If $$f:A\rightarrow B $$ is surjective then

no two elements of $$A$$ have the same image in $$B$$

every element of $$A$$ has an image in $$B$$

$$A$$ and $$B$$ are finite non empty sets

If $$a,b\in A, a*b\in A$$ then

$$*$$ is a unary operation in $$A$$

If $$*$$ is a binary operation in $$A$$ then

$$A$$ is not closed under $$*$$

$$A$$ is not closed under $$+$$

not a binary operation on $$R$$

not a binary operation in $$ E$$

If $$ f: R->R$$ is defined by $$f(x) = |x|$$, then

$$f^{-1}_{}(x) = \dfrac{1}{|x|}$$

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

MCQ Questions for Class 12 Commerce Maths Relations And Functions Quiz 1

$f(x) =x+\tan x$ and

$f$ is inverse of

$g$ , then

$g'(x)$ is equal to

0%
$\dfrac{1}{1+[g(x)-x]^2}$
0%
$\dfrac{1}{2-[g(x)+x]^2}$
0%
$\dfrac{1}{2+(x-g(x))^2}$
0%
$None\ of\ these$

Explanation

Since,

$f=g^{-1}$

$\implies f(g(x))=x$

$\Rightarrow f(g(x))=g(x)+\tan(g(x))=x$

$\tan (g(x))=x-g(x)$

Differentiating both sides with respect to

$x$

$g'(x)+\sec ^2(g(x))g'(x)=1$

$g'(x)=\dfrac{1}{1+\sec ^2(g(x))}=\dfrac{1}{2+\tan ^2(g(x))}$

$g'(x)=\dfrac{1}{2+(x-g(x))^2}$

Hence, option

$C$ is correct answer.

$f:R\rightarrow R$ is a function defined by

$f(x)=10x-7$ . If

$g=f^{-1}$ then

$g(x)=$

0%
$\dfrac{1}{10x-7}$
0%
$\dfrac{1}{10x+7}$
0%
$\dfrac{x+7}{10}$
0%
$\dfrac{x-7}{10}$

Explanation

Given

$f(x)=10x-7$ and

$g=f^{-1}$
Let

$f\left ( x \right )=10x-7=y$

$\Rightarrow x=\dfrac{y+7}{10}=f^{-1}\left ( y \right )=g\left ( y \right )$

$\therefore g\left ( x \right )=\dfrac{x+7}{10}$

A constant function

$f:A\rightarrow B$ will be one-one if

0%
$n (A) = n(B)$
0%
$n(A) = 1$
0%
$n (B) = 1$
0%
$n (A) < n (B)$

Explanation

Given f is a constant functions.

$\Rightarrow$ range of f is

$\left \{ c \right \}(say)$
Since f is one-one

$\Rightarrow$ domain of A should also contain
one element.

$\therefore n(A)=1.$

$f:\mathbb{N} \rightarrow \mathbb{N}$ and

$f(x) = x^{2}$ then the function is

0%
not one to one function
0%
one to one function
0%
into function
0%
none of these

Explanation

Given :

$f(x)=x^2$

$\because x^2>0 \implies f(x)>0$ for every

$x\in \mathbb{N}$

Let' find domain and range of

$f(x)$

$x =1\longrightarrow f(x)=1$

$x =2\longrightarrow f(x)=4$

$x =3\longrightarrow f(x)=9$

$x=4\longrightarrow f(x)=16$

$x =5\longrightarrow f(x)=25$

Here we get that, no two elements of the domain has the same image and no element of co-domain is the image of more than one element in the domain.

$\therefore$

$f$ is one-one.

Function

$f:X\rightarrow Y$ is onto if, for any

$y\in {Y}$ there exist

$x\in {X}$ such that

$f(x)=y$

Let's prove that

$f(x)=x^2$ is not onto.

Let's take an example

$y=3\in \mathbb{N}$

$\implies f(x)=y$ ............ by definition of onto function

$\implies x^2=3$

$\implies x=\sqrt{3}\ or\ -\sqrt3$ which does not belong to

$\mathbb{N}$

Hence, for

$y=3\in \mathbb{N}$ there does not exist any

$x\in \mathbb{N}$ such that

$f(x)=y$ .

$\therefore$

$f$ is not onto.

Hence,

$f$ in only one-one.

$f(x)=1$ , if

$x$ is rational and

$f(x)=0$ , if

$x$ is irrational
then

$(fof) (\sqrt{5})=$

0%
$0$
0%
$1$
0%
$\sqrt{5}$
0%
$\dfrac{1}{\sqrt{5}}$

Explanation

Given,

$f(x)=\begin{cases}1 & \text {; if x is rational} \\ 0 & \text{; if x is irrational}\end{cases}$

Consider

$x=\sqrt{5}$ ,

$\text{(f o f)}(\sqrt{5})=f(f(\sqrt{5}))$

$=f(0)$ ..........

$[\because \sqrt{5}$ is irrational

$]$

$=1$ ..........

$[\because 0$ is rational

$]$

$\therefore \text{(f o f)}(\sqrt{5})=1$
Hence, option B is correct.

$f(x) = 3x + 2, g(x) = x^2 + 1$ , then the value of

$(fog) (x^2 +1)$ is

0%
$3x^4 + 6x^2 + 8$
0%
$3x^4 + 3x + 4$
0%
$6x^4 + 3x^2 + 2$
0%
$3x^2 + 6x + 2$

Explanation

We have

$f(x) = 3x + 2, g(x) = x^2 + 1$

$fog(x) = f[g(x)] =3(x^2+1)+2= 3x^2 + 5$
so

$(fog) (x^2 +1) = 3(x^2 + 1)^2 + 5 =3(x^4+2x^2+1)+5= 3x^4+ 6x^2 + 8$

$f:A\rightarrow B$ is surjective then

0%
no two elements of $A$ have the same image in $B$
0%
every element of $A$ has an image in $B$
0%
every element of $B$ has at least one pre-image in $A$
0%
$A$ and $B$ are finite non empty sets

Explanation

Surjective means onto function.
co domain

$=$ Range
So every element of

$B$ has at least one pre-image in

$A$ .

$f:(0,\infty )\rightarrow (0,\infty )$ is defined by

$f(x)=x^{2}$ , then

$f^{-1}(x)=$

0%
$\sqrt{x}$
0%
$\dfrac{1}{\sqrt{x}}$
0%
Not invertible
0%
$\dfrac{2}{\sqrt{x}}$

Explanation

$f$ is defined from

$\left ( 0,\infty \right )\rightarrow \left ( 0,\infty \right )$

$\therefore f$ is onto and one - one.
Let

$f\left ( x \right )=x^{2}=y$

$\Rightarrow x=f^{-1}\left ( y \right )$
Also,

$x^2=y\Rightarrow x=\sqrt{y}$ as

$x> 0$

$\therefore f^{-1}\left ( y \right )=\sqrt{y}$

Hence,

$f^{-1}\left ( x \right )=\sqrt{x}$

$f:R\rightarrow R , g:R\rightarrow R$ and

$f(x)= \sin x$ ,

$g(x)=x^{2}$ then

$fog(x)=$

0%
$x^{2}+\sin x$
0%
$x^{2}\sin x$
0%
$\sin^{2}x$
0%
$\sin x^{2}$

Explanation

$fog\left ( x \right )=f\left(g ( x \right ))=f (x^{2})$

$=\sin x^{2}$

Find the value of

$\displaystyle \left( g\circ f \right) \left( 6 \right)$ if

$\displaystyle g\left( x \right) ={ x }^{ 2 }+\frac { 5 }{ 2 }$ and

$\displaystyle f\left( x \right) =\frac { x }{ 4 } -1$ .

0%
2.75
0%
3
0%
3.5
0%
8.625

Explanation

$f(x)=\dfrac {x}{4}-1$ and

$g(x)=x^2+\dfrac {5}{2}$

$\therefore f(6)=\dfrac{6}{4}-1=\dfrac{3}{2}-1=\dfrac{1}{2}$

$\therefore (gof)(6)=g(\dfrac{1}{2})=\left(\dfrac{1}{2}\right)^2+\dfrac{5}{2}=\dfrac{11}{4}=2.75$

The first component of all ordered pairs is called

0%
Range
0%
Domain
0%
Function
0%
None of these

The second component of all ordered pairs of a relation is

0%
Range
0%
Domain
0%
mapping
0%
none of these

A ______ maps elements of one set to another set.

0%
order
0%
set
0%
relation
0%
function

Explanation

A relation map elements of one set to another set

$R:A\to B$

Elements in A is mapped to elements in set B.

Suppose y is equal to the sum of two quantities of which one varies directly as x and the other inversely as x If y = 6 when x = 4 and

$\displaystyle y=\frac{10}{3}$ when x = 3 then what is the relation between x and y?

0%
$\displaystyle y=x+\frac{4}{x}$
0%
$\displaystyle y+2x=\frac{4}{x}$
0%
$\displaystyle y=2x+\frac{8}{x}$
0%
$\displaystyle y=2x-\frac{8}{z}$

Explanation

(A)If

$y=X+\frac{4}{x}\Rightarrow 4+\frac{4}{4}=3 \neq 6=6$

(B)

$y+2x=\frac{4}{x}\Rightarrow 6+2(4)=\frac{4}{4}\Rightarrow 6+8\neq 1$

(C)

$y=2x+\frac{8}{x}\Rightarrow 6=2(4)+\frac{8}{4}\Rightarrow 6\neq 8+2$

(D)

$y=2x-\frac{8}{x}\Rightarrow 6=2(4)-\frac{8}{4}=8-2=6$ If x=4 and y=6

$y=2x-\frac{8}{x}\Rightarrow \frac{10}{3}=2\times 3-\frac{8}{3}\Rightarrow \frac{10}{3}=\frac{10}{3}$ If x=3 and y=

$y=\frac{10}{3}$

Then option (D)

$Y=2x-\frac{8}{x}$

If X is brother of the son of Y's son. How is X related to Y?

0%
Son
0%
Brother
0%
Cousin
0%
Grandson
0%
Uncle

In the group

$G = \left \{1, 5, 7, 11\right \}$ under

$\otimes_{12}$ the value of

$7\otimes_{12} 11^{-1}$ is equal to

(

$\otimes_{12}$ : under multiplication modulo

$12$ )

0%
$5$
0%
$7$
0%
$11$
0%
$1$

Explanation

Clearly

$11^{-1} = 11(\because 11\otimes_{12} 11 = 1)$

$\therefore 7 \otimes_{12} 11^{-1} = 7\otimes_{12} 11 = 5$

What is the relation for the statement "A is taller than B"?

0%
is taller than
0%
A is taller
0%
B is taller
0%
is less than

x varies directly as y and inversely as the square of z. When y = 4 and z is 14 x =If y = 16 and z = 7 what is x?

0%
180
0%
160
0%
280
0%
200

Explanation

Given x varies directly as y and inversely as the square of z. When y = 4 and z is 14 x = 10. If y = 16 and z = 7

Then

$x=k\frac{y}{z^{2}}$

$\Rightarrow 10=k\times \frac{4}{14\times14}$

$\Rightarrow k=490$

Thus

$x=490\times \frac{16}{7\times 7}=160$

$f: R \rightarrow R$ and

$g: R \rightarrow R$ are defined by

$f(x) =2x +3, g(x)=x^2 + 7$ , what are the values of

$x$ such that

$g(f(x))=8$ ?

0%
$1, 2$
0%
$-1, 2$
0%
$-1, -2$
0%
$1, -2$

Explanation

Given,

$f(x)=2x+3$

Also given

$g(x)={ x }^{ 2 }+7$

Therefore,

$g[f(x)]={ \left( f(x) \right) }^{ 2 }+7$

$\Rightarrow g[f(x)]={ \left( 2x+3 \right) }^{ 2 }+7$

Thus

${ \left( 2x+3 \right) }^{ 2 }+7=8\\ \Rightarrow { \left( 2x+3 \right) }^{ 2 }=1\\ \Rightarrow 2x+3=\pm 1\\ \Rightarrow 2x=\pm 1-3\\ \Rightarrow 2x=-2,-4\\ \Rightarrow x=-1,-2$

Option C is correct.

Find the correct expression for

$\displaystyle f\left( g\left( x \right) \right)$ given that

$\displaystyle f\left( x \right) =4x+1$ and

$\displaystyle g\left( x \right) ={ x }^{ 2 }-2$

0%
$\displaystyle -{ x }^{ 2 }+4x+1$
0%
$\displaystyle { x }^{ 2 }+4x-1$
0%
$\displaystyle 4{ x }^{ 2 }-7$
0%
$\displaystyle 4{ x }^{ 2 }-1$
0%
$\displaystyle 16{ x }^{ 2 }+8x-1$

Explanation

Given,

$g(x)=x^2-2$

$f(g(x))=f(x^2-2)$

Also given,

$f(x)=4x+1$

$\therefore f(x^2-2)=4(x^2-2)+1$

$\Rightarrow 4x^2-8+1$

$\Rightarrow 4x^2-7$

$a,b\in A, a*b\in A$ then

0%
$*$ is a unary operation in $A$
0%
$a * b = b * a$
0%
$*$ is a binary operation in $A$
0%
$a * b \neq b * a$

Explanation

$a,b\in A,a*b\in A,$ then

$*$ is a binary operation in A.

$f(x) = \sqrt {x^{2} - 3x + 6}$ and

$g(x) = \dfrac {156}{x +17}$ , find the value of the composite function

$g(f(4))$ .

0%
$5.8$
0%
$7.4$
0%
$7.7$
0%
$8.2$
0%
$10.3$

Explanation

$f(x)=\sqrt { x^{ 2 }-3x+6 } ,\quad g(x)=\frac{ 156 }{ x+17 }\\ g(f(4))=\quad ?\\ f(4)=\sqrt { 10 } \\ g(f(4))=g(\sqrt { 10 } )=\quad 3.16\\ g(f(4))=7.7$

Squaring a given number is a

0%
relation in some set
0%
relation
0%
unary operation
0%
binary operation

$*$ is a binary operation in

$A$ then

0%
$A$ is closed under $*$
0%
$A$ is not closed under $*$
0%
$A$ is not closed under $+$
0%
$A$ is closed under $-$

Explanation

$*$ is a binary in

$A$ ,then A is closed under

$*$

$+$ is

0%
binary operation on $R$
0%
not a binary operation on $R$
0%
a binary operation in $Q^c$
0%
not a binary operation in $E$

$*$ is said to be commutative in

$A$ for all

$a,b\in A$

0%
$a + b = b + a$
0%
$a * b = b * a$
0%
$a - b = b - a$
0%
$a * b \neq b * a$

Explanation

$(B) a*b = b*a$

here

$*$ is commutative.

$f: R \rightarrow R$ and

$g: R \rightarrow R$ are defined by

$f(x) =3x -4$ , and

$g(x)=2 + 3x$ , find

$(g^{-1}\, of^{-1})(5)$ .

0%
$1$
0%
$\dfrac {1}{2}$
0%
$\dfrac {1}{3}$
0%
$\dfrac {1}{5}$

Explanation

Given,

$f(x)=3x-4$

Let

$y=3x-4$

$\Rightarrow x=\dfrac { y+4 }{ 3 } \\ \Rightarrow f^{ -1 }\left( x \right) =\dfrac { x+4 }{ 3 }$

Also given

$g(x)=2+3x$

$\Rightarrow y=2+3x\\ \Rightarrow x=\dfrac { y-2 }{ 3 } \\ \Rightarrow g^{ -1 }\left( x \right) =\dfrac { x-2 }{ 3 }$

Thus

$g^{ -1 }\left( \text{of}^{ -1 } \right) \left( x \right) =\dfrac { \left( \dfrac { x+4 }{ 3 } \right) -2 }{ 3 }$

$\Rightarrow g^{ -1 }\left( \text{of}^{ -1 } \right) \left( x \right) =\dfrac { x-2 }{ 9 } \\ \Rightarrow g^{ -1 }\left( \text{of}^{ -1 } \right) \left( 5 \right) =\dfrac { 5-2 }{ 9 } =\dfrac { 1 }{ 3 }$

So, option C is correct.

For what value of x is

$fog = gof$ if

$f(x)=x - 2$ and

$g(x)=x^3+3$ ?

0%
$\dfrac{-2}{3}$
0%
$-1$
0%
$\dfrac{3}{2}$
0%
$\dfrac{-3}{2}$

Explanation

$f(x)=x-2\\ g(x)={ x }^{ 3 }+3\\ f[g(x)]=g(x)-2\\ \Rightarrow fog={ x }^{ 3 }+3-2={ x }^{ 3 }+1\\ g[f(x)]={ \{ f(x)\} }^{ 3 }+3\\ \Rightarrow gof={ (x-2) }^{ 3 }+3\\ fog=gof\\ { x }^{ 3 }+1={ (x-2) }^{ 3 }+3\\ { x }^{ 3 }-2={ (x-2) }^{ 3 }\\ { x }^{ 3 }-2={ x }^{ 3 }-8-6x(x-2)\\ { x }^{ 3 }-2={ x }^{ 3 }-8-6{ x }^{ 2 }+12x\\ \Rightarrow { x }^{ 2 }+1+2x=0\\ \Rightarrow { (x+1) }^{ 2 }=0\\ \Rightarrow x=-1$

So none of the above options are correct

Find number of all such functions

$y = f(x)$ which are one-one?

0%
$0$
0%
$3^{5}$
0%
$^{5}P_{3}$
0%
$5^{3}$

Explanation

One-one functions are those in which each element in the domain has a unique image in the range of the function.

Here

$B$ is the co-domain of the function which has lesser number of elements than the domain itself. This shows that it is not possible for the function to be one-one.

The inverse of the function

$y=\cfrac { { 2 }^{ x } }{ 1+{ 2 }^{ x } }$ is

0%
$x=\log _{ 2 }{ \cfrac { 1 }{ 1-{ 2 }^{ y } } }$
0%
$x=\log _{ 2 }{ \left( 1-\cfrac { 1 }{ y } \right) }$
0%
$x=\log _{ 2 }{ \left( \cfrac { 1 }{ 1-y } \right) }$
0%
$x=\log _{ 2 }{ \left( \cfrac { y }{ 1-y } \right) }$

Explanation

Given,

$y=\cfrac { { 2 }^{ x } }{ 1+{ 2 }^{ x } } \Rightarrow y+{ 2 }^{ x }y={ 2 }^{ x }$

$\Rightarrow y={ 2 }^{ x }(1-y)\Rightarrow { 2 }^{ x }=\cfrac { y }{ 1-y }$
Taking log on both sides at base

$2$ , we get

$\log _{ 2 }{ \left( { 2 }^{ x } \right) } =\log _{ 2 }{ \cfrac { y }{ 1-y } }$

$\Rightarrow x=\log _{ 2 }{ \cfrac { y }{ 1-y } }$

$\therefore$ Inverse is

$\log _{ 2 }{ \cfrac { y }{ 1-y } }$

$f : R - \left \{\dfrac {3}{5}\right \}\rightarrow R - \left \{\dfrac {3}{5}\right \}; f(x) = \dfrac {3x + 1}{5x - 3}$ , then ___________.

0%
$f^{-1} (x) = 2f(x)$
0%
$f^{-1} (x) = f(x)$
0%
$f^{-1} (x) = -f(x)$
0%
$f^{-1} (x)$ does not exists

Explanation

$f(x) = \dfrac {3x + 1}{5x - 3}$ ....

$(i)$
We know that if

$f(x) = \dfrac {ax + b}{cx + d}$ and

$a + d = 0$ , then

$f(x) = f^{-1}(x)$

From

$(i)$ , we get

$a=3,b=1,c=5,d=-3$

$a+d=3-3=0$

Hence,

$f^{-1}(x)=f(x)$

Suppose that

$g\left( x \right) =1+\sqrt { x }$ and

$f\left( g\left( x \right) \right) =3+2\sqrt { x } +x$ , then

$f\left( x \right)$ is

0%
$1+2{ x }^{ 2 }$
0%
$2+{ x }^{ 2 }$
0%
$1+x$
0%
$2+x$

Explanation

$g(x)=1+\sqrt{x}$

$(g(x))^{2}=1+2\sqrt{x}+x$

$2+(g(x))^{2}=3+2\sqrt{x}+x$

$\therefore f(x)=2+x^{2}$

The number of real linear functions

$f(x)$ satisfying

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

0%
$0$
0%
$4$
0%
$5$
0%
$2$

Explanation

Let

$f(x) = ax+b$

$f(f(x))=a(ax+b)+b = a^2x+ab+b$

Given,

$f(f(x))=x+f(x)$

$\implies a^2x+ab+b = x+ ax+b$

$\implies (a^2-a-1)x+ab=0$

Since above equation is valid for all values of x, we have

$a^2-a-1 =0$ and

$ab=0$

$a^2-a-1 =0$

$a=\dfrac{ 1\pm \sqrt{1+4}}{2}$

$\implies a=\dfrac{ 1\pm \sqrt{5}}{2}$

$ab=0$

Since

$a$ is non-zero,

$\implies b=0$

$f(x) =\dfrac{ 1+ \sqrt{5}}{2}x$ and

$f(x) =\dfrac{ 1- \sqrt{5}}{2}x$ satisfy the condition.

Hence the answer is option (D)

Let A = {0, 1} and N the set of all natural numbers. Then the mapping

$f : N \rightarrow A$ defined by

$f(2n - 1) = 0, f (2n) = 1 \forall n \epsilon N$
is many-one onto.

0%
True
0%
False

Explanation

Let

$A={0,1}$ and N the set of all natural numbers. Then the mapping

$f:N \longrightarrow A$ defines by

$f(2n-1)=0,$

$f(2x)=1$

$\forall$

$n$

$\epsilon$

$N$ is many-one onto. The above statement is absolutely true.

$D$ be subset of the set of all rational numbers which can be expressed as terminating decimals, then

$D$ is closed under the binary operations of:

0%
addition, subtraction and division
0%
addition, multiplication and division
0%
addition, subtraction and multiplication
0%
subtraction, multiplication and division

Explanation

Division of rational number that can be expressed as terminating decimal may not be rational.

$\therefore$

$D$ is closed under addition, subtraction, and multiplication.

$a\ast b={ a }^{ 3 }+{ b }^{ 3 }$ on

$z$ , then

$\left( 1\ast 2 \right) \ast 0=........$

0%
$0$
0%
$729$
0%
$81$
0%
$27$

Explanation

$a*b=a^{3}+b{3}$

$1*2=1^{3}+2{3}=1+8=9$

$(1*2)*0=9*0=9^{3}+0^{3}=729+0=729$

State True or False.

Let

$f : R \rightarrow R$ be defined by

$f (x) = cos (5x + 2)$ . Then

$f$ is invertible.

0%
True
0%
False

Explanation

$f\left( x \right) =\cos\left( 5x+2 \right)$

$x=\left[ { \cos }^{ -1 }\left( f\left( x \right) \right) -2 \right] \dfrac { 1 }{ 5 }$

As,

$-1\le f\left( x \right)\le 1$

So, we can't put other values which makes it INTO function.

So,

$f\left( x \right) =cos\left( 5x+2 \right)$ is not invertible.

If a language of natural numbers has a binary regularly of

$0$ and

$1$ , then which one of the following strings represents the natural number

$7$ ?

0%
$1$
0%
$101$
0%
$110$
0%
$111$

The number of binary operations on

$\left\{ 1,2,3,4 \right\}$ is ______.

0%
${ 4 }^{ 2 }$
0%
${ 4 }^{ 8 }$
0%
${ 4 }^{ 3 }$
0%
${ 4 }^{ 16 }$

Explanation

$\{1,2,3,4\}$ there are 4 elements.

The number of binary operations in a set with n elements

$=n^{(n\times n)}$

Here

$n=4$

So the number of binary operations in a set with n elements

$=4^{(4\times 4)}=4^{16}$

$f: R->R$ is defined by

$f(x) = |x|$ , then

0%
$f^{-1}_{}(x) = -x$
0%
$f^{-1}_{}(x) = \dfrac{1}{|x|}$
0%
The function $f^{-1}_{}(x)$ does not exist
0%
$f^{-1}_{}(x) = \dfrac{1}{x}$

Explanation

For a function

$f(x)$ to be invertible, the function must be one-one and onto.

The range of

$f(x) = |x|$ is

$[0, \infty)$ , while the co-domain of

$f(x)$ is given as

$\mathbb{R}$ . Hence

$f(x)$ is not onto.

Also, since

$f(x) = f(-x)$ ,

$f(x)$ is also not one-one in its domain.

Hence, f(x) is not invertible, ie, the function

$f^{-1}(x)$ does not exist.

Option C is the right answer.

$a \times b =2 a - 3b + ab$ , then

$3 \times 5+5\times 3$ is equal to

0%
$22$
0%
$24$
0%
$26$
0%
$28$

Explanation

$a\times b = 2a-3b+ab$

find

$3\times 5+5\times 3$

$3\times 5[a = 3, b = 5]$

$\therefore 3\times 5 = 2(3)-3(5)+3(5)$

$3\times 5= 6$

$5\times 3 [a = 5, b = 3]$

$\therefore 5\times 3 = 2(5)-3(3)+(5)(3)$

$= 10-9+15$

$5\times 3= 16$

$\therefore 3\times 5+5\times 3 = 6+16$

$3\times 5+5\times 3= 22$

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

$R$ .

0%
$z, z$
0%
$z^+, z$
0%
$z, z^-$
0%
None of these

Explanation

Given:

$R=\{ (a, b) : a, b, \in z, a-b \text{ is an integer}\}$

As difference of integers are also integers so,

Domain of

$R = z$

Range of

$R = z$ , as

$a-b$ spans the whole integer values.

$x \times y = x^{2}+y^{2}-xy$ then the value of

$9 \times 11$ is :

0%
$93$
0%
$103$
0%
$113$
0%
$121$

Explanation

$\textbf{Hint: Put}$

$\boldsymbol{x=9, y=11}$

$\textbf{in the given expression and then simplify to get the result.}$

Correct Option:

$\textbf{B}$

Solution:

$\textbf{Step 1: Put}$

$\boldsymbol{x=9, y=11}$

$\textbf{in the given expression}$

We have given,

$x \times y = x^{2}+y^{2}-xy$

By putting

$x=9, y=11$ , we get

$9\times 11 = 9^{2}+11^{2}-9(11)$

$\textbf{Step 2: Simplify the expression in R.H.S}$

$\Rightarrow9 \times 11 = 81+121-99$

$\Rightarrow 9\times 11 = 202-99$

$\Rightarrow 9\times 11 = 103$

$\textbf{Hence, the correct option is B}$

Let

$f(x)={x}^{3}-6{x}^{2}+15x+3$ . Then,

0%
$f(x)> 0$ for all $x\in R$
0%
$f(x)> f(x+1)$ for all $x\in R$
0%
$f(x)$ is invertible
0%
$f(x)< 0$ for all $x\in R$

Explanation

$f(x)=x^3-6x^2+15x+3$

$f'(x)=3x^2-12x+15$

$=3(x^2-4x+5)$

$=3(x^2-4x+4+1)$

$f'(x)=3(x-2)^2+{3} > 0$

Therefore

$f(x)$ is strictly increasing function

$\Rightarrow f^{-1}(x)$ exists

Hence

$f(x)$ is a invertible function.

The number of binary operation on {1, 2, 3... n} is..

0%
$2^n$
0%
$n^2$
0%
$n^3$
0%
$n^{2n}$

Explanation

we have general formula,

for the

$n$ number of series, the number of binary operation is given by,

$2^n$

Read the following information and answer the three items that follow :
Let

$f(x) = x^2 + 2x - 5$ and

$g(x) = 5x + 30$
Consider the following statements:

$f[g(x)]$ is a polynomial of degree 3.

$g[g(x)]$ is a polynomial of degree 2.
Which of the above statements is/are correct ?

0%
1 only
0%
2 only
0%
Both 1 and 2
0%
Neither 1 nor 2

Explanation

Neither 1 nor 2

Given,

$f(x)=x^2+2x-5$

$g(x)=5x+30$

(i)

$f[g(x)]$

$=f[5x+30]$

$=(5x+30)^2+2(5x+30)-5$

upon solving the above equation, we get,

$f[g(x)]=25x^2+310x+955$

degree

$=2$

(ii)

$g[g(x)]$

$=g[5x+30]$

$=5(5x+30)+30$

$=25x+150+30$

$=25x+180$

$g[g(x)]=25x+180$

degree

$=1$

Let

$f(x)=\cfrac { 1 }{ 1-x }$ . Then

$\left\{ f\circ \left( f\circ f \right) \right\} (x)$

0%
$x$ for all $x\in R$
0%
$x$ for all $x\in R-\left\{ 1 \right\}$
0%
$x$ for all $x\in R-\left\{ 0,1 \right\}$
0%
none of these

Explanation

$f(x)=\dfrac{1}{1-x}$ for

$x\in R-\{1\}$

$\{fof\}(x)=\dfrac{1}{1-\dfrac{1}{1-x}}=\dfrac{1}{\dfrac{1-x-1}{1-x}}=\dfrac{x-1}{x}=1-\dfrac{1}{x}$ for

$x\in R - \{0,1\}$

$\{fofof\}(x)=\dfrac{1}{1-fof(x)}=\dfrac{1}{1-\dfrac{x-1}{x}}=\dfrac{1}{\dfrac{x-x+1}{x}}=x$ for

$x\in R-\{0,1\}$

Read the following information and answer the three items that follow :
Let

$f(x) = x^2 + 2x - 5$ and

$g(x) = 5x + 30$
What are the roots of the equation

$g[f(x)] = 0$ ?

0%
$1, -1$
0%
$-1, -1$
0%
$1, 1$
0%
$0, 1$

Explanation

Given,

$f(x)=x^2+2x-5$

$g(x)=5x+30$

$g[f(x)]=0$

$g[x^2+2x-5]=0$

$5(x^2+2x-5)+30=0$

$5x^2+10x-25+30=0$

$5x^2+10x+5=0$

$x^2+2x+1=0$

$(x+1)^2=0$

$\therefore x=-1,-1$

Read the following information and answer the three items that follow :
Let

$f(x) = x^2 + 2x - 5$ and

$g(x) = 5x + 30$
If

$h(x) = 5f(x) - xg (x)$ , then what is the derivative of

$h(x)$ ?

0%
$-40$
0%
$-20$
0%
$-10$
0%
$0$

Explanation

Given,

$f(x)=x^2+2x-5$

$g(x)=5x+30$

$h(x)=5f(x)-xg(x)$

$=5(x^2+2x-5)-x(5x+30)$

$=5x^2+10x-25-5x^2-30x$

$h(x)=-20x-25$

$\dfrac{d}{dx}[h(x)]=\dfrac{d}{dx}(-20x-25)$

$=-20$

The number of one-one functions that can be defined from

$A=\{4,8,12,16\}$ to

$B$ is

$5040,$ then

$n(B)=$

0%
$7$
0%
$8$
0%
$9$
0%
$10$

Explanation

The first element

$4$ of

$A$ can be mapped to any of the

$n$ elements in

$B$
Similarly second element

$8$ of

$A$ can be mapped to any of

$(n-1)$ elements in

$B$

$\Rightarrow$ Total no of one - one functions is

$n(n-1)(n-2)(n-3)=5040$

$\Rightarrow n=10$

CBSE Questions for Class 12 Commerce Maths Relations And Functions Quiz 1 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Relations And Functions Quiz 1 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.