Class 11 Commerce Applied Mathematics - Functions

If $f(x)=\sqrt{x+2}+\dfrac{1}{\log_{10} (1-x)}$ . Find the domain of definition of $f(x)$ .

We have

$\sqrt{x+2}$ is defined for

$x+2\ge 0$ or,

$x\ge -2$ ........(1).

Now

$\dfrac{1}{\log_{10}(1-x)}$ is defined for

$\log_{10}(1-x)\ne0$ or,

$x\ne1$ ......(2) and

$(1-x)\gt 0$ or,

$x\lt 1$ .......(3).

Combining (1),(2) and (3) we get the domain of definition for

$f(x)$ and it is

$[-2,0)\cap(0,1)$ .

Find the domains of definition of the following functions.
$\displaystyle y \, = \, \frac{\sqrt{6x \, - \, x^2 \, - \, 5}}{5^{x \, - \, 2} \, - \, 1}.$

$y = \dfrac{\sqrt{6x-x^{2}-5}}{5^{x-2}-1}, \dfrac{\sqrt{p(x)}}{Q(x)}$

Domain

$\Rightarrow p(x)\geq 0$ and

$Q(x)\neq 0$

$6x-x^{2}-5\geq 0$

multiply -1 both side

$x^{2}-6x+5\leq 0$

$(x-5) (x-1) \leq 0$

$x\epsilon [1,5]...(i)$

and

$5^{x-2}-1\neq 0$

$5^{x-2}\neq 1$

$5^{x-2}\neq 5^{0}$

$x-2\neq 0$

$x\neq 2$

$x\epsilon (-\infty ,\infty )-[2]...(ii)$

taking intersection of both values of function

Domain

$= x\epsilon [1,5]-{\left \{ 2 \right \}}$

1127053_886661_ans_f6c67e5697314878beaf71ee275af9be.jpg

Let $f(x)=\sqrt{x+2}+\dfrac{1}{\log_{10}(1-x)}$ . Then find the domain of $f(x)$ .

$f(x)=\sqrt{x+2}+\dfrac{1}{\log_{10}(1-x)}$ .

Now

$\sqrt{x+2}$ is defined for

$x+2\ge 0\Rightarrow x\ge -2$ .....(1).

Again

$\dfrac{1}{\log_{10}(1-x)}$ is defined for

$\log_{10}(1-x)\ne 0$ and

$(1-x)\gt 0$ .

or,

$x\ne 0$ and

$x\lt 1$ .

So the domain for

$f(x)$ is

$[-2,0)\cap(0,1)$ .

Find the domains of definition of the following functions.
$\displaystyle y \, = \, \sqrt{\frac{3^x \, - \, 4^x}{2x^2 \, - \, x \, - \, 8}}.$

R.E.F image

Domain of function

$\displaystyle y \, = \, \sqrt{\frac{3^x \, - \, 4^x}{2x^2 \, - \, x \, - \, 8}}, y = \dfrac{\sqrt{p(x)}}{q(x)}$

$p(x)\geq 0$

$3^{x}\geq 4^{x}$

$(\dfrac{3}{4})^{x}\geq 1$

use

$a^{x}\geq (a)^{0}$

$a< 1$

$x\leq 0$

$(\dfrac{3}{4})^{x}\geq (\dfrac{3}{4}^{0})$

$x\leq 0$

$(-\infty ,0)...(i)$

$2(x)> 0$

$2x^{2}-x-8> 0$

$(x-a)(x-B)> 0$

$a = \dfrac{1-\sqrt{65}}{4}, B = \dfrac{1+\sqrt{65}}{4}$

$x\epsilon \left(-\infty ,\dfrac{1-\sqrt{65}}{4},\right ) \cup \left(\dfrac{1+\sqrt{65}}{4},\infty \right)...(ii)$

Using intersection of both domain

$x\epsilon \left(-\infty ,\dfrac{1-\sqrt{65}}{4}\right)$

1127081_886659_ans_1cdcb267e23f4044aa781938216c0fad.jpg

If $f(x)=\dfrac {2x+3}{3x-2}$
Prove that $fof$ is an identity function

$\begin{array} { *{ 20 }{ l } }{ f\left( x \right) =\dfrac { { 2x+3 } }{ { 3x-2 } } } \\ { f\circ f\left( x \right) =f\left( { f\left( x \right) } \right) =f\left( { \dfrac { { 2x+3 } }{ { 3x-2 } } } \right) } \\ { =\dfrac { { 2\left( { \dfrac { { 2x+3 } }{ { 3x-2 } } } \right) +3 } }{ { 3\left( { \dfrac { { 2x+3 } }{ { 3x-2 } } } \right) -2 } } } \\ { =\dfrac { { 2x+6+9x-6 } }{ { 6x+9-6x+4 } } } \\ { =\dfrac { { 13x } }{ { 13 } } } \\ { =x } \\ { f\circ f\left( x \right) =x } \end{array}$

$f\circ f$ is an identity function

Find the domain and range of :-
(i) $\sqrt x - 1$
(ii) $(x-1)$

$\sqrt(x)-1$

$x\epsilon[0,\infty)$ is the domain

range is

$[-1,\infty)$

ii)

$x-1$

domain is

$(-\infty,\infty)$

range is

$(-\infty,\infty)$

A relation R is defined on the set Z of integers as follows: (x, y) $\in R' x^2+y^2= 25$
Express R and $R^{-1}$ as the sets of ordered pairs and hence find their respective domains.

$y= \pm \sqrt{25 -x^2}$
where both x and y are integers.
Cleraly for

$x= 0,\pm 3, \pm 4, \pm 5$ , y will be integers as

$y = \pm 5, \pm 4, \pm 3, 0$

$\therefore R = {(0, \pm 5), (\pm 3, \pm 4), (\pm 4, \pm 3), (\pm 5, 0) }$

$R^{-1} ={ ( \pm 5, 0), ( \pm 4, \pm 3), (\pm 3, \pm 4), (0, \pm 5)}$
Each R and

$R^{-1}$ consists of 2+ 4 + 4 +2 = 12 ordered pairs.
Domain of R ={

$0, \pm 3, \pm 4, \pm 5$ } = domain of

$R^{-1}$ .

Find the domain and range of $f(x) = \sqrt{x - 5}$

$f(x)=\sqrt{x-5}$

For

$f(x)$ to be defined, the term under the Square-root should be greater than or equal to zero.

$x-5\geq 0$

$x\geq 5$

So, the domain is

$[5,\infty )$

Now, for

$x-5\geq 0$

$\sqrt{x-5}\geq 0$

$y\geq 0$

So, the range is

$[0,\infty )$

The domain of $f(x)=\sqrt{\sin ^{-1}(\log _{2}x)}$ is

Find the domain of

$\sin^{-1}\left[log_2\left(\cfrac{x}{2}\right)\right]$

$log_2\left(\cfrac{x}{2}\right)$ is defined for

$\cfrac{x}{2}>0$

$x>0$

Also domain of

$\sin^{-1}x$ is

$[-1,1]$

when

$x=1$ then

$log_2\left(\cfrac{x}{2}\right)$ becomes

$-1$

when

$x=4$ then

$log_2\left(\cfrac{x}{2}\right)$ becomes

$1$

so, domain

$[1,4]$

Find the domain of the function:
$f(x)=\dfrac{1}{\sqrt{x^2-4x}}$ .

The function will be defined if

$x^2-4x>0$

or,

$x(x-4)>0$

or,

$x>0$ and

$x>4$ or,

$x<0$ and

$x<4$

or,

$x\in (4,\infty)$ or,

$x\in (-\infty,0)$ .

Domain is

$(-\infty,0)\cup(4,\infty)$ .

Find the domain of the function $f(x)=\dfrac{x^2}{\sqrt{(a-x)(x-b)}}$ , $b>a$ .

Given the function is

$f(x)=\dfrac{x^2}{\sqrt{(a-x)(x-b)}}$ with

$b>a$ .

Now the function

$f(x)$ will be defined if

$(a-x)(x-b)>0$

or,

$(x-a)(x-b)<0$

or,

$x\in (a,b)$ .

So domain is

$(a,b)$ .

Solve:
$x\sqrt {1 + {x^2}} + \log \left( {x + \sqrt {{x^2} + 1} } \right).$

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

$x+\sqrt{1+x^{2}}> 0$

$\Rightarrow x\geq 0$ ___(1)

$x\sqrt{1+x^{2}}+log (\frac{(x+\sqrt{1+x^{2}})(x+\sqrt{1+x^{2}})}{(x-\sqrt{1+x^{2}})})$

$\Rightarrow x\sqrt{1+x^{2}}+log(\frac{x^{2}-1-x^{2}}{x\sqrt{1+x^{2}}})$

$\Rightarrow x\sqrt{1+x^{2}}+log(\frac{1}{\sqrt{1+x^{2}}-x})$

$\frac{1}{\sqrt{1+x^{2}}-x}> 0$

$\Rightarrow \sqrt{1+x^{2}}-x> 0$

$\sqrt{1+x^{2}}> x$

$\Rightarrow 1+x^{2}> x^{2}$

$\rightarrow$ possible for all

$x\epsilon R$ ___(2)

$\therefore$ From (1) and (2)

$x\epsilon [0,\infty ]$

1112230_1189902_ans_53270268b9fe4c2cabceff429689e3c5.jpeg

Find the domain of the following

(i) $f\left( x \right) = \sqrt {\left[ {{x^2}} \right] - 1} + \sqrt {3 - \left[ {{x^2}} \right]}$

$\sqrt { \left[ x ^ { 2 } \right] - 1 } + \sqrt { 3 - \left[ x ^ { 2 } \right] }$

$\Rightarrow \quad \left[ x ^ { 2 } \right] - 1\geq 0 + 3 - \left[ x ^ { 2 } \right] \geq 0$

$\Rightarrow x \in ( - \infty , - 1 ] \cup [ 1,0 )$

$f \quad x \in [ - \sqrt { 3 } , \sqrt { 3 } ]$

$\Rightarrow x \in [ \sqrt { 3 } , \infty )$

Find the domain of function $\sqrt{25-x^2}$

The function is defined only when

$25-x^2\geq0\\x^2\leq25\\|x|\leq 5\\x \in [-5,5]$

Find the domain of the following real functions :
(i) $f\left( x \right)=\sqrt{16-x^2}$

The function is defined only when

$16-x^2\geq0\\x^2\leq16\\|x|\leq 4\\x \in [-4,4]$

The domain of Function $\sqrt{9-x^2}$

The function is defined only when

$9-x^2\geq0\\x^2\leq9\\|x|\leq 3\\x \in [-3,3]$

The value of $f(3)$ if $f(x)=\dfrac1{x^2-x}$

$f(x)=\dfrac1{x^2-x}\\f(3)=\dfrac1{3^2-3}=\dfrac16$

Find the Domain of the following function:
$y=\dfrac{x}{\sqrt{1+x}}$

The function

$f(x)=\dfrac{x}{\sqrt{1+x}}$ is defined only when

$1+x>0\\x>-1\\x \in (-1,0)$

Find the domain for the following functions :
(a) $y=1-\log _{ 10 }{ x }$ (b) $y=\frac { 1 }{ \sqrt { { x }^{ 2 }-4x } }$

$\mathrm{(a)}y=1-\log_{10}x\implies x>0$

So the domain is

$x\in (0,\infty)$

$\mathrm{(b)} y=\dfrac{1}{\sqrt{x^2-4{x}}}\implies x^2-4{x}>0$

$\implies x(x-4)>0\implies x\in (-\infty,0)\cup (4,\infty)$

Find the domain of $y=\dfrac{\sqrt{x-25}} {x+5}$

The function

$y=\dfrac{\sqrt{x-25}}{x+5}$ is defined only when

$x+5\neq0\\x\neq-5$ and

$x-25\geq 0\\x\geq25$

So the domain of function is

$x\in[25,\infty)$

Find the domain of the function $f(x)=\dfrac{1}{\sqrt{2x^2+5x+2}}$ .

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

Therefore,

$2x^2+5x+2 >0$

$(2x+1)(x+2) >0$

$2x+1>0$ and

$x+2>0$

$2x+1<0$ and

$x+2<0$

$x<-2$ or

$x>\dfrac{-1}{2}$

$x \in (-\infty, -2) \cup (\dfrac{-1}{2}, \infty)$ .

Find the domain and range of the real function $f(x)=\frac{1}{1-x^{2}}$

The function is not defined when

$1-x^2=0\\1=x^2\\x=\sqrt 1\\x=\pm 1$

So the domain of function is

$R-\{\pm 1\}$

Let

$\dfrac {1}{1-x^2}=y\\1=y-yx^2\\yx^2=y-1\\x^2=\dfrac {y-1}{y}\\x=\sqrt {\dfrac{y-1}{y}}$

So the Range is

$R-[0,1)$

Find the domain of
$f(x)=\sqrt { 4 - 9 x ^ { 2 } }$

Given

$f(x)=\sqrt{4-9 x^2}$

$\implies 4-9 x^2\ge 0$

$\implies 9 x^2-4\le 0$

$\implies ( x-\frac{3}{2})(x+\frac{3}{2})\le 0$

$\implies x\in [-3/2,3/2]$

The domain of

$f(x)$ is

$[-1.5,1.5]$

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

The function

$f(x)=\dfrac{1}{1+x^2}$ is only defined when

$1+x^2\neq0\\x^2\neq-1\\x\in R$

Let

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

Find the domain of $f(x)=\sqrt{9-x^{2}}$

Given

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

$\implies 9-x^2 \ge 0$

$\implies x^2-9\le 0$

$\implies x\in [-3,3]$

The domain of

$f(x)$ is

$[-3,3]$

Find $f(5)$ if $f(x)=x^2+3x+2$

$f(x)=x^2+3x+2\\f(5)=5^2+3(5)+2\\f(5)=25+15+2=42$

Find the domain and the range of the following function:
$f(x) =\dfrac{x-3}{2x+1}$

The function

$f(x)=\dfrac{x-3}{2x+1}$ is defined only when

$2x+1\neq0\\x\neq\dfrac{-1}{2}$

So domain is

$R-\left\{\dfrac{-1}{2}\right\}$

Let

$y=\dfrac{x-3}{2x+1}\\(2x+1)y=x-3\\2xy+y=x-3\\y+3=x(1-2y)\\x=\dfrac{y+3}{1-2y}$

This function is defined only when

$1-2y\neq0\\y\neq \dfrac{1}{2}$

So the range is

$R-\left\{\dfrac{1}{2}\right\}$

The domain of function $\sqrt{x^2-4}$ is

the function is

$\sqrt{x^2-4}$

The function is defined only when

$x^2-4\geq 0\\x^2\geq4\\|x|\geq 2\\x\epsilon \infty- [-2,2]$

Find the domain and the inverse of the following functions:
$f(x)=\dfrac{x^{2}}{1+x^{2}}$

The function

$f(x)=\dfrac{x^2}{1+x^2}$ is only defined when

$1+x^2\neq0\\x^2\neq-1\\x\in R$

Let

$y=\dfrac{x^2}{1+x^2}\\x^2=y(1+x^2)\\x^2=y+x^2y\\x^2-x^2y=y\\x^2(1-y)=y\\x^2=\dfrac{y}{1-y}\\x=\sqrt{\dfrac{y}{1-y}}\\implies f^{-1}(x)=\sqrt{\dfrac{x}{1-x}}$

Find the domain & range of the following function.
$f\left(x\right)=\dfrac{{x}^{2}-3x+2}{{x}^{2}+x-6}$

1306962_1481985_ans_066ac8f511044cb699bba5670ee06d94.jpeg

Find the domain of each of the following real value functions real variable:
$f(x)=\dfrac {1}{\sqrt {x^2 -1}}$

1417239_1664675_ans_7d3f781c36f641e3a9ed92757ed46794.jpg

Find the domain of each of the following real valued functions of real variable:
$f(x)=\dfrac {3x-2}{x+1}$

1417232_1664659_ans_f22da4d31ed04aec978cf0bddd270ffe.jpg

Find the domain of each of the following real valued functions of real variable:
$f(x)=\dfrac {1}{x}$

1417198_1664656_ans_4cab6b16db91431889a09610ed85d566.jpg

Find the domain of each of the following real value functions real variable:
$f(x)=\sqrt {x-2}$

1417237_1664674_ans_c4591978840945468bd09b5bf6273301.jpg

Find the domain of:

$\sec^{-1} x - \tan^{-1} x$ .

We know that the domain of

$\sec^{-1}{x}$ is

$\left(-\infty,-1\right]\cup\left[1,\infty\right)$

the domain of

$\tan^{-1}{x}$ is

$\left(-\infty,\infty\right)$

$\therefore\,$ Domain of

${\sec}^{-1}{x}-{\tan}^{-1}{x}$ is intersection of both result.

$=\left(-\infty,-1\right]\cup\left[1,\infty\right)$

Find the domain of each of the following real valued functions of real variable:
$f(x)=\dfrac {2x+1}{x^2 -9}$

1417234_1664668_ans_ab89c60ab5154de8bece5c8cee40f139.jpg

Find the domain of each of the following real valued functions of real variable:
$f(x)=\dfrac {1}{x-7}$

1417199_1664658_ans_2f2a5207fd7049b9b14a6e38258bc1eb.jpg

Given $A=\left\{2,3,4 \right\}, B=\left\{2,5,6,7 \right\}$ . Construct an example of each of the following.
a mapping from $A$ to $B$ which is not injective.

Let

$g:A\to B$ denote a mapping such that

$g=\left\{(2,2),(3,5),(4,5)\right\}$ which is not an injective mapping.

Given $A=\left\{2,3,4 \right\}, B=\left\{2,5,6,7 \right\}$ . Construct an example of each of the following.
a mapping from $A$ to $B$ .

Let

$h:B\to A$ denote a mapping such that

$h=\left\{(2,2),(5,3),(6,4),(7,4)\right\}$ which is mapping from

$A$ to

$B$ .

If the mappings, $f$ and $g$ are given by $f=\left\{(1,2),(3,5),(4,1)\right\}$ and $g=\left\{(2,3),(5,1),(1,3)\right\}$ .

Given that

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

And

$g=\left\{(2,3),(5,1),(1,3)\right\}$ .

Now,

$fog(2)=f\left\{g(2)\right\} =f(3)=5$

$fog(5)=f\left\{g(5)\right\}=f(1)=2$

$fog (1)=f\left\{g(1)\right\}=f(3)=5$

$fog =\left\{(2,5),(5,2),(1,5)\right\}$

Give an example of a map
which is not one-one but onto.

Let

$f:N\to N$ given by

$f(1)=f(2)=1$ and

$f(x)=x-1$ for every

$x > 2$ is onto but not one-one

$f$ is not one-one as

$f(1)=f(2)=1$ . But

$f$ is onto.

Give an example of a map
which is one-one but not onto.

Let

$f:N\to N$ be mapping define by

$f(x)=2x$

which is one-one

For

$f(x)=f(x_2)$

$\Rightarrow \ 2x_1 =2x_2$

Further

$f$ is not onto as for

$1\ \in N$ , there does not exist aby

$x$ in

$N$ such that

$f(x)=2x+1$

Give an example of a map
which is neither one-one but nor onto.

The mapping

$f:R\to R$ define as

$f(x)==x^2$ , is neither one-one not onto.

Given $A=\left\{2,3,4 \right\}, B=\left\{2,5,6,7 \right\}$ . Construct an example of each of the following.
an injective mapping from $A$ to $B$ .

Let

$f:A\to B$ donate a mapping

$f=\left\{(x,y); y=x+3\right\}$

i.e,

$f=\left\{(2,5),(3,-6),(4,7)\right\} f=\left\{(2,5),(3,6),(4,7)\right\}$ which is an injective mapping.

If domain for $y=\cos^{-1}\left (\dfrac {1-2|x|}{3}\right )+\log_{|x-1|}x$ is $x\in (a, b)\cup (c, d)$ . Find $a+b+c+d$

Here,

$\cos^{-1}\left (\dfrac {1-2|x|}{3}\right )$ exists, if

$-1\leq \dfrac {1-2|x|}{3} \leq 1$

$\Rightarrow -3 \leq 1 - 2 |x| \leq 3\Rightarrow -4 \leq -2 |x| \leq 2$

$\Rightarrow 2\geq |x| \geq -1\Rightarrow -2 \leq x \leq 2$ .....(i)

Also,

$\log_{|x-1|}x$ exists, if

$x > 0$ and

$|x-1| > 0$

$x > 0$ and

$x\in R-\left \{1\right \}$ ....(ii)

Hence, from Eqs. (i) and (ii), we get

$x\in (0, 1)\cup (1, 2)$

$\Rightarrow a+b+c+d=4$

Let $\displaystyle y= \sqrt{\frac{\left ( x+1 \right )\left ( x-3 \right )}{x-2}}.$
Find all the real values of x for which y takes real values.

$\displaystyle y= \sqrt{\frac{\left ( x+1 \right )\left ( x-3 \right )}{x-2}}.$

For

$y$ to be real, the term under the square root should be greater or equal to 0

$\displaystyle \frac { \left( x+1 \right) \left( x-3 \right) }{ x-2 } \ge 0$

Case I: For this, the numerator should be greater than equal to 0 and denominator should be greater than 0

i.e

$(x+1)(x-3) \ge 0$ and

$x-2 >0$

$\Rightarrow x\in (-\infty,-1)\cup [3,\infty]$ and

$x>2$

So, the solution set is

$[3,\infty)$

Case II: For this, the numerator should be less than equal to 0 and denominator should be less than 0

i.e

$(x+1)(x-3) \le 0$ and

$x-2 <0$

$\Rightarrow x\in [-1,3]$ and

$x<2$

So, the solution set is

$[-1,2)$

Hence, the domain is

$[-1,2) \cup [3,\infty)$

Find the number of all onto functions from the set $\left \{1, 2, 3, ...., n\right \}$ to itself

Taking set

$\{ 1,2,3\}$

Since

$f$ is onto,all elements of

$\{ 1,2,3\}$ have unique pre-image.

Total no. of one-one function

$=3\times 2\times 1=6$

eg:-Since

$f$ is onto,all elements of

$\{ 1,2,3\}$ have unique pre-image.

total no. of onto functions

$=n\times n-1\times n-2\times 2\times 1 =n!$

1036396_458708_ans_e60d2495fb1245c7995a426cc9735767.png

Find the domain of the real valued square root function $f$ given below
$f(x)=\sqrt{(2x)}$

$f\left( x \right) =\sqrt { 2x }$
As

$\sqrt{2x} >0$

Where

$x>0$

Its domain is

$(0,\infty)$

If the domain of the real valued function $f$ given below
$f(x)=\sqrt{\dfrac{-x+1}{x+3}}$ is $(-\infty,-x)\cup(-x,1]$ .Find $x$

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

Domain of

$f(x)=[-\infty,x)\cup (x,1]$

$f(x)=\sqrt{-R}$ and

$\sqrt{R}_{0}$ is not possible , whre

$R$ is the set of real numbers.

So,

$x+3$ should not be zero

$x+3=0$

$x=-3$

$\sqrt{-x+1}$ should not be

$\sqrt{-R}$

So, x should be less or equal to 1 and less than -3 and greater than -3 upto 1.

Domain of

$f(x)=[-\infty ,x)\cup(-x,1]$

So,

$x=-3$

Find the domain of the real valued rational function $f$ given below
$f(x)=1/({x}^{3}+{x}^{2}-2x)$

for domain, we have to find out the values at which function is defined

${ x }^{ 3 }+{ x }^{ 2 }-2x\quad \neq 0\\ x\neq -2,0,1$

$x\in (-\infty ,-2)\cup (-2,0)\cup (0,1)\cup (1,\infty )$

If the domain of $f(x)=\sqrt{2x+1}$ is $\Big[ \dfrac{-1}{m},\infty\Big)$ .Find $m$

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

$2x+1\geq 0$

$2x\geq -1$

$x\geq \dfrac{-1}{2}$

Domain of

$f(x)=[\cfrac{-1}{2},\infty)$

So,

$m=2$

The domain of the real valued sine function $f(x)=2\sin{(x-1)}$ is $(-\infty,\infty)$ . If true enter 1 else0.

Domain of

$x-1$ is all real numbers i.e.

$R$

And we know that domain of

$\sin f(x)$ is same as domain of

$f(x)$

Hence domain of

$f(x)=2\sin(x-1)$ is

$(-\infty, \infty)$

Find the domain of the real valued exponential function $f$ given below
$f(x)={e}^{(x-4)}$

$f(x)=e^{(x-4)}$

Domain of

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

Because exponential function has both positive and negative integers.

Show that, if $f : R - \left \{ \dfrac{7}{5} \right \}\rightarrow R -\left \{ \dfrac{3}{5} \right \}$ is defined by f (x) = $\dfrac{3x + 4}{5x - 7}$ and $g : R -\left \{ \dfrac{3}{5} \right \}\rightarrow R -\left \{ \dfrac{7}{5} \right \}$ is defined by g(x) = $\dfrac{7x + 4}{5x - 3}$ ,
then log = $l_A$ and gof = $l_B$ , where $l_A$ and $l_B$ are called identity functions on sets A and B.

$f(x)=\dfrac{3x+4}{5x-7}, g(x)=\dfrac{7x+4}{5x-3}$

Note Heat

$f(x)$ and

$g(x)$ are both one one and onto in the given domain and co-domain hence both are unvertible

$fog(x)=g(g(x))=\dfrac{3\left(\dfrac{7x+4}{5x-3}\right)+4}{5\left(\dfrac{7x+4}{5x-3}\right)-7}=\dfrac{21x+12+20x-12}{35x+20-35x+21}=\dfrac{41x}{41}=x=I_{A}$

$gof (x)=g(f(x))=\dfrac{7\left(\dfrac{3x+4}{5x-7}\right)+4}{5\left(\dfrac{3x+4}{5x-7}\right)-3}=\dfrac{21x+28+20x-28}{15x+20-15x+21}=\dfrac{41x}{41}=x=I_{B}$

Hence proved

Find the domains of definition of the following functions.
$\displaystyle y \, = \, \sqrt{\sin \, x} \, + \, \sqrt{16 \, - \, x^2}.$

Domain of

$y=\sqrt{\sin{x}}+\sqrt{16-x^2}$

$\sin{x}$ and

$16-x^2$ is under root it should be greater than zero.

Thus

$\sin{x}\gt0$ and

$16-x^2\gt0$

$\sin{x}$ is greater than

$0$ when

$x \epsilon [0,\pi]$ and

$x^2\lt16$

thus

$x\epsilon[0,\pi]$ and

$x\lt4$

Thus the domain of

$y$ is

$x\epsilon[0,1]$

Let $A$ be a set of $n$ distinct elements. Then, the total number of distinct functions from $A$ to $A$ is .... and out of these .... are onto functions.

A is set of n different elements.

$A$ ={ ${x_1,x_2,x_3,.............x_n}$ }

For one element there are n functions possible.

So for n elements ,

No. of functions possible $=n⋅n⋅n⋅n………..n$ [n times]

∴ number of functions from A to A $=n^n$

Onto functions: when every element of the image set has an inverse image in the first set.

If ,for $x_1$ there are n functions possible.

For $x_2$ there are (n-1) functions possible

Similarly for $x_3,x_4,.......x_n$

Number of Onto functions are $=n\times(n-1) \times (n-2).............1=n!$

∴ number of Onto functions from $A$ to $A =n!$

Let $f:D\rightarrow R$ , where $D$ is the domain of $f$ . Find the inverse of $f$ , if it exists
$f\left( x \right) ={ 1-2 }^{ -x }$
$f\left( x \right) ={ \left( 4-{ \left( x-7 \right) }^{ 3 } \right) }^{ 1/5 }$
$f\left( x \right) ={ \ell n\left( x+\sqrt { { 1+x }^{ 2 } } \right) }$

$(1)$

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

$y=1-2^{-x}$

for

$f^{-1}(x)$ , Put

$x=y$ and

$y=x$

$x=1-2^y$

$-x+1=2^y$

$-y=\log_2(1-x)$

$y=\log_2(1/(1-x))=f^{-1}(x)$

$(2)$

$y=(4-(x-7)^3)^{1/5}$

interchange

$x$ and

$y$

$x=(4-(y-7)^3)^{1/5}$

$(y-7)^3=4-x^5$

$y=7+(4-x^5)^{1/3}=f^{-1}(x)$

$(3)$

$y=\ln (x+\sqrt{(1+x)}$

interchange

$x$ and

$y$

$x=\ln (y+\sqrt{(1+y)}$

$e^x=y+\sqrt{(1+y^2}$

$(e^x-y)^2=1+y^2$

$y=\cfrac{e^{2x}-1}{2e^x}=f^{-1}(x)$

Find the domains of definition of the following functions.
$\displaystyle y \, = \, \frac{x}{\sqrt{x^2 \, - \, 5x \, + \, 6}}.$

R.E.F image

Find domains of functions

$y = \dfrac{x}{\sqrt{x^{2}-5x+6}}$ for

$\dfrac{1}{\sqrt{y(x)}},y(x)> 0$

$x^{2}-5x+6> 0$

$(x-2) (x-3) >0$

$x\epsilon (-\infty ,2)\cup _{1}(3,\infty )$

domain of function

$x\epsilon (-\infty ,2)\cup (3,\infty )$

1126415_886662_ans_1ca3304071264cf0bb326f2e7055e355.jpg

Let $f:\mathbb{N}\to \mathbb{N}$ be defined by $f(n)=\begin{cases}\dfrac{n+1}{2}, &\mbox{if}\ n\ \mbox{is odd}\\ \dfrac{n}{2}, &\mbox{if}\ n\ \mbox{is even}\end{cases}$ . $n\in \mathbb{N}$
State whether the function $f$ is bijective. Justify your answer.

For

$n$ being odd,

let

$n=2K+1$

where

$K$ is odd.

now,

$\cfrac{n+1}{2}=K+1$ is even.

And when

$K$ is even.

$\cfrac{n+1}{2}=K+1$ is odd.

Also

$n=$ even

$=2K$

for

$K=$ even.

$\cfrac{n}{2}=K+1$ is even.

for

$K=$ odd.

$\cfrac{n}{2}=K+1$ is odd.

Thus every value of set

$N$ is covered in range as well as domain.

$f(x)$ is bijective

If $:R\rightarrow R$ defined by $f(x)=1+{ x }^{ 2 }$ , then show that $f$ is neither $1-1$ nor onto

Let

${x}_{1}=1$ ,

${x}_{2}=-1$ (any type of this example)

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

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

$f({ x }_{ 1 })=f({ x }_{ 2 })\quad$
but

${ x }_{ 1 }\neq { x }_{ 2 }$

$\therefore$

$f$ is not

$1-1$
To show

$f$ is not onto
Let

$-11\in R$ (co domaine) no

$x$ in

$R$ (domaine)
Such that

$f(x)=-11$
hence Range of

$f\ne$ co domain of

$f$

$\therefore$

$f$ is not on to.

Find the domain of $f\left( x \right) = \sqrt {\left( {\left[ x \right] - 1} \right)} + \sqrt {\left( {4 - \left[ x \right]} \right)}$
(where[] represents the greater integer function)

$f(x)=\sqrt { [x]-1 } +\sqrt { 4-[x] } \\ [x]-1\ge 0\\ [x]\ge 1\\ x\epsilon [1,\infty )\quad -(i)\\ 4-[x]\ge 0\\ [x]\le 4\\ x<5\quad -(ii)\\ from(i)\& (ii)\\ x\epsilon [1,5)$

show that $f : A \rightarrow B$ and $g : B \rightarrow C$ are onto, then $gof : A \rightarrow$ is also onto:

Since

$g:B \rightarrow C$ is on-to.

Let

$a\epsilon C$ then there will a pre-image of

$a$ in

$B$ .

Let

$P$ is the preimage.

$\therefore$ we can say

$g(P)=a$ .

Similarly,

$f:A\rightarrow B$ is onto.

$\because P\epsilon B$ ,then there is a pre-image in

$A$ .

Let

$x$ be the pre-image of

$P$

$\therefore f\left( x \right)=P\\ \therefore gof=g\left( f\left( x \right) \right)\\ gof=g\left( P \right)\\ gof=a$

So,for every

$x$ in

$A$ there is an image in

$C$ .

$gof$ is an onto function.

Let $f: [ \alpha, \infty) \rightarrow [x, \infty), f(x) = x^2 -6x +5$ then find three possible value of $'\alpha'$ if $f(x)$ is onto.

$f\left( x \right)={ x }^{ 2 }-6x+5$

By graph we can say

$x\epsilon [3,\infty )$ then,

$f\left( x \right) \epsilon [-4,\infty )$

when

$x\epsilon [5,\infty )$ then

$f\left( x \right) \epsilon [0,\infty )$

$\therefore$ Possible value of

$\alpha$ only

$3\&5$

1039471_1015801_ans_f69fa40f73b94d40a765382fd62ac2bc.png

Let $f(x) = \sqrt{2 -x -x^{2}}$ and $g(x) = cos x$ . Check given statement is true or not?
I. Domain of $f((g(x))^{2})$ = Domain of $f(g(x))$

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

$g\left(x\right)=(\cos x)$

Now,

$y_{1}=f\left(g\left(x\right)\right)^{2}=f\left(cos^{2}x\right)$

$y_{1}=\sqrt{2-\cos^{2}x-\cos^{4}x}$

For

$y_{1}$ to exist,

$2-\cos^{2}x-\cos^{4}x\ge 0$

$\Rightarrow(x-0)(x-\pi)\le 0$

$\Rightarrow x\in[0,\pi]$

Now,

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

$=f\left(\cos x\right)$

$=\sqrt{2-\cos x-\cos^{2}}$

For existance of

$y_{2},$

$2-\cos x-\cos^{2}x\ge0$

$\Rightarrow \cos^{2}x+\cos x-2\le 0$

$\Rightarrow(\cos x-1)\le0$

$\Rightarrow x\in R$ (as

$\cos x\in [-1,1]\forall x\in R$ )

So,

$D(y_{1}\neq D(y_{2})$

This solution is false.

Find the domain of $f(x)=\sqrt{\sin x}+\sqrt{16-x^2}$ .

$f{\left( x \right)} = \sqrt{\sin{x}} + \sqrt{16 - {x}^{2}}$

Let

$f{\left( x \right)} = {f}_{1}{\left( x \right)} + {f}_{2}{\left( x \right)}$

Here,

${f}_{1}{\left( x \right)} = \sqrt{\sin{x}}$

${f}_{2}{\left( x \right)} = \sqrt{16 - {x}^{2}}$

Now,

Domain of

${f}_{2}{\left( x \right)}$ , i.e.,

$\sqrt{16 - {x}^{2}}$ -

$16 - {x}^{2} \ge 0$

$\Rightarrow {x}^{2} \le 16$

$\Rightarrow -4 \le x \le 4$

$\Rightarrow$ Domain of

${f}_{2}{\left( x \right)}$ is

$\left[ -4, 4 \right]$ .

Domain of

${f}_{1}{\left( x \right)}$ , i.e.,

$\sqrt{\sin{x}}$ -

$\sin{x} \ge 0$

$\Rightarrow 0 \le \sin{x} \le 1$

$\Rightarrow \sin^{-1}{0} \le x \le \sin^{-1}{1}$

$\Rightarrow 2n \pi \le x \le \left( 2n + 1 \right) \pi$

$\Rightarrow$ Domain of

${f}_{1}{\left( x \right)}$ is

$\left[ 2n \pi, \left( 2n + 1 \right) \pi \right]$

For

$n = 0 \Rightarrow 0 \le x \le \pi$

For

$n = 1 \Rightarrow 2 \pi \le x \le 3 \pi$

Now domain of the function

$f{\left( x \right)}$ will be-

Domain of

$f{\left( x \right)} \equiv$ Domein of

${f}_{1}{\left( x \right)} \cap$ Domain of

${f}_{2}{\left( x \right)}$

$\therefore$ Domain of

$f{\left( x \right)} \equiv -4 \le 0 \le x \le \pi \le 4$

$\Rightarrow$ Domain of

$f{\left( x \right)} \equiv 0 \le x \le \pi$

Find the domain of $f(x) = \log_{10}\log_{2}\log_{\pi/2}(\tan^{-1}x)^{-1}$

We must have

$\log_{2}\log_{\pi/2}(tan^{-1}x)^{-1} > 0$
or

$\log_{\pi/2}(tan^{-1}x)^{-1} > 1$
or

$0 < (tan^{-1} x)^{-1} < \frac{2}{\pi}$
or

$\frac{\pi}{2} < tan^{-1}x < \infty$ , which is not possible. Hence the domain is

$\phi$

Let $f: N \rightarrow Y$ be a function defined by $f(x) = 4x^2+ 12x+15$ , where $Y=$ range of $f.$ Show that $f$ is invertible and find the inverse of $f$ .

Solution:-

$f : N \rightarrow Y$

$f \left( x \right) = 4{x}^{2} + 12 x + 15$

A function is invertible if the function is one-one and onto.

Let

${x}_{1} , {x}_{2} \in N$ , such that

$f \left( {x}_{1} \right) = f \left( {x}_{2} \right)$

$4{{x}_{1}}^{2} + 12 {x}_{1} + 15 = 4{{x}_{2}}^{2} + 12 {x}_{2} + 15$

$\Rightarrow \; \left( {{x}_{1}}^{2} - {{x}_{2}}^{2} \right) + 3 \left( {x}_{1} - {x}_{2} \right) = 0$

$\Rightarrow \; \left( {x}_{1} - {x}_{2} \right) \left( {x}_{1} + {x}_{2} + 3 \right) = 0$

$\because \; {x}_{1}, {x}_{2} \in N$

$\Rightarrow \; {x}_{1} + {x}_{2} + 3 \ne 0$

$\therefore \; {x}_{1} = {x}_{2}$

Thus,

$f \left( x \right)$ is one-one.

The function is onto if there exist x in N such that

$f \left( x \right) = y$

$\therefore \; 4{x}^{2} + 12 x + 15 = y$

$\Rightarrow \; 4{x}^{2} + 12 x + \left( 15 - y \right) = 0$

Here,

$a = 4$

$b = 12$

$c = 15 - y$

From quadratic formula,

$x = \cfrac{-b \pm \sqrt{{b}^{2} - 4ac}}{2a}$ , we have

$x = \cfrac{-12 \pm \sqrt{{12}^{2} - 4\times 4 \times \left( 15 - y \right)}}{2 \times 4}$

$\Rightarrow \; x = \cfrac{-3 \pm \sqrt{y - 6}}{2}$

$\because \; x \in N$

$\therefore \; x = \cfrac{-3 + \sqrt{y - 6}}{2}$

${f}^{-1}{\left( x \right)} = \cfrac{-3 + \sqrt{x - 6}}{2}; \; x \ge 6$

If $f: R\rightarrow R$ be the function defined by $f(x) = 4x^3+7$ , then show that $f$ is a bijection.

A function is bijective if it is both one-one and onto.

The function is

$f: R\rightarrow R$ such that

$f(x) = 4x^3+7$

Onto. Let

$x_1, x_2 \in A$

$f(x_1)= f(x_2) \forall x_1, x_2\in R$

$\Rightarrow 4x_1^3+7 = 4x_2^3+7$

$\Rightarrow 4x_1^3= 4x_2^3\Rightarrow x_1^3- x_2^3=0$

$\Rightarrow (x_1-x_2)(x_1^2+ x_1x_2+ x_2^2)=0$ [Using

$a^3-b^3= (a-b) ( a^2+ ab +b^2)$ ]

$\Rightarrow (x_1- x_2)\left\{\left[ x_1+\dfrac{x_2}{2}\right]^2+\dfrac{3x_2^2}{4}\right\}=0$

$\Rightarrow x_1 -x_2=0,$ because

$\left[ x_1+\dfrac{x_2}{2} \right]^2+\dfrac{3x_2^2}{4} \neq 0$

Since

$f(x_1) = f(x_2) \Rightarrow x_1=x_2$ ( as proved above ), therefore

$f(x)$ is a one-one function.

960548_1041842_ans_5ea5d998bbd5450abe95fcd91bc68f26.png

Find the domain for which the function $f(x)=3{x}^{2}-2x$ and $g(x)=9x-6$ are equal.

Consider the given function.

$f\left( x \right)=3{{x}^{2}}-2x$

$g\left( x \right)=9x-6$

If $f\left( x \right)=g\left( x \right)$ , then

$3{{x}^{2}}-2x=9x-6$

$3{{x}^{2}}-11x+6=0$

$3{{x}^{2}}-2x-9x+6=0$

$x\left( 3x-2 \right)-3\left( 3x-2 \right)=0$

$\left( x-3 \right)\left( 3x-2 \right)=0$

$x=\dfrac{2}{3},3$

Therefore,

Required domain

$=\left\{ \dfrac{2}{3},3 \right\}$

Evaluate:
$\displaystyle \int \sin^2 x \cos^2 x dx$

$I=\displaystyle \int \sin^2 x \cos^2 x\ dx$

$I=\dfrac {1}{4} \displaystyle \int (2\sin x \cos x)^2 dx$

$I=\dfrac {1}{4} \displaystyle \int (\sin 2x)^2 dx$

$I=\dfrac {1}{4}\displaystyle \int \sin^2 2x\ dx$

we know that

$\cos 4x=\cos 2 \cdot 2x=1-2\sin^2 x$

$\sin^2 2x=\dfrac {1-\cos 4x}{2}$

Therefore,

$I=\displaystyle \int \dfrac {1}{4} \left (\dfrac {1-\cos 4x}{2}\right)dx$

$I=\dfrac {1}{8} \left [\displaystyle \int (1-\cos 4s) dx \right]$

$I=\dfrac {1}{8} \left [x-\dfrac {\sin 4x}{4}\right]+C$

Simplify $\dfrac{1}{1+a^{m-n}}+\dfrac{1}{1+a^{n-m}}$

We have,

$\dfrac{1}{1+a^{m-n}}+\dfrac{1}{1+a^{n-m}}$

$=\dfrac{1}{1+\dfrac{a^m}{a^n}}+\dfrac{1}{1+\dfrac{a^n}{a^m}}$

$=\dfrac{a^n}{a^m+a^n}+\dfrac{a^m}{a^m+a^m}$

$=\dfrac{a^n+a^m}{a^m+a^n}=1$

Hence, this is the answer.

Find the domain of
$f\left( x \right) = \sqrt {\left( {\left[ x \right] - 1} \right)} + \sqrt {\left( {4 - \left[ x \right]} \right)}$
(where [ ] represents the greatest integer function).

$\begin{array}{l}\left[ x \right] - 1 \ge 0\\4 - \left[ x \right] \ge 0\\ = 1 \le \left[ x \right] \le 4\\ = 1 \le x < 5\\x \in \left( {1,5} \right)\end{array}$

Find the domain of $f(x) = \sqrt{x + 2} + {log_{10}}(1 - x)$

$f(x) = \sqrt{x+2}+log_{10}(1-x)$

Domain of

$\sqrt{x+2}$

$x+2> 0$

$x> -2 ...(1)$

Domain of

$log_{10}(1-x)$

$1-x> 0$

$x< 1 ...(2)$

From the conditions (1) and (2), we

can say that the Domain of

$f(x)$ is

$x\epsilon (-2,1)$

1175941_1070980_ans_83b6eb6aec524d70bdcf8cebc3004f46.jpg

The domain of definition of the function $y(x)$ given by $2^x+2^y=2$ is

For domain of

$x$ let's assume

$y$ as a function of

$x:$

so,

$2-2^x>0$

$x^2<2$

$x<1;$

Therefore solution is set form we have

$x\epsilon(-\infty,1)$

If $f:R\rightarrow S,defined\ by\ f\left( x \right) =\sin { x } -\sqrt { 3 } \cos { x } +1$ , is onto, then find the set $S$ ?

$f\left( x \right) =sinx\sqrt { 3cosx } +1$

$2\left( \dfrac { 1 }{ 2 } sinx-\dfrac { 1 }{ 2 } \sqrt { 3 } cosx \right) +1$

$2\left( sin\left( x-\dfrac { \pi }{ 3 } \right) \right) +1$

$-1\le sinx-\dfrac { \pi }{ 3 } \le 1$

$-2\le 2sinx-\dfrac { \pi }{ 3 } \le 2$

$-1\le 2sinx-\dfrac { \pi }{ 3 } +1\le 3$

$-1,3$

Domain of $f\left( x \right) = \dfrac{{{x^2} + 3x + 5}}{{{x^2} - 5x + 4}}$ is

To find domain.

equate denominator

$x^2-5x+4$ to zero. and find the value of x.

$x^2-5x+4=0$

$x^2-4x-x+4=0$

$x(x-4)-1(x-4)=0$

$(x-1)(x-4)=0$

$x=1,4$

This means that, when

x =

1 or 4, we have the

$x^2-5x+4$

Implying that

$f\left( x \right) =\ \dfrac { { x }^{ 2 }+3x+5 }{ 0 }$ which is undifined .

so , Domain of function =

$R-$ {1,4}.

Show that $f: N→N$ = $x-1$ , if $x$ is even
is both one-one and onto

The question is

Show that

$f:N\rightarrow\,N$ , given by

$f\left(x\right)=\begin{cases} x+1 ;if\,x\,is\,odd \\ x-1; if\,x\,is\,even \end{cases}$

For one-one

Case:

$I$ When

${x}_{1},{x}_{2}$ are odd natural number.

$\therefore\,f\left({x}_{1}\right)=f\left({x}_{2}\right)$

$\Rightarrow\,{x}_{1}+1={x}_{2}+1\,\,\,for\,all\,{x}_{1},{x}_{2}\in\,N$

$\Rightarrow\,{x}_{1}={x}_{2}$

$\therefore\,f$ is one-one.

Case:

$II:$ When

${x}_{1},{x}_{2}$ are even natural number.

$\therefore\,f\left({x}_{1}\right)=f\left({x}_{2}\right)$

$\Rightarrow\,{x}_{1}-1={x}_{2}-1$

$\Rightarrow\,{x}_{1}={x}_{2}$

$\therefore\,f$ is one-one

Case

$III:$ When

${x}_{1}$ is odd and

${x}_{2}$ is an even number.

$f\left({x}_{1}\right)=f\left({x}_{2}\right)\Rightarrow\,{x}_{1}+1={x}_{2}-1$

$\Rightarrow\,{x}_{2}-{x}_{1}=2$

which is never possible as the difference of odd and even number is always an odd number.

Hence in this case

$f\left({x}_{1}\right)\neq\,f\left({x}_{2}\right)$

Thus,

$f$ is one-one.

Case

$IV:$ When

${x}_{1}$ is even and

${x}_{2}$ is odd natural number.

$f\left({x}_{1}\right)=f\left({x}_{2}\right)\Rightarrow\,{x}_{1}-1={x}_{2}+1$

$\Rightarrow\,{x}_{1}-{x}_{2}=2$

which is never possible as the difference of odd and even number is always an odd number.

Hence in this case

$f\left({x}_{1}\right)\neq\,f\left({x}_{2}\right)$

Thus,

$f$ is one-one.

For onto:

$\therefore\,f\left(x\right)=x+1$ if

$x$ is odd

$=x-1$ if

$x$ is even.

$\Rightarrow\,$ For every even number

$y$ of co-domain there exists an odd number

$y-1$ in domain and for every odd number

$y$ of co-domain there exists evert number

$y+1$ in domain.

Thus,

$f$ is an onto fucntion.

Hence

$f$ is one-one onto function.

Let $f:R \to R$ be defined by i) $f\left( x \right) = x + 1$ ii) $f\left( x \right) = x + \left| x \right|$ . Determine whether or not $f$ is onto.

Onto means in

$f:R$ that range is

$R$

$f(x)=x+1$

learly x can take any value and range is

$R$

Range is

$R$

Function is onto

(ii)

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

$fx<0$

else

$x>0$

$f(x)=2x$

Range is

$[0,\infty)$

Not onto function

The domain of $\frac{1}{\sqrt{x-x^{2}}}+\sqrt{3x-1-2x^{2}}$ is:

Consider the problem

$\begin{array}{l} x-{ x^{ 2 } }=0 \\ \Rightarrow x\left( { 1-x } \right) =0 \\ \Rightarrow x=0,1 \end{array}$

And

$\begin{array}{l} 3x-1-2{ x^{ 2 } }=0 \\ \Rightarrow -2{ x^{ 2 } }+3x-1=0 \\ \Rightarrow 2{ x^{ 2 } }-3x+1=0 \\ \Rightarrow 2{ x^{ 2 } }-2x-x+1=0 \\ \Rightarrow 2x\left( { x-1 } \right) -1\left( { x-1 } \right) =0 \\ \Rightarrow \left( { x-1 } \right) \left( { 2x-1 } \right) =0 \\ \Rightarrow x=1,x=\frac { 1 }{ 2 } \end{array}$

$3x - 1 - 2{x^2} < 0\,when\,x \in \left( { - \infty ,2} \right) \cup \left( {1,\infty } \right)$

and

$x - {x^2} \le 0\,when\,x \in \left( { - \infty ,0} \right] \cup \left[ {1,\infty } \right)$

Therefore, Domain of given function is

$R - \left\{ {\left\{ {\left( { - \infty ,2} \right) \cup \left( {1,\infty } \right)} \right\} \cup \left\{ {\left( { - \infty ,0} \right] \cup \left[ {1,\infty } \right)} \right\}} \right\}$

The domain of the function $\dfrac{x}{\sqrt{x^2-3x+2}}$ is

Given,let

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

function is well defined when

$x^{2}-3x+2> 0$

$\Rightarrow (x-1)(x-2)> 0$

$\therefore x\epsilon (-\infty ,1) \bigcup (2,\infty )$

1073173_1183476_ans_825528170b43461d96f13b56b9661b1e.png

If $f(x + y) = f(xy) \forall x, y\epsilon R$ then prove that $f$ is a constant function.

Let

$x$ be any real and

$y = 0$ .

Then

$f(x+0) = f(x×0)$

$\Rightarrow f\left( x \right) = f\left( 0 \right) \forall x \in R$

Hence the function is a constant function as it is the property of constant function.

The domain of the function $\sqrt{x^{2}-5x+6}+\sqrt{2x+8-x^{2}}$ is

$\begin{array}{l} f\left( x \right) =\sqrt { { x^{ 2 } }-5x+6 } +\sqrt { 2x+8-{ x^{ 2 } } } \\ Domain\, of\, f\left( x \right) ={ x^{ 2 } }-5x+\ge 0\& 2x+8-{ x^{ 2 } }\ge 0 \\ =\left( { x-2 } \right) \left( { x-3 } \right) \ge 0\& { x^{ 2 } }-2x-8\le 0 \\ x\in (-\infty ,2]\cup [3,\infty )\& \left( { x-4 } \right) \left( { x+2 } \right) \le 0 \\ x\in (-\infty ,2]\cup [3,\infty )\& x\in \left[ { -2,4 } \right] \\ \Rightarrow \left\{ { -2 } \right\} \cup \left[ { 3,4 } \right] \end{array}$

The domain of $f(x)=\dfrac{x^2-8}{x-2}$

$f(x)=\dfrac{x^2-8}{x-2}$

The function is not definred when

$x-2=0\\x=2$

So domin is

$R-\{2\}$

f(x) and g(x) are linear function such that for all x, f(g(x)) and g(f(x)) are Identity functions. If f(0) = 4 and g(5) = 17, compute f(2006)

$f(g(x))$ and

$g(f(x))$ are identity functions

$\implies f^{-1}(x)=g(x)$

$f(x)$ is linear function

$\implies f(x)=ax+b\\\implies g(x)=\dfrac{x-b}{a}$

$f(0)=4\\a(0)+b=4\\b=4$

$g(5)=17\\\dfrac{5-4}{a}=17\\a=\dfrac{1}{17}$

$f(2006)=\dfrac{1}{17}2006+4\\=118+4\\122$

Solve: $\dfrac { \left| x-1 \right| }{ x+2 } <1$

$\left|x-1\right|<x+2$

$x-1<x+2$ and

$-x+1<x+2$

$\Rightarrow -1<2$ and

$-2x<1$

$\Rightarrow x<\dfrac{-1}{2}$

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

Find the domain of $f(x)=\sqrt { \cos\left(\sin{x}\right) }$

Outer function

$=\sqrt{x}$

Middle function

$=\cos x$

Inner function

$=\sin x$

For a value of x to produce a real output, the value in the

$\sqrt{}$ must be

$\geq 0$ . i.e.,

$\cos(\sin x)\geq 0$

For

$\cos()$ to give a value

$\geq 0$ , the input of

$\cos ()$ must be in the interval:

$\left[-\dfrac{\pi}{2}+2k\pi , \dfrac{\pi}{2}+2k\pi\right]$

where k

$\in$ Z, must include

$\left[-\dfrac{5\pi}{2}, \dfrac{-3\pi}{2}\right], \left[-\dfrac{\pi}{2}, \dfrac{pi}{2}\right]$

$\left[\dfrac{3\pi}{2}, \dfrac{5\pi}{2}\right]$

This means since must be in one of those intervals.

$\sin x$ will never be

$> 1$ or

$< -1$ , so we can ignore all previous intervals except

$\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ which contains all possible outputs of

$\sin x$ .

The only restriction is that

$\sin x$ must be in the interval

$\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ which happens to be true for any real number x.

So domain is all real numbers.

1237251_1500461_ans_966d06ed51de4e68b19f42a606fed4db.PNG

Find the domain $\dfrac { { x }^{ 3 }-5x+3 }{ { x }^{ 2 }-1 }$

$f(x)$ is not defined if

${ x }^{ 2 }-1=0$

$\Rightarrow { x }^{ 2 }=1\quad \Rightarrow \quad x=1$ or

$-1$

$\therefore$ Domain of

$f\left( x \right) =R-\left\{ -1,1 \right\}$

Find domain of the function $f\left( x \right) =\sqrt { 4-\sqrt { 5+2x } }$ .

consider

$f(x) = \sqrt{4-\sqrt{5+2x}}$

We must have

$4 - \sqrt{5 + 2x} \ge 0$ and

$5+2x \ge 0$

$\sqrt{5 + 2x} \le 4$ and

$x \ge \dfrac{-5}{2}$

$5+2x \le 16$

$5x \le 11$

$x \le \dfrac{11}{2}$

Hence domain of

$f(x)$ will be

$\left[\dfrac{-5}{2}, \dfrac{11}{2}\right]$

1515096_1472551_ans_07376fa2f091479b85b398a07fca2ccf.png

Find the solution of the inequation $\left| 2 ^ { x } - 1 \right| + | 4 - 2 | \leq 3$

$|2^{x}-1|+|4-2|\leq 3$

$|2^{x}-1|+2\leq 3$

$|2^{x}-1|\leq 1$

$x\geq 0$

$\Rightarrow 2^{x}-1\leq 1$

$2^{x}\leq 2$

$x\leq 1$

$\therefore x \epsilon [0,1]...(1)$

$x< 0$

$\therefore 1-2^{x}\leq 1$

$0\leq 2^{x}$

$\Rightarrow x\epsilon R$

$\therefore x\epsilon (-\infty ,0)...(2)$

$\therefore (1) \cup (2) \Rightarrow$

$x\epsilon (-\infty ,1)$ Ans

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Find the largest possible domain for the real valued function given by $f\left( x \right) = \dfrac{{\sqrt {9 - {x^2}} }}{{\sqrt {{x^2} - 1} }}.$

Given

$\dfrac{\sqrt{9-x^{2}}}{\sqrt{x^{2}{-1}}}$

Domain of

$9 - x^{2}$

$9 - x ^{2} \geq 0$

$9 \geq x^{2}$

$x^{2} \leq 9$

$x \leq \pm 3$

Domain of

$\sqrt{x^{2}-1}$

$x^{2} - 1 >$

$x^{2} > 1$

$x > \pm 1$

$\therefore Domain \,of \dfrac{\sqrt{9-x^{2}}}{x^{2}-1} is -1<x\leq 3$

Find the largest possible domain for the real-valued function given by
$f(x)=\dfrac{\sqrt{9-x^2}}{\sqrt{x^2-1}}$

1206564_1505598_ans_e259a808cb7642e0a38848d587f53599.png

Write the domain of the real function $f(x)=\sqrt{\left[ x \right] -x}$ .

$f(x)=\sqrt{[x]-x}$ ......... [ Given ]

$[x]$ is the greatest integer function.

$[x]\le x,\forall_x\in R$

$\Rightarrow$

$[x]-x\le 0,\forall_x\in R$

$\Rightarrow$

$\sqrt{[x]-x}$ does not exist for any

$x\in R.$

$\therefore$ Domain

$=\phi$

Write the domain of the real function $f(x)=\sqrt{x-\left[ x \right] }$ .

$f(x)=\sqrt{x-[x]}$ ......... [ Given ]

The domain will be real only if

$x-[x]\ge 0$

$\Rightarrow$ Domain of

$f=x$

$\forall_x\in R$

Since,

$f(x)=x-[x]=x.$ for all

$x\in R$

$\therefore$ Domain of

$f=R$

If $f:A\rightarrow B$ and $g:B\rightarrow C$ are onto functions, then show that $g\circ f$ is an onto function.

$f:A\rightarrow B$ and

$g:B\rightarrow C$ are onto. ......... [ Given ]

Then,

$g\circ f:A\rightarrow C$

Let us take an element

$z$ in the co-domain.

Now,

$z$ is in

$C$ and

$g:B\rightarrow C$ is onto.

So, there exists some element

$y$ in

$B$ , such that,

$\Rightarrow$

$g(y)=z$ ---- ( 1 )

Now,

$y$ is in

$B$ and

$f:A\rightarrow B$ is onto.

So, there exists some

$x$ in

$A$ , such that,

$\Rightarrow$

$f(x)=y$ ----- ( 2 )

From ( 1 ) and ( 2 ),

$\Rightarrow$

$z=g(y)=g(f(x))=(g\circ f)(x)$

So,

$z=(g\circ f)(x)$ , where

$x$ is in

$A.$

$\therefore$

$g\circ f$ is onto.

Write the domain of the real function $f(x)=\cfrac{1}{\sqrt{\left[ x \right] -x}}$ .

Case

$1:$

$x>0$

$|x|=x$

$\Rightarrow$

$\dfrac{1}{\sqrt{|x|-x}}=\dfrac{1}{\sqrt{x-x}}$

$=\dfrac{1}{0}$

$=\infty$

Case

$2:$

When

$x<0$

$|x|=-x$

$\Rightarrow$

$\dfrac{1}{\sqrt{|x|-x}}=\dfrac{1}{\sqrt{-x-x}}$

$=\dfrac{1}{\sqrt{-2x}}$ ........[ Exists because when

$x<0,\,-2x>0$ ]

$\Rightarrow$

$f(x)$ is defined when

$x<0$

So, domain

$=(-\infty,0)$

Write the domain of the real function $f$ defined by $f(x)=\sqrt{25-{x}^{2}}$ .

We have,

$f(x)=\sqrt{25-x^2}$

The function is defined only when,

$25-x^2\ge 0$

$\Rightarrow$

$25-x^2= 0$

$\Rightarrow$

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

$\Rightarrow$

$(5+x)(5-x)= 0$

$\Rightarrow$

$5+x=0$ and

$5-x=0$

$\Rightarrow$

$x\in [-5,5]$ [ Since,

$x\in R$ ]

So, domain of the given function is

$[-5,5].$

Find the domain of each of the following real value functions real variable:
$f(x)=\sqrt {9-x^2}$

1418844_1664677_ans_08ea7f40a19e43c4a5722363bc1ce1cc.jpg

Find the domain and range of each of the following real value functions:
$f(x)=\dfrac {ax-b}{cx-d}$

1416799_1664683_ans_8ff011d145a84ef695e0015501c51de5.jpg

Find the domain and range of each of the following real value functions:
$f(x)=\dfrac {ax-b}{bx-a}$

1418847_1664682_ans_5cf79353a5634b46b051a8b7f2017413.jpg

Find $f(g(x)). f(x) = 5x^2 - 2x + 3\ and\ g(x) = 4x^2 + 7x - 5.$

$f(x)=5x^2−2x+3$

$g(x)=4x^2+7x−5$

$f(g(x))=f(4x^2+7x−5)$

$f(g(x))=5(4x^2+7x−5)2−2(4x^2+7x−5)+3$

$=80x^4+280x^3+45x^2−350x+125−8x^2−14x+10+3$

$=80x^4+280x^3+45x^2−8x^2−350x−14x+125+10+3$

$f(g(x))=80x^4+280x^3+37x^2−364x+138$

Let $f(x)=2x+5$ and $g(x)=x^2+x$ . Describe $f-g$
Find the domain in each case.

1418664_1664708_ans_97447a9ec5814041b497c4269e1b23a5.jpg

Let $f(x)=2x+5$ and $g(x)=x^2+x$ . Describe $fg$
Find the domain in each case.

1418665_1664709_ans_480ab657865749c89d21d4e47e387c8d.jpg

Find the domain and range of each of the following real value functions:
$f(x)=\sqrt {x-3}$

1416804_1664688_ans_a6aa2ef02d534175bc287eaecc5a9674.jpg

Let $f(x)=2x+5$ and $g(x)=x^2+x$ . Describe $f/g$
Find the domain in each case.

1419010_1664712_ans_df00498f36e74eb8867650bcce16172d.jpg

Let $f(x)=2x+5$ and $g(x)=x^2+x$ . Describe $f+g$
Find the domain in each case.

1418662_1664707_ans_cc6e297eb4cb44769477a79ba0f173cc.jpg

Give an example of a function which is onto but not one-one.

1544073_1705486_ans_a979307f4f8443b99b2b64965f201fb6.png

Let $f =\{ (0, -5),(1,-2),(2,1),(3,4),(4,7) \}$ be linear function from $Z$ into $Z$ . Write an expression for $f.$

1858781_1708511_ans_402624429b164eaa9cc37dda3b567b96.jpeg

Let $f: R \rightarrow R : f(x) = \dfrac {x}{c} ,$ where c is constant.
Find $(c^2 f)(x)$

1859801_1708725_ans_953ecfe32cb04c3693f9af52118ecf9e.jpeg

Let $f: R \rightarrow R : f(x) = \dfrac {x}{c} ,$ where $c$ is constant.
Find $\left( \dfrac {1}{cf}\right)(x)$

1861444_1708730_ans_35db0aeb6b054064969a8e4f8b4fd8f6.jpeg

Let $f: R \rightarrow R : f(x) = \dfrac {x}{c} ,$ where $c$ is constant.
Find $(cf)(x)$ .

1859796_1708723_ans_0f35236472934f31a617fb27bee084e0.jpeg

Are the following set of ordered pairs function? If so, examine whether the mapping is injective or surjective.
$\left\{(x,y):x \text{ is a person },y\text{ is the mother of } x \right\}$ .

Given set of ordered pair is

$\left\{(x,y):x \text{ is a person },y\text{ is the mother of } x \right\}$

Clearly each person

$x$ has only one biological mother. So the above set of ordered pair is a function.

Now, more than one person may have same mother. So the function is many-one and surjective.

Are the following set of ordered pairs function? If so, examine whether the mapping is injective or surjective.
$\left\{(a,b):a \text{ is a person } ,b \text{ is an ancestor of } a \right\}$ .

Given set of ordered pair is

$\left\{(a,b):a \text{ is a person } ,b \text{ is an ancestor of } a \right\}$ .

Here, each element of the domain does not have a unique image since a person can have many ancestors. So, it does not represent a function.

The function 't' which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by $t (c) = \frac{9C}{5} + 32$ . Find
(i) t (0)

The given function is t(c) =

$\frac{9C}{5} + 32$ .

Therefore,

(i) t (0) =

$\frac{9 \times 0}{5} + 32 = 0 + 32 = 32$

The function 't' which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by $t (c) = \frac{9C}{5} + 32$ . Find
(ii) t (28)

The given function is t(c) =

$\frac{9C}{5} + 32$ .

Therefore,

(ii) t (28) =

$\frac{9 \times 28}{5} + 32 = \frac{252 + 160}{5} = \frac{412}{5} = 82.4$

The function 't' which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by $t (c) = \frac{9C}{5} + 32$ . Find
(iii) t (-10)

The given function is t(c) =

$\frac{9C}{5} + 32$ .

Therefore,

(iii) t (-10) =

$\frac{9 \times (-10)}{5} + 32 = 9 \times (-2) + 32 = -18 + 32 = 14$

The function $'t'$ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by $t (c) = \dfrac{9C}{5} + 32$ . Find the value of $C$ , when $t (C) = 212.$

The given function is

$t(c) = \dfrac{9C}{5} + 32$ .

Therefore,

It is given that

$t (c) = 212$

$\therefore 212 = \dfrac{9C}{5} + 32$

$\Rightarrow \dfrac{9C}{5} = 212 - 32$

$\Rightarrow \dfrac{9C}{5} = 180$

$\Rightarrow 9C = 180 \times 5$

$\Rightarrow C = \dfrac{180 \times 5}{9} = 100$

Thus, the value of

$t$ , when

$t (c) = 212$ , is

$100.$

Is the function $\sin (\sin^{-1}n)$ bijective? Which is defined on $[-1,1]$ to $[-1,1]$

Let

$f(n)= \sin (\sin^{-1} n)$
let

$n_1 n_2 \in [-1,1]$

$f(n_1)=f(n_2) \Rightarrow \sin (\sin^{-1} n_1)=\sin [\sin^ 1 (n_2)]$

$\Rightarrow n_1=n_2$
hence

$f$ is one-one
let

$y=\sin (\sin^{-1}n)=n$

$\forall y \in [-1,1] \exists [-1,1] f(n)=n$
hence

$f$ is onto
since

$f$ is one-one and onto it is bijective.

Write the domain and range of the function $f(x)=x^{3}+1$

Find the domain by finding where the equation is defined. The range is the set of values that correspond with the domain.

$Domain: (-\infty,\infty),\{x/x\epsilon R\}$

$Range: (-\infty,\infty),\{y/y\epsilon R\}$

1708917_1812399_ans_006edfc7899146a3ae1b33b509a6ff5b.png

Examine which of the following is / are functions :
{( 1 ,a)(2 , b)( 1 ,b)(2 ,a)}

{( 1 ,a)(2 , b)( 1 ,b)(2 ,a)}
It is not a function because element 1 corresponds to two elements a and b .

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

$f(x)=|x-1|$

$f(x)$ is real for all

$x\in R$

$\therefore$ Domain of

$f=R$
Also

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

$\therefore$ Range of

$f=[0, \infty)$ .

Define an identity function and draw its graph also find its domain and range.

The function

$f:M \to R; f(x)=x$ for all

$\parallel x\in R$ is called an identity function on

$R$ .

$x$	$-2$	$-1$	$0$	$1$	$2$
$f(x)$	$-2$	$-1$	$0$	$1$	$2$

Domain of

$f=R$
Range of

$f=R$

1955164_1862326_ans_25086fe496c74176aaed61b989ceff0c.png

Examine which of the following is / are functions :
{(a , 0)(b ,0), (c ,1)( d , 1)}

{(a , 0)(b ,0), (c ,1)( d , 1)}
It is a function because under this each elements corresponds to one and only one element.

Examine which of the following is / are functions :
{(1 ,2),(2 ,3),(3 , 4), ( 2 , 1)}

{(1 ,2),(2 ,3),(3 , 4), ( 2 , 1)}
It is not a function because element 2 corresponds to two elements 3 and 1.

Define a constant function and draw its graph also find its domain and range.

Let

$c$ be a fixed real number. Then, the function

$f: R \to R, f(x)=c$ for all

$x\in R$ is called the constant function.

$x$	$-2$	$-1$	$0$	$1$	$2$
$f(x)=c$	$c$	$c$	$c$	$c$	$c$

$f(x)$ is defined for all real number,

$\therefore$ Domain

$=M$
Range

$=\left\{ c \right\}$
1955165_1862328_ans_6f95994b84b643339e022a7c4332d741.png

Examine which of the following is / are functions :
{( a ,a)(b , b)( c ,c)}

{( a ,a)(b , b)( c ,c)}
It is a function because first element of ordered pair set is not same.

Examine which of the following is / are functions :
{( a ,b)}

{( a ,b)}
It is a function because a corresponds to b

Examine which of the following is / are functions :
$\left \{ ( x ,y): x , y \in R , y^{2} = x \right \}$

$\left \{ ( x ,y): x , y \in R , y^{2} = x \right \}$
Here

$y^{2} = x \Rightarrow y = \pm \sqrt{x} and if x = 4 then y = \pm 2$ Hence, element of y is related with 2 and - 2 so , it is not a function.

Examine which of the following is / are functions :
$\left \{ (x , y) : x , y \in R , x = y^{3} \right \}$

$\left \{ (x , y) : x , y \in R , x = y^{3} \right \}$
It is a function because

$y = x ^{1 /3} , \forall x \in R$ unique image is the set B.

Examine which of the following is / are functions :
{( 4 , 1)(4,2),(4 , 3), ( 4 , 4)}

{( 4 , 1)(4,2),(4 , 3), ( 4 , 4)}
It is not a function because first element or ordered pair set is same.

Examine which of the following is / are functions :
{( 1, 4)(2 , 4),(3 , 4), ( 4 , 4 )}

{( 1, 4)(2 , 4),(3 , 4), ( 4 , 4 )}
It is a function because first elements of ordered pair set is unequal .

Examine which of the following is / are functions :
$\left \{ ( x , y) : x , y \in R , x^{2} = y \right \}$

$\left \{ ( x , y) : x , y \in R , x^{2} = y \right \}$
It is a function because for

$y = x^{2}$ , each real value of x there is a unique image in R for each elements of R

A function $R\rightarrow R$ is defined by $\displaystyle f\left ( x \right )=\frac{\alpha x^{2}+6x-8}{\alpha +6x-8x^{2}}$ . If $f$ is an onto function for $\alpha \in [P, Q]$ then $P + Q$ is

Since,

$f:R\rightarrow R$ is an onto mapping.

$\therefore$ Range of f=R

$\Rightarrow$

$\displaystyle \frac{\alpha x^{2}+6x-8}{\alpha +6x-8x^{2}}$ assumes all real values of x.
Let

$\displaystyle y=\frac{\alpha x^{2}+6x-8}{\alpha +6x-8x^{2}}$
Then, y assumes all real values for real values of x.

$\Rightarrow$

$\alpha y+6xy-8x^{2}y=\alpha x^{2}+6x-8$ ,

$\forall y\epsilon R$

$\Rightarrow$

$x^{2}\left ( \alpha +8y \right )+6x\left ( 1-y \right )-\left ( 8+\alpha y \right )=0$ ,

$\forall y\epsilon R$
we know, above equation assumes all real values

$\Rightarrow$

$D\geq 0$
So,

$36\left ( 1-y \right )^{2}+4\left ( \alpha +8y \right )\left ( 8+\alpha y \right )\geq 0$

$\Rightarrow$

$4\left [ 9\left ( 1-2y+y^{2} \right )+\left ( 8\alpha +\alpha ^{2}y+64y+8\alpha y^{2} \right ) \right ]\geq 0$

$\Rightarrow$

$\left [ 9-18y+9y^{2}+8\alpha +\alpha ^{2}y+64y+8\alpha y^{2} \right ]\geq 0$

$\Rightarrow$

$\left [ y^{2}\left ( 8\alpha +9 \right )+y\left ( \alpha ^{2}+46 \right )+\left ( 8\alpha +9 \right ) \right ]\geq 0$
We know, if

$ax^{2}+bx+c> 0\forall x$ and a> 0

$\Rightarrow$ D< 0
So,

$\left ( \alpha ^{2}+46 \right )^{2}-4\left ( 8\alpha +9 \right )\left ( 8\alpha +9 \right )\leq 0$ and

$\left ( 8\alpha +9 \right )> 0$

$\Rightarrow$

$\left ( \alpha ^{2}+46 \right )^{2}-\left [ 2\left ( 8\alpha +9 \right ) \right ]^{2}\leq 0$ and

$\alpha > -9/8$

$\Rightarrow$

$\left ( \alpha ^{2}+46-16\alpha -18 \right )\left ( \alpha ^{2}+46+16\alpha +18 \right )\leq 0$ and

$\alpha > -9/8$

$\left ( \alpha ^{2}-16\alpha +28 \right )\left ( \alpha ^{2}+16\alpha +64 \right )\leq 0$ and

$\alpha > -9/8$

$\Rightarrow$

$\left ( \alpha -14 \right )\left ( \alpha -2 \right )\left ( \alpha +8 \right )^{2}\leq 0$ and

$\alpha > -9/8$

$\alpha \epsilon \left [ 2, 14 \right ]\cup \left \{ -8 \right \}$ and

$\alpha > -9/8$
Thus,

$\alpha \epsilon \left [ 2, 14 \right ]$

Find the domain of the real valued linear function $f$ given below
$f(x)=x+1$

1305100_488797_ans_896d446e86b34824bbdd258abf20d1c9.jpeg

Find the domain of the real valued linear function $f$ given as below:
$f(x)=\dfrac{(x-1)}{(x-3)}$

Given,

$f\left( x \right) =\cfrac { x-1 }{ x-3 }$
Thus,

$f\left( x \right)$ is real if

$x-3\neq 0$

$\implies x\neq 3$

$\therefore$ Domain

$=R-\left\{ 3 \right\}$

$=\left( -\infty ,\infty \right) -\left\{ 3 \right\}$ .

Find the domain of the real valued logarithmic function $f$ given below
$f(x)=\ln({x}^{2}-9)$

$f(x)=\ln({ x }^{ 2 }−9)$
Thus,

$f(x)$ is real valued if

$\left( { x }^{ 2 }-9 \right) >0$

$\implies { x }^{ 2 }>9$ and

$\left| x \right| >3$

$\implies x<-3$ or

$x>3$

So, domain of given function is given by

$\left( -\infty ,-3 \right) \cup \left( 3,\infty \right)$ .

Indicate the domain of definition of the function and simplify the given expression.
$\displaystyle\, y = x\frac{1 + 2(x + 4)^{-0.5}}{2 - (x + 4)^{0.5}} + (x + 4)^{0.5} + 4(x + 4)^{-0.5}$

$y=x.\dfrac { 1+2{ \left( x+4 \right) }^{ -0.5 } }{ 2-{ \left( x+4 \right) }^{ 0.5 } } +{ \left( x+4 \right) }^{ 0.5 }+4{ \left( x+4 \right) }^{ -0.5 }$

$=x.\dfrac { 1+\dfrac { 2 }{ \sqrt { x+4 } } }{ 2-\sqrt { x+4 } } +\sqrt { x+4 } +\dfrac { 4 }{ \sqrt { x+4 } }$

$=\dfrac { x.\left( \sqrt { x+4 } +2 \right) }{ 2\sqrt { x+4 } -\left( x+4 \right) } +\sqrt { x+4 } +\dfrac { 4 }{ \sqrt { x+4 } }$

$=\dfrac { x.\left( \sqrt { x+4 } +2 \right) }{ 2\sqrt { x+4 } \left( 1-\sqrt { x+4 } \right) } +\dfrac { x+8 }{ \sqrt { x+4 } }$

$=\dfrac { x.\left( \sqrt { x+4 } +2 \right) +2\left( x+8 \right) \left( 1-\sqrt { x+4 } \right) }{ 2\sqrt { x+4 } \left( 1-\sqrt { x+4 } \right) }$

$=\dfrac { x\left( \sqrt { x+4 } \right) +2x+2x+16-2x\sqrt { x+4 } -16\sqrt { x+4 } }{ 2\sqrt { x+4 } \left( 1-\sqrt { x+4 } \right) }$

$=\dfrac { 4x+16-\sqrt { x+4 } \left( x+16 \right) }{ 2\sqrt { x+4 } \left( 1-\sqrt { x+4 } \right) }$

For domain

$\sqrt { x+4 }>0$

$\Rightarrow x+4>0$

$\Rightarrow x>-4$

and

$1-\sqrt { x+4 }\neq0$

$\sqrt { x+4 }\neq 1$

$x+4\neq 1$

$x\neq 1-4$

$x\neq -3$

$\therefore x>-4$ and

$x\neq -3$

$x\in \left(-4,\infty\right)-\left(-3\right)$

Hence, the answer is

$x\in \left(-4,\infty\right)-\left(-3\right) .$

Let $f=\left\{(1,1),(2,3),(0,-1),(-1,3)\right\}$ be a linear function from $z$ to $z$ , find $f(x)$ where $x$ is the set of integers.

Since

$f$ is a linear function.

Let

$f\left(x\right)=mx+c$

	$x$	$y$	$y=mx+c$
$\left(1,1\right)$	$1$	$1$	$1=m\times 1+c\Rightarrow 1=m+c$
$\left(2,3\right)$	$2$	$3$	$3=m\times 2+c\Rightarrow 3=2m+c$

Solving

$m+c=1$ .....

$(1)$

$2m+c=3$ .......

$(2)$

Equation

$(2)-(1)$

$\Rightarrow 2m+c-\left(m+c\right)=3-1$

$\Rightarrow 2m+c-m-c=2$

$\Rightarrow m=2$

Put

$m=2$ in

$(1)$

$m+c=1$

$\Rightarrow 2+c=1$

$\Rightarrow c=1-2=-1$

So,

$f\left(x\right)=mx+c=2x-1$

Find the domain of $\sqrt[3]{5x-3}$

2041225_1456455_ans_7baf57235b3b414a8ad4bb01649e7993.jpg

Find the domain of $\cfrac{1}{((x-1)(x-2)}$ .

2041218_1456430_ans_b1f8db45d5ee45d796607e361e1aa0c4.jpg

The domain of the real-valued function $f\left( x \right) =\log_{10}{\sqrt{\cfrac{(3-x)(x+2)}{(x+1)(x-4)}}}$ does not contain the intervals $(-\infty,-5)$ and $(5,6)$ .

2042509_1472071_ans_9314c454c19b47ae8127beed4fedbc45.jpg

find the domain of x
$\dfrac{x^4-3x^3+2x^2}{x^2-x-30} > 0$ .

1304426_1581503_ans_713d4bafbb3248268c45c53fdfddb343.jpg

Give an example of a function which is neither one-one nor onto.

1544065_1705483_ans_77757d7965eb4bacb58d0d7e84509f74.jpg

Show that the function $f:R\rightarrow R:f(x)=x^2$ is neither one-one nor onto.

1544528_1705707_ans_6037ff93a8f74b2cb17368e0c88472da.jpg

Show that the function $f: R\rightarrow R:f(x)=\sin x$ is neither one-one nor onto.

one-one function is a function that comprises individuality that never maps discrete elements of its domain to the equivalent element of its codomain. We can say, every element of the codomain is the image of only one element of its domain.

We know that

$\sin(0)=0$ and

$\sin(\pi)=0$ .

Thus,

$0$ and

$\pi$ have the same image.

So, f is many-one.

Into Function : Function f from set A to set B is Into function if at least set B has a element which is not connected with any of the element of set A.

Range

$(f)=[-1, 1]\subset R$ . Hence, f is into.

So, f is neither one-one nor onto.

Let $X = \{ 1, 2, 3 ,4\}, Y = \{1,5,9, 11, 15 , 16 \}$ and $f = \{ (1,5) , (2,9), ( 3,1), (4,5), 92, 11)\}$ . Are the following true?
(i) $f$ is a relation from $X$ to $Y$ .
(ii) $f$ is function from $X$ to $Y$ . Justify your answer in each case.

1853185_1708486_ans_464212bdbb194ac4892ab9031ed4b08f.jpg

Number of integers in the domain of function, satisfying $f(x) + f(x^{-1}) = \dfrac{x^3 + 1}{x}$ , is

1768487_1754013_ans_697276db4eae41a08a8bb704e1c748c3.jpg

If $f ( x + \frac {1}{x} ) = (x^2 +\frac {1}{x^2} )$ for all $x \epsilon R - \left\{0 \right\}$ then write an expression for $f(x)$ .

Given

$f\left( x+\frac { 1 }{ x } \right) ={ x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } }$

$\therefore f\left( x+\frac { 1 }{ x } \right) ={ x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } +2-2$

$\therefore f\left( x+\frac { 1 }{ x } \right) =\left[ { x }^{ 2 }+\left( 2\times \left( x \right) \times \frac { 1 }{ x } \right) +\frac { 1 }{ { x }^{ 2 } } \right] -2$

$\therefore f\left( x+\frac { 1 }{ x } \right) =\left[ { x }^{ 2 }+\left( 2\times \left( x \right) \times \frac { 1 }{ x } \right) +{ \left( \frac { 1 }{ x } \right) }^{ 2 } \right] -2$

We know that

${ a }^{ 2 }+2ab+b^{ 2 }={ \left( a+b \right) }^{ 2 }$

$\therefore f\left( x+\frac { 1 }{ x } \right) ={ \left( x+\frac { 1 }{ x } \right) }^{ 2 }-2$

Replace

$x+\frac { 1 }{ x }$ by x , we get,

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

If $f(x) =$ sin log $_e \left ( \frac{\sqrt{4 - x^2}}{1 - x} \right )$ , then the domain of $f(x)$ is_ and its range is _______. (IIT-JEE, 1985)

1806473_1754606_ans_893fc0b89c8c47ed9ebbae7141b1374b.jpeg

Functions - Class 11 Commerce Applied Mathematics - Extra Questions

If f(x)=√x+2+1log10(1−x)f(x)=\sqrt{x+2}+\dfrac{1}{\log_{10} (1-x)}. Find the domain of definition of f(x)f(x).

Find the domains of definition of the following functions.y=√6x−x2−55x−2−1.\displaystyle y \, = \, \frac{\sqrt{6x \, - \, x^2 \, - \, 5}}{5^{x \, - \, 2} \, - \, 1}.

Let f(x)=√x+2+1log10(1−x)f(x)=\sqrt{x+2}+\dfrac{1}{\log_{10}(1-x)}. Then find the domain of f(x)f(x).

Find the domains of definition of the following functions.y=√3x−4x2x2−x−8.\displaystyle y \, = \, \sqrt{\frac{3^x \, - \, 4^x}{2x^2 \, - \, x \, - \, 8}}.

If f(x)=2x+33x−2f(x)=\dfrac {2x+3}{3x-2} Prove that foffof is an identity function

Find the domain and range of :-(i) √x−1\sqrt x - 1(ii) (x−1)(x-1)

A relation R is defined on the set Z of integers as follows: (x, y)∈R′x2+y2=25\in R' x^2+y^2= 25 Express R and R−1R^{-1} as the sets of ordered pairs and hence find their respective domains.

Find the domain and range of f(x)=√x−5f(x) = \sqrt{x - 5}

The domain of f(x)=√sin−1(log2x)f(x)=\sqrt{\sin ^{-1}(\log _{2}x)} is

Find the domain of the function:f(x)=1√x2−4xf(x)=\dfrac{1}{\sqrt{x^2-4x}}.

Find the domain of the function f(x)=x2√(a−x)(x−b)f(x)=\dfrac{x^2}{\sqrt{(a-x)(x-b)}}, b>ab>a.

Solve:x√1+x2+log(x+√x2+1).x\sqrt {1 + {x^2}} + \log \left( {x + \sqrt {{x^2} + 1} } \right).

Find the domain of the following (i)f(x)=√[x2]−1+√3−[x2]f\left( x \right) = \sqrt {\left[ {{x^2}} \right] - 1} + \sqrt {3 - \left[ {{x^2}} \right]}

Find the domain of function √25−x2\sqrt{25-x^2}

Find the domain of the following real functions :(i)f(x)=√16−x2f\left( x \right)=\sqrt{16-x^2}

The domain of Function √9−x2\sqrt{9-x^2}

The value of f(3)f(3) if f(x)=1x2−xf(x)=\dfrac1{x^2-x}

Find the Domain of the following function:y=x√1+xy=\dfrac{x}{\sqrt{1+x}}

Find the domain for the following functions :(a)y=1−log10xy=1-\log _{ 10 }{ x } (b)y=1√x2−4xy=\frac { 1 }{ \sqrt { { x }^{ 2 }-4x } }

Find the domain of y=√x−25x+5y=\dfrac{\sqrt{x-25}} {x+5}

Find the domain of the function f(x)=1√2x2+5x+2f(x)=\dfrac{1}{\sqrt{2x^2+5x+2}}.

Find the domain and range of the real function f(x)=11−x2f(x)=\frac{1}{1-x^{2}}

Find the domain of f(x)=√4−9x2f(x)=\sqrt { 4 - 9 x ^ { 2 } }

Find the domain and the inverse of the following functions:f(x)=11−x2f(x)=\dfrac{1}{1-x^{2}}

Find the domain of f(x)=√9−x2 f(x)=\sqrt{9-x^{2}}

Find f(5)f(5) if f(x)=x2+3x+2f(x)=x^2+3x+2

Find the domain and the range of the following function:f(x)=x−32x+1f(x) =\dfrac{x-3}{2x+1}

The domain of function √x2−4\sqrt{x^2-4} is

Find the domain and the inverse of the following functions:f(x)=x21+x2f(x)=\dfrac{x^{2}}{1+x^{2}}

Find the domain & range of the following function.f(x)=x2−3x+2x2+x−6f\left(x\right)=\dfrac{{x}^{2}-3x+2}{{x}^{2}+x-6}

Find the domain of each of the following real value functions real variable:f(x)=1√x2−1f(x)=\dfrac {1}{\sqrt {x^2 -1}}

Find the domain of each of the following real valued functions of real variable:f(x)=3x−2x+1f(x)=\dfrac {3x-2}{x+1}

Find the domain of each of the following real valued functions of real variable:f(x)=1xf(x)=\dfrac {1}{x}

Find the domain of each of the following real value functions real variable:f(x)=√x−2f(x)=\sqrt {x-2}

Find the domain of: sec−1x−tan−1x\sec^{-1} x - \tan^{-1} x.

Find the domain of each of the following real valued functions of real variable:f(x)=2x+1x2−9f(x)=\dfrac {2x+1}{x^2 -9}

Find the domain of each of the following real valued functions of real variable:f(x)=1x−7f(x)=\dfrac {1}{x-7}

Given A={2,3,4},B={2,5,6,7}A=\left\{2,3,4 \right\}, B=\left\{2,5,6,7 \right\}. Construct an example of each of the following.a mapping from AA to BB which is not injective.

Given A={2,3,4},B={2,5,6,7}A=\left\{2,3,4 \right\}, B=\left\{2,5,6,7 \right\}. Construct an example of each of the following.a mapping from AA to BB.

If the mappings, ff and gg are given by f={(1,2),(3,5),(4,1)}f=\left\{(1,2),(3,5),(4,1)\right\} and g={(2,3),(5,1),(1,3)}g=\left\{(2,3),(5,1),(1,3)\right\}.

Give an example of a mapwhich is not one-one but onto.

Give an example of a mapwhich is one-one but not onto.

Give an example of a mapwhich is neither one-one but nor onto.

Given A={2,3,4},B={2,5,6,7}A=\left\{2,3,4 \right\}, B=\left\{2,5,6,7 \right\}. Construct an example of each of the following.an injective mapping from AA to BB.

If domain for y=cos−1(1−2|x|3)+log|x−1|xy=\cos^{-1}\left (\dfrac {1-2|x|}{3}\right )+\log_{|x-1|}x is x∈(a,b)∪(c,d)x\in (a, b)\cup (c, d). Find a+b+c+da+b+c+d

Let y=√(x+1)(x−3)x−2.\displaystyle y= \sqrt{\frac{\left ( x+1 \right )\left ( x-3 \right )}{x-2}}.Find all the real values of x for which y takes real values.

Find the number of all onto functions from the set {1,2,3,....,n}\left \{1, 2, 3, ...., n\right \} to itself

Find the domain of the real valued square root function ff given belowf(x)=√(2x)f(x)=\sqrt{(2x)}

If the domain of the real valued function ff given belowf(x)=√−x+1x+3f(x)=\sqrt{\dfrac{-x+1}{x+3}} is (−∞,−x)∪(−x,1](-\infty,-x)\cup(-x,1].Find xx

Find the domain of the real valued rational function ff given belowf(x)=1/(x3+x2−2x)f(x)=1/({x}^{3}+{x}^{2}-2x)

If the domain of f(x)=√2x+1f(x)=\sqrt{2x+1} is [−1m,∞)\Big[ \dfrac{-1}{m},\infty\Big) .Find mm

The domain of the real valued sine function f(x)=2sin(x−1)f(x)=2\sin{(x-1)} is (−∞,∞)(-\infty,\infty). If true enter 1 else0.

Find the domain of the real valued exponential function ff given belowf(x)=e(x−4)f(x)={e}^{(x-4)}

Find the domains of definition of the following functions.y=√sinx+√16−x2.\displaystyle y \, = \, \sqrt{\sin \, x} \, + \, \sqrt{16 \, - \, x^2}.

Let AA be a set of nn distinct elements. Then, the total number of distinct functions from AA to AA is .... and out of these .... are onto functions.

Find the domains of definition of the following functions.y=x√x2−5x+6.\displaystyle y \, = \, \frac{x}{\sqrt{x^2 \, - \, 5x \, + \, 6}}.

Let f:N→Nf:\mathbb{N}\to \mathbb{N} be defined by f(n)={n+12,if n is oddn2,if n is evenf(n)=\begin{cases}\dfrac{n+1}{2}, &\mbox{if}\ n\ \mbox{is odd}\\ \dfrac{n}{2}, &\mbox{if}\ n\ \mbox{is even}\end{cases}.n∈Nn\in \mathbb{N}State whether the function ff is bijective. Justify your answer.

If :R→R:R\rightarrow R defined by f(x)=1+x2f(x)=1+{ x }^{ 2 }, then show that ff is neither 1−11-1 nor onto

Find the domain of f(x)=√([x]−1)+√(4−[x])f\left( x \right) = \sqrt {\left( {\left[ x \right] - 1} \right)} + \sqrt {\left( {4 - \left[ x \right]} \right)} (where[] represents the greater integer function)

show that f:A→B f : A \rightarrow B and g:B→C g : B \rightarrow C are onto, then gof:A→gof : A \rightarrow is also onto:

Let f:[α,∞)→[x,∞),f(x)=x2−6x+5f: [ \alpha, \infty) \rightarrow [x, \infty), f(x) = x^2 -6x +5 then find three possible value of ′α′'\alpha' if f(x)f(x) is onto.

Let f(x)=√2−x−x2f(x) = \sqrt{2 -x -x^{2}} and g(x)=cosxg(x) = cos x. Check given statement is true or not?I. Domain of f((g(x))2)f((g(x))^{2}) = Domain of f(g(x))f(g(x))

Find the domain of f(x)=√sinx+√16−x2f(x)=\sqrt{\sin x}+\sqrt{16-x^2}.

Find the domain of f(x)=log10log2logπ/2(tan−1x)−1f(x) = \log_{10}\log_{2}\log_{\pi/2}(\tan^{-1}x)^{-1}

Let f:N→Yf: N \rightarrow Y be a function defined by f(x)=4x2+12x+15f(x) = 4x^2+ 12x+15, where Y=Y= range of f.f. Show thatf f is invertible and find the inverse of ff.

If f:R→Rf: R\rightarrow R be the function defined by f(x)=4x3+7f(x) = 4x^3+7, then show that ff is a bijection.

Find the domain for which the function f(x)=3x2−2xf(x)=3{x}^{2}-2x and g(x)=9x−6g(x)=9x-6 are equal.

Evaluate:∫sin2xcos2xdx\displaystyle \int \sin^2 x \cos^2 x dx

Simplify 11+am−n+11+an−m\dfrac{1}{1+a^{m-n}}+\dfrac{1}{1+a^{n-m}}

Find the domain of f(x)=√([x]−1)+√(4−[x])f\left( x \right) = \sqrt {\left( {\left[ x \right] - 1} \right)} + \sqrt {\left( {4 - \left[ x \right]} \right)} (where [ ] represents the greatest integer function).

Find the domain of f(x)=√x+2+log10(1−x)f(x) = \sqrt{x + 2} + {log_{10}}(1 - x)

The domain of definition of the function y(x)y(x) given by 2x+2y=22^x+2^y=2 is

If f:R→S,defined by f(x)=sinx−√3cosx+1f:R\rightarrow S,defined\ by\ f\left( x \right) =\sin { x } -\sqrt { 3 } \cos { x } +1, is onto, then find the set SS ?

Domain of f(x)=x2+3x+5x2−5x+4f\left( x \right) = \dfrac{{{x^2} + 3x + 5}}{{{x^2} - 5x + 4}} is

Show that f:N→Nf: N→N = x−1x-1, if xx is even is both one-one and onto

Let f:R→Rf:R \to R be defined by i) f(x)=x+1f\left( x \right) = x + 1 ii) f(x)=x+|x|f\left( x \right) = x + \left| x \right|. Determine whether or not ff is onto.

The domain of 1√x−x2+√3x−1−2x2\frac{1}{\sqrt{x-x^{2}}}+\sqrt{3x-1-2x^{2}} is:

The domain of the function x√x2−3x+2\dfrac{x}{\sqrt{x^2-3x+2}} is

If f(x+y)=f(xy)∀x,yϵRf(x + y) = f(xy) \forall x, y\epsilon R then prove that ff is a constant function.

If $f(x)=\sqrt{x+2}+\dfrac{1}{\log_{10} (1-x)}$ . Find the domain of definition of $f(x)$ .

Find the domains of definition of the following functions.
$\displaystyle y \, = \, \frac{\sqrt{6x \, - \, x^2 \, - \, 5}}{5^{x \, - \, 2} \, - \, 1}.$

Let $f(x)=\sqrt{x+2}+\dfrac{1}{\log_{10}(1-x)}$ . Then find the domain of $f(x)$ .

Find the domains of definition of the following functions.
$\displaystyle y \, = \, \sqrt{\frac{3^x \, - \, 4^x}{2x^2 \, - \, x \, - \, 8}}.$

If $f(x)=\dfrac {2x+3}{3x-2}$
Prove that $fof$ is an identity function

Find the domain and range of :-
(i) $\sqrt x - 1$
(ii) $(x-1)$

A relation R is defined on the set Z of integers as follows: (x, y) $\in R' x^2+y^2= 25$
Express R and $R^{-1}$ as the sets of ordered pairs and hence find their respective domains.

Find the domain and range of $f(x) = \sqrt{x - 5}$

The domain of $f(x)=\sqrt{\sin ^{-1}(\log _{2}x)}$ is

Find the domain of the function:
$f(x)=\dfrac{1}{\sqrt{x^2-4x}}$ .

Find the domain of the function $f(x)=\dfrac{x^2}{\sqrt{(a-x)(x-b)}}$ , $b>a$ .

Solve:
$x\sqrt {1 + {x^2}} + \log \left( {x + \sqrt {{x^2} + 1} } \right).$

Find the domain of the following

(i) $f\left( x \right) = \sqrt {\left[ {{x^2}} \right] - 1} + \sqrt {3 - \left[ {{x^2}} \right]}$

Find the domain of function $\sqrt{25-x^2}$

Find the domain of the following real functions :
(i) $f\left( x \right)=\sqrt{16-x^2}$

The domain of Function $\sqrt{9-x^2}$

The value of $f(3)$ if $f(x)=\dfrac1{x^2-x}$

Find the Domain of the following function:
$y=\dfrac{x}{\sqrt{1+x}}$

Find the domain for the following functions :
(a) $y=1-\log _{ 10 }{ x }$ (b) $y=\frac { 1 }{ \sqrt { { x }^{ 2 }-4x } }$

Find the domain of $y=\dfrac{\sqrt{x-25}} {x+5}$

Find the domain of the function $f(x)=\dfrac{1}{\sqrt{2x^2+5x+2}}$ .

Find the domain and range of the real function $f(x)=\frac{1}{1-x^{2}}$

Find the domain of
$f(x)=\sqrt { 4 - 9 x ^ { 2 } }$

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

Find the domain of $f(x)=\sqrt{9-x^{2}}$

Find $f(5)$ if $f(x)=x^2+3x+2$

Find the domain and the range of the following function:
$f(x) =\dfrac{x-3}{2x+1}$

The domain of function $\sqrt{x^2-4}$ is

Find the domain and the inverse of the following functions:
$f(x)=\dfrac{x^{2}}{1+x^{2}}$

Find the domain & range of the following function.
$f\left(x\right)=\dfrac{{x}^{2}-3x+2}{{x}^{2}+x-6}$

Find the domain of each of the following real value functions real variable:
$f(x)=\dfrac {1}{\sqrt {x^2 -1}}$

Find the domain of each of the following real valued functions of real variable:
$f(x)=\dfrac {3x-2}{x+1}$

Find the domain of each of the following real valued functions of real variable:
$f(x)=\dfrac {1}{x}$

Find the domain of each of the following real value functions real variable:
$f(x)=\sqrt {x-2}$

Find the domain of:

$\sec^{-1} x - \tan^{-1} x$ .

Find the domain of each of the following real valued functions of real variable:
$f(x)=\dfrac {2x+1}{x^2 -9}$

Find the domain of each of the following real valued functions of real variable:
$f(x)=\dfrac {1}{x-7}$

Given $A=\left\{2,3,4 \right\}, B=\left\{2,5,6,7 \right\}$ . Construct an example of each of the following.
a mapping from $A$ to $B$ which is not injective.

Given $A=\left\{2,3,4 \right\}, B=\left\{2,5,6,7 \right\}$ . Construct an example of each of the following.
a mapping from $A$ to $B$ .

If the mappings, $f$ and $g$ are given by $f=\left\{(1,2),(3,5),(4,1)\right\}$ and $g=\left\{(2,3),(5,1),(1,3)\right\}$ .

Give an example of a map
which is not one-one but onto.

Give an example of a map
which is one-one but not onto.

Give an example of a map
which is neither one-one but nor onto.

Given $A=\left\{2,3,4 \right\}, B=\left\{2,5,6,7 \right\}$ . Construct an example of each of the following.
an injective mapping from $A$ to $B$ .

If domain for $y=\cos^{-1}\left (\dfrac {1-2|x|}{3}\right )+\log_{|x-1|}x$ is $x\in (a, b)\cup (c, d)$ . Find $a+b+c+d$

Let $\displaystyle y= \sqrt{\frac{\left ( x+1 \right )\left ( x-3 \right )}{x-2}}.$
Find all the real values of x for which y takes real values.

Find the number of all onto functions from the set $\left \{1, 2, 3, ...., n\right \}$ to itself

Find the domain of the real valued square root function $f$ given below
$f(x)=\sqrt{(2x)}$

If the domain of the real valued function $f$ given below
$f(x)=\sqrt{\dfrac{-x+1}{x+3}}$ is $(-\infty,-x)\cup(-x,1]$ .Find $x$

Find the domain of the real valued rational function $f$ given below
$f(x)=1/({x}^{3}+{x}^{2}-2x)$

If the domain of $f(x)=\sqrt{2x+1}$ is $\Big[ \dfrac{-1}{m},\infty\Big)$ .Find $m$

The domain of the real valued sine function $f(x)=2\sin{(x-1)}$ is $(-\infty,\infty)$ . If true enter 1 else0.

Find the domain of the real valued exponential function $f$ given below
$f(x)={e}^{(x-4)}$

Find the domains of definition of the following functions.
$\displaystyle y \, = \, \sqrt{\sin \, x} \, + \, \sqrt{16 \, - \, x^2}.$

Let $A$ be a set of $n$ distinct elements. Then, the total number of distinct functions from $A$ to $A$ is .... and out of these .... are onto functions.

Find the domains of definition of the following functions.
$\displaystyle y \, = \, \frac{x}{\sqrt{x^2 \, - \, 5x \, + \, 6}}.$

Let $f:\mathbb{N}\to \mathbb{N}$ be defined by $f(n)=\begin{cases}\dfrac{n+1}{2}, &\mbox{if}\ n\ \mbox{is odd}\\ \dfrac{n}{2}, &\mbox{if}\ n\ \mbox{is even}\end{cases}$ . $n\in \mathbb{N}$
State whether the function $f$ is bijective. Justify your answer.

If $:R\rightarrow R$ defined by $f(x)=1+{ x }^{ 2 }$ , then show that $f$ is neither $1-1$ nor onto

Find the domain of $f\left( x \right) = \sqrt {\left( {\left[ x \right] - 1} \right)} + \sqrt {\left( {4 - \left[ x \right]} \right)}$
(where[] represents the greater integer function)

show that $f : A \rightarrow B$ and $g : B \rightarrow C$ are onto, then $gof : A \rightarrow$ is also onto:

Let $f: [ \alpha, \infty) \rightarrow [x, \infty), f(x) = x^2 -6x +5$ then find three possible value of $'\alpha'$ if $f(x)$ is onto.

Let $f(x) = \sqrt{2 -x -x^{2}}$ and $g(x) = cos x$ . Check given statement is true or not?
I. Domain of $f((g(x))^{2})$ = Domain of $f(g(x))$

Find the domain of $f(x)=\sqrt{\sin x}+\sqrt{16-x^2}$ .

Find the domain of $f(x) = \log_{10}\log_{2}\log_{\pi/2}(\tan^{-1}x)^{-1}$

Let $f: N \rightarrow Y$ be a function defined by $f(x) = 4x^2+ 12x+15$ , where $Y=$ range of $f.$ Show that $f$ is invertible and find the inverse of $f$ .

If $f: R\rightarrow R$ be the function defined by $f(x) = 4x^3+7$ , then show that $f$ is a bijection.

Find the domain for which the function $f(x)=3{x}^{2}-2x$ and $g(x)=9x-6$ are equal.

Evaluate:
$\displaystyle \int \sin^2 x \cos^2 x dx$

Simplify $\dfrac{1}{1+a^{m-n}}+\dfrac{1}{1+a^{n-m}}$

Find the domain of
$f\left( x \right) = \sqrt {\left( {\left[ x \right] - 1} \right)} + \sqrt {\left( {4 - \left[ x \right]} \right)}$
(where [ ] represents the greatest integer function).

Find the domain of $f(x) = \sqrt{x + 2} + {log_{10}}(1 - x)$

The domain of definition of the function $y(x)$ given by $2^x+2^y=2$ is

If $f:R\rightarrow S,defined\ by\ f\left( x \right) =\sin { x } -\sqrt { 3 } \cos { x } +1$ , is onto, then find the set $S$ ?

Domain of $f\left( x \right) = \dfrac{{{x^2} + 3x + 5}}{{{x^2} - 5x + 4}}$ is

Show that $f: N→N$ = $x-1$ , if $x$ is even
is both one-one and onto

Let $f:R \to R$ be defined by i) $f\left( x \right) = x + 1$ ii) $f\left( x \right) = x + \left| x \right|$ . Determine whether or not $f$ is onto.

The domain of $\frac{1}{\sqrt{x-x^{2}}}+\sqrt{3x-1-2x^{2}}$ is:

The domain of the function $\dfrac{x}{\sqrt{x^2-3x+2}}$ is

If $f(x + y) = f(xy) \forall x, y\epsilon R$ then prove that $f$ is a constant function.

The domain of the function $\sqrt{x^{2}-5x+6}+\sqrt{2x+8-x^{2}}$ is