definations of the following functions:f(x)=√1−√1−x2f(x)=\sqrt {1-\sqrt {1-x^{2}}}

Let A={1,2,3,4}A = \left\{ {1,2,3,4} \right\} and RR is a relation on AA such that R={(a,b):a=2b}R = \left\{ {\left( {a,b} \right):a = 2b} \right\}

If A={a,b,c}A = \{a, b, c\} , B={1,2}B = \{1, 2\} , then find A×B,B×AandA×AA \times B , \, B \times A \, and \, A \times A

If the relation R:A→BR: A \rightarrow B where A={1,2,3,4},B={1,3,5}A = \{1, 2, 3, 4\}, B = \{1, 3, 5\} is defined by R={(x,y):x<y,x∈A,y∈B}R = \{(x, y): x < y, x \in A, y \in B\} then find RR and R−1R^{-1}.

If P={a,b}P = \{a , b\} and Q={x,y,z}Q = \{x, y, z\}, show that P×Q≠Q×P.P \times Q \neq Q \times P.

Let A={1,−1}A=\{1,-1\}. Then find A×AA\times A.

Given R={(x,y):x,y∈W,x2+y2=25}R = \{(x, y): x, y \in W, x^2 + y^2 = 25\}, where WW is the set of all whole numbers. Find the domain and range of RR.

If R={(x,y)|y=2x+7|R = \{(x, y)|y = 2x + 7|, where x∈Rx \in R and −5≤x≤5}-5 \le x \le 5\} is relation, find the domain of RR.

If f(x)=x+(1/x)f(x) = x + (1/x), show that {f(x)}3=f(x3)+3f(1/x)\{f(x)\}^3 = f(x^3) + 3f(1/x)

If A={0,1}A = \{0, 1\} and B={1,2,3}B = \{1, 2, 3\}, show that A×B≠B×AA \times B \neq B \times A

Let A={1,2,3}A = \{1, 2, 3\}. Find the number of relations from AA to A.A.

The Cartesian product P×PP\times P has 1616 elements among which two elements are (a,b)(a,b) and (c,d)(c,d). Find the set PP and the remaining elements of P×PP\times P

If 1|a|>1b\dfrac{1}{{\left| a \right|}} > \dfrac{1}{b}, then |a|<b\left| a \right| < b, where a & b are non-zero real numbers.

Let A = {1,2} and B = {3,4}. Find the number of relations from A to B.

Find the domain and range of the following real functions: i) f(x)=−|x|f(x) = -|x|ii) f(x)=√9−x2f(x) = \sqrt{9 - x^2}

If f(x)=x3−1x3f\left( x \right) = {x^3} - \frac{1}{{{x^3}}}, prove that f(x)+f(1x)=0f\left( x \right) + f\left( {\frac{1}{x}} \right) = 0

Find domain and range of R={(a,b)/a∈N,a<5 andb=2}R = \{ (a, b) / a \in N, \, \, \, a < 5 \, \, \text{ and} \, \, b = 2\}

If A={3,4}andB={1,2}A=\left\{ 3,4 \right\} \quad and\quad B=\left\{ 1,2 \right\} then of A×BA\times B is .

Determine the domain and the range of the relation RR defined by R={(x+1,x+5):x∈{0,1,2,3,4,5}}R=\{ \left( x+1,\quad x+5 \right) :x\in \{ 0,\quad 1,\quad 2,\quad 3,\quad 4,\quad 5\}\}

Show that the relation RR in the set ZZ of integers given by R={(a,b):2 divides a−b}R=\left\{(a,b):2\ divides\ a-b\right\}

If p(x)=ix2+5x+8 p(x)=ix^2+5x+8, then p(−1)p{(-1)}=

The value of xx satisfying the equation −3(x+6)=24-3(x+6)=24 ?

Let f:R→Rf:R\rightarrow R be defined as f(x)=10x+7f(x)=10x+7. Find the function g:R→Rg:R\rightarrow R such that gof=fog=IRgof=fog=I_{R}.

If f(x)=sin([x]π)x2+x+1f(x)=\cfrac{\sin{([x]\pi)}}{{x}^{2}+x+1}, where [x][x] denotes the integral part of xx. Show that f(x)f(x) is a constant function.

If A={1,2}A=\left\{1,2\right\}, form the set A×A×AA\times A\times A.

If A={1,2},B={x:x∈Nandx2−9=0},A = \left\{ {1,2} \right\},B = \left\{ {x:x \in N\,\,and\,\,{x^2} - 9 = 0} \right\}, Find A×B.A \times B.

If A×B={(a,1),(b,3),(a,3),(b,1),(a,2),(b,2)}A\times B=\left\{\left(a,1\right),\left(b,3\right),\left(a,3\right),\left(b,1\right),\left(a,2\right),\left(b,2\right)\right\}.Find AA and BB

Find xx and yy if (x+3,5)=(6,2x+y)\left(x+3,5\right)=\left(6,2x+y\right)

Let RR be relation from QQ to QQ defined by R={(a,b):a,b ∈ Q and a−b ∈ Z}R=\left\{(a,b):a,b\ \in \ Q\ and\ a-b\ \in \ Z\right\}. Show that :(a,b) ∈ R(a,b)\ \in \ R implies that (b,a) ∈ R(b,a)\ \in \ R.

Find the relations between aa and bb, so tat the function ff defined by:f(x)={ax+1,bx+3,if x≤3if x>3f\left( x \right)=\begin{cases} ax+1, \\ bx+3, \end{cases}\begin{matrix} if\ x\le 3 \\ if\ x>3 \end{matrix}is continuous at x=3x=3.

If A=\left\{a,b\right\}A=\left\{a,b\right\} and B=\left\{1,2,3\right\}B=\left\{1,2,3\right\} find \left(A\times B\right)\cap\left(B\times\,A\right)\left(A\times B\right)\cap\left(B\times\,A\right)

The cartesian product A\times AA\times A has 99 elements among which are found \left(-1,0\right)\left(-1,0\right) and \left(0,1\right)\left(0,1\right).Find the set AA and the remaining elements of A\times AA\times A

If A=\left\{1,3,5,6\right\}A=\left\{1,3,5,6\right\} and B=\left\{2,4\right\}B=\left\{2,4\right\}.Find A\times\,BA\times\,B and B\times \,AB\times \,A

Let RR be relation from QQ to QQ defined by R=\left\{(a,b):a,b\ \in \ Q\ and\ a-b\ \in \ Z\right\}R=\left\{(a,b):a,b\ \in \ Q\ and\ a-b\ \in \ Z\right\}. Show that :(a,b)\ \in \ R(a,b)\ \in \ R and (b,c)\ \in R(b,c)\ \in R implies that (a,c)\ \in \ R(a,c)\ \in \ R.

Find the values of aa and bb if \left(3a-2,b+3\right)=\left(2a-1,3\right)\left(3a-2,b+3\right)=\left(2a-1,3\right)

If A=\left\{a,b\right\}A=\left\{a,b\right\} and B=\left\{1,2,3\right\}B=\left\{1,2,3\right\} find A\times AA\times A and B\times\,BB\times\,B

List the relation R defined by R=\left \{ (a,b):a\leq b^{3} \right \}R=\left \{ (a,b):a\leq b^{3} \right \} in Roaster form. a,b \in Na,b \in N

If A=\left\{a,b\right\}A=\left\{a,b\right\} and B=\left\{1,2,3\right\}B=\left\{1,2,3\right\} find A\times BA\times B and B\times\,AB\times\,A

Let R=\left\{ \left( a,{ a }^{ 3 } \right) :\text{a is a prime number less than }\ 5 \right\} R=\left\{ \left( a,{ a }^{ 3 } \right) :\text{a is a prime number less than }\ 5 \right\} be a relation. Find the range of RR.

Write the domain of the relation RR defined on the set \mathbb{Z}\mathbb{Z} of integers as follows:(a,b)\in R\Leftrightarrow {a}^{2}+{b}^{2}=25(a,b)\in R\Leftrightarrow {a}^{2}+{b}^{2}=25

If R=\left\{ (x,y):x+2y=8 \right\} R=\left\{ (x,y):x+2y=8 \right\} is a relation on NN, then write the range of RR.

If R=\left\{ (x,y):{ x }^{ 2 }+{ y }^{ 2 }\le 4;x,y\in \mathbb{Z} \right\} R=\left\{ (x,y):{ x }^{ 2 }+{ y }^{ 2 }\le 4;x,y\in \mathbb{Z} \right\} is a relation on \mathbb{Z}\mathbb{Z}, write the domain of RR.

Write the identity relation on set A=\left\{ a,b,c \right\} A=\left\{ a,b,c \right\} .

Find the domain and range of each of the following real value functions:f(x)= |x-1|f(x)= |x-1|

There are four men and six women on the city councils. If one council number is selected for a committee at random, how likely that it is a woman

Let the relation RR be defined in NN by aRbaRb if 2a+3b=302a+3b=30. Then R=R= ....

Let the relation RR be defined on the set A=\left\{ 1,2,3,4,5 \right\} A=\left\{ 1,2,3,4,5 \right\} by R={(a,b):a^2b^2 < 8}.R={(a,b):a^2b^2 < 8}. Then RR is given by ______.

For the set A=\left\{1,2,3\right\}A=\left\{1,2,3\right\}, define a relation RR in the set AA as follows:R=\left\{(1,1),(2,2),(3,3),(1,3)\right\}R=\left\{(1,1),(2,2),(3,3),(1,3)\right\}Write the ordered pairs to be added to RR to make it the smallest equivalence relation.

Let RR relation defined on the set of natural number NN as follows:R=\left\{(x,y): x\in \ N, y\in \ N, 2x+y=41\right\}R=\left\{(x,y): x\in \ N, y\in \ N, 2x+y=41\right\}. Find the domain and rang of the relation RR. Also verify RR is reflexive, symmetric and transitive.

If RR be a relation << from A=\{1,2,3,4\}A=\{1,2,3,4\} to B=\{1,3,5\}B=\{1,3,5\} i.e., (a,b) \in(a,b) \in R if a<ba<b, then find ROR^{-1}ROR^{-1}

If f: R\rightarrow Rf: R\rightarrow R defined by f(x)=x^2+1f(x)=x^2+1, then the values of f^{-1}(17)f^{-1}(17) are a and b, then a+b =?

For each set of ordered pairs below, state whether it is a function or not give reasons(i){(3, 2), (4, 2), (5, 2)}if a function enter 1 else 0

If x, y\in \left [ 0, 2\pi \right ]x, y\in \left [ 0, 2\pi \right ], then find the total number of ordered pairs \left ( x, y \right )\left ( x, y \right ) satisfying the equation \sin x\cos y=1\sin x\cos y=1

Use elements of set P = {2 , 3} to find the number of all possible ordered pairs.

Given A = \{5, 6, 7\}A = \{5, 6, 7\} and B = \{3, 4\}B = \{3, 4\}. Form all possible ordered pairs and write the total number of ordered pairs formed so that both the components are from B.

Let A = \{1, 2\}A = \{1, 2\} and B = \{2, 3, 4\}B = \{2, 3, 4\}.If B \times B= \{(2,2), (2,3),(2, 4), (3,2), (3,3), (3,4) , (4,2), (4,3), (4,4)\} B \times B= \{(2,2), (2,3),(2, 4), (3,2), (3,3), (3,4) , (4,2), (4,3), (4,4)\}. Please enter 11 or else 00.

Let A = \{1, 2\}A = \{1, 2\} and B = \{2, 3, 4\}B = \{2, 3, 4\}, then B \times A=\{(2,1),(2,2), (3,1), (3,2),(4,1),(4,2)\}B \times A=\{(2,1),(2,2), (3,1), (3,2),(4,1),(4,2)\}. If true enter 11, or else enter 00.

$\displaystyle f\left ( x \right )=1$ and $\phi \left ( x \right )=\sin ^{2}x+\cos ^{2}x.$
Show that both functions are equal?

State whether the given statement is true or false. If true enter $1$ or else enter $0$ .
$(x, y)$ is an ordered pair, whose first component is $x$ and the second component is $y$ .

The cartesian product $A\times A$ has $9$ elements among which are found $(-1, 0)$ and $(0, 1)$ . Find the set $A$ and the remaining elements of $A\times A$ .

If $x$ is real, then find the solution set of $\sqrt{x+1}+\sqrt{x-1}=1$ .

Use the elements of set A ={x, y, z} to form all possible ordered pairs.
Find the number of elements in the ordered pair.

What is the $y-$ coordinate of point $R$ ?

If $\displaystyle \left ( \frac{x}{3}+1,y-\frac{2}{3} \right )=\left ( \frac{5}{3},\frac{1}{3} \right )$ find the values of x and y

Find the number of ordered pairs in $R o R$ ?

$A = \{1, 2, 3, 5\}$ and $B = \{4, 6, 9\}$ . Define a relation $R$ from $A$ to $B$ by $R = \{(x, y):$ the difference between $x$ and $y$ is odd $\displaystyle x\in A, y\in B\}$ . Write $R$ in roster form

Let $A = \{1, 2, 3, 4, 6\}$ and $R$ be the relation on $A$ defined by $\{(a, b): a, \displaystyle b\in A\ , b$ is exactly divisible by $a\}$
(i) Write $R$ in roster form
(ii) Find the domain of $R$
(iii) Find the range of $R$

$(x, y)$ is an ordered pair whose components are $x$ and $y$ .

Let $A$ and $B$ be two sets such that $n(A) = 3$ and $n(B) = 2$ . If $(x, 1), (y, 2), (z, 1)$ are in $\displaystyle A\times B$ find $A$ and $B$ where $x, y$ and $z$ are distinct elements

Suppose $3$ bulbs are selected at random from a lot. Each bulb is tested and classified as defective $(D)$ or non-defective $(N)$ . Write the sample space of this experiment?

Let $A =\{x, y, z\}$ and $B = \{1, 2\}$ . Find the number of relations from $A$ to $B$ .

Determine the domain and range of the relation $R$ defined by $R = \{(x, x + 5) : \displaystyle x\in \{0, 1, 2, 3, 4, 5\}\}$

Write the relation $R = \{(x,\displaystyle x^{3}): x\mbox{ is a prime number less than 10}\}$ in roster form

If $k$ is given a real constant $f(x)$ is a real quadratic in ' $x$ such that $f(x + k) = f(-x)$ . The coefficient of $x^2$ is $1$ . Find $f(x)$ .

Let R be a relation from A = {1,2,3,4} to B = {1,3,5} i.e. (a,b) $\epsilon$ R if a < b then find R o $R^{-1}$

If $G=\left\{ 7,8 \right\}$ and $H=\left\{ 5,4,2 \right\}$ , find $H\times G$ .

Let $S=\{1,2,3\}$ and a relation $R$ is defined by $aRb$ iff $a=b$ then find $R$ .

If $f(x)=\log_e\sec x$ and $\phi(x)=\log_e\tan x$ then prove that $e^{2f(x)}-e^{2\phi(x)}=1$ .

Find the domain of definations of the following functions:
$f(x)=\sqrt {1-\sqrt {1-x^{2}}}$

Let $A = \left\{ {1,2,3,4} \right\}$ and $R$ is a relation on $A$ such that $R = \left\{ {\left( {a,b} \right):a = 2b} \right\}$

If $A = \{a, b, c\}$ , $B = \{1, 2\}$ , then find $A \times B , \, B \times A \, and \, A \times A$

If the relation $R: A \rightarrow B$ where $A = \{1, 2, 3, 4\}, B = \{1, 3, 5\}$ is defined by $R = \{(x, y): x < y, x \in A, y \in B\}$ then find $R$ and $R^{-1}$ .

If $P = \{a , b\}$ and $Q = \{x, y, z\}$ , show that $P \times Q \neq Q \times P.$

Let $A=\{1,-1\}$ . Then find $A\times A$ .

Given $R = \{(x, y): x, y \in W, x^2 + y^2 = 25\}$ , where $W$ is the set of all whole numbers. Find the domain and range of $R$ .

If $R = \{(x, y)|y = 2x + 7|$ , where $x \in R$ and $-5 \le x \le 5\}$ is relation, find the domain of $R$ .

If $f(x) = x + (1/x)$ , show that $\{f(x)\}^3 = f(x^3) + 3f(1/x)$

If $A = \{0, 1\}$ and $B = \{1, 2, 3\}$ , show that $A \times B \neq B \times A$

Let $A = \{1, 2, 3\}$ . Find the number of relations from $A$ to $A.$

The Cartesian product $P\times P$ has $16$ elements among which two elements are $(a,b)$ and $(c,d)$ . Find the set $P$ and the remaining elements of $P\times P$

If $\dfrac{1}{{\left| a \right|}} > \dfrac{1}{b}$ , then $\left| a \right| < b$ , where a & b are non-zero real numbers.

Find the domain and range of the following real functions:
i) $f(x) = -|x|$
ii) $f(x) = \sqrt{9 - x^2}$

If $f\left( x \right) = {x^3} - \frac{1}{{{x^3}}}$ , prove that $f\left( x \right) + f\left( {\frac{1}{x}} \right) = 0$

Find domain and range of
$R = \{ (a, b) / a \in N, \, \, \, a < 5 \, \, \text{ and} \, \, b = 2\}$

If $A=\left\{ 3,4 \right\} \quad and\quad B=\left\{ 1,2 \right\}$ then of $A\times B$ is .

Determine the domain and the range of the relation $R$ defined by $R=\{ \left( x+1,\quad x+5 \right) :x\in \{ 0,\quad 1,\quad 2,\quad 3,\quad 4,\quad 5\}\}$

Show that the relation $R$ in the set $Z$ of integers given by
$R=\left\{(a,b):2\ divides\ a-b\right\}$

If $p(x)=ix^2+5x+8$ , then $p{(-1)}$ =

The value of $x$ satisfying the equation $-3(x+6)=24$ ?

Let $f:R\rightarrow R$ be defined as $f(x)=10x+7$ . Find the function $g:R\rightarrow R$ such that $gof=fog=I_{R}$ .

If $f(x)=\cfrac{\sin{([x]\pi)}}{{x}^{2}+x+1}$ , where $[x]$ denotes the integral part of $x$ . Show that $f(x)$ is a constant function.

If $A=\left\{1,2\right\}$ , form the set $A\times A\times A$ .

If $A = \left\{ {1,2} \right\},B = \left\{ {x:x \in N\,\,and\,\,{x^2} - 9 = 0} \right\},$ Find $A \times B.$

If $A\times B=\left\{\left(a,1\right),\left(b,3\right),\left(a,3\right),\left(b,1\right),\left(a,2\right),\left(b,2\right)\right\}$ .Find $A$ and $B$

Find $x$ and $y$ if $\left(x+3,5\right)=\left(6,2x+y\right)$

Let $R$ be relation from $Q$ to $Q$ defined by $R=\left\{(a,b):a,b\ \in \ Q\ and\ a-b\ \in \ Z\right\}$ .
Show that :
$(a,b)\ \in \ R$ implies that $(b,a)\ \in \ R$ .

Find the relations between $a$ and $b$ , so tat the function $f$ defined by:
$f\left( x \right)=\begin{cases} ax+1, \\ bx+3, \end{cases}\begin{matrix} if\ x\le 3 \\ if\ x>3 \end{matrix}$
is continuous at $x=3$ .

If $A=\left\{a,b\right\}$ and $B=\left\{1,2,3\right\}$ find $\left(A\times B\right)\cap\left(B\times\,A\right)$

The cartesian product $A\times A$ has $9$ elements among which are found $\left(-1,0\right)$ and $\left(0,1\right)$ .Find the set $A$ and the remaining elements of $A\times A$

If $A=\left\{1,3,5,6\right\}$ and $B=\left\{2,4\right\}$ .Find $A\times\,B$ and $B\times \,A$

Let $R$ be relation from $Q$ to $Q$ defined by $R=\left\{(a,b):a,b\ \in \ Q\ and\ a-b\ \in \ Z\right\}$ .
Show that :
$(a,b)\ \in \ R$ and $(b,c)\ \in R$ implies that $(a,c)\ \in \ R$ .

Find the values of $a$ and $b$ if $\left(3a-2,b+3\right)=\left(2a-1,3\right)$

If $A=\left\{a,b\right\}$ and $B=\left\{1,2,3\right\}$ find $A\times A$ and $B\times\,B$

List the relation R defined by $R=\left \{ (a,b):a\leq b^{3} \right \}$ in Roaster form. $a,b \in N$

If $A=\left\{a,b\right\}$ and $B=\left\{1,2,3\right\}$ find $A\times B$ and $B\times\,A$

Let $R=\left\{ \left( a,{ a }^{ 3 } \right) :\text{a is a prime number less than }\ 5 \right\}$ be a relation. Find the range of $R$ .

Write the domain of the relation $R$ defined on the set $\mathbb{Z}$ of integers as follows:
$(a,b)\in R\Leftrightarrow {a}^{2}+{b}^{2}=25$

If $R=\left\{ (x,y):x+2y=8 \right\}$ is a relation on $N$ , then write the range of $R$ .

If $R=\left\{ (x,y):{ x }^{ 2 }+{ y }^{ 2 }\le 4;x,y\in \mathbb{Z} \right\}$ is a relation on $\mathbb{Z}$ , write the domain of $R$ .

Write the identity relation on set $A=\left\{ a,b,c \right\}$ .

Find the domain and range of each of the following real value functions:
$f(x)= |x-1|$

Let the relation $R$ be defined in $N$ by $aRb$ if $2a+3b=30$ . Then $R=$ ....

Let the relation $R$ be defined on the set $A=\left\{ 1,2,3,4,5 \right\}$ by $R={(a,b):a^2b^2 < 8}.$ Then $R$ is given by ______.

For the set $A=\left\{1,2,3\right\}$ , define a relation $R$ in the set $A$ as follows:
$R=\left\{(1,1),(2,2),(3,3),(1,3)\right\}$
Write the ordered pairs to be added to $R$ to make it the smallest equivalence relation.

Let $R$ relation defined on the set of natural number $N$ as follows:
$R=\left\{(x,y): x\in \ N, y\in \ N, 2x+y=41\right\}$ . Find the domain and rang of the relation $R$ . Also verify $R$ is reflexive, symmetric and transitive.

If $R$ be a relation $<$ from $A=\{1,2,3,4\}$ to $B=\{1,3,5\}$ i.e., $(a,b) \in$ R if $a<b$ , then find $ROR^{-1}$

If $f: R\rightarrow R$ defined by $f(x)=x^2+1$ , then the values of $f^{-1}(17)$ are a and b, then a+b =?

For each set of ordered pairs below, state whether it is a function or not give reasons
(i){(3, 2), (4, 2), (5, 2)}
if a function enter 1 else 0

If $x, y\in \left [ 0, 2\pi \right ]$ , then find the total number of ordered pairs $\left ( x, y \right )$ satisfying the equation $\sin x\cos y=1$

Given $A = \{5, 6, 7\}$ and $B = \{3, 4\}$ . Form all possible ordered pairs and write the total number of ordered pairs formed so that both the components are from B.

Let $A = \{1, 2\}$ and $B = \{2, 3, 4\}$ .
If $B \times B= \{(2,2), (2,3),(2, 4), (3,2), (3,3), (3,4) , (4,2), (4,3), (4,4)\}$ .
Please enter $1$ or else $0$ .

$B \times A=\{(2,1),(2,2), (3,1), (3,2),(4,1),(4,2)\}$ . If true enter $1$ , or else enter $0$ .

Let $A = \{1, 2\}$ and $B = \{2, 3, 4\}$ , then $A \times A = \{ (1,1), (1,2),(2,1), (2,2)\}$ . If true enter $1$ , or else enter $0$ .

Let $A = \{1, 2\}$ and $B = \{2, 3, 4\}$ , then $A \times B = \{ (1,2), (1,3),(1, 4), (2,2), (2,3), (2,4) \}$ . If true enter $1$ , or else enter $0$ .

Given $A = \{5, 6, 7\}$ and $B = \{3, 4\}$ . Form all possible ordered pairs and write the total number of ordered pairs formed. So that :
Both the components are from A
Find the total number of such pairs.

If $A = \{x, y\}$ and $B = \{y, x\}$ ; then $A \times B = ?$
What are the number of elements of such set?

If $A =$ { $x$ : $x \,\,\in\,\, W$ , $3 \leq x < 6$ }, $B = \{3, 5, 7\}$ and $C = \{2, 4\}$ ; find : $(A - B) \times C$