Class 12 Engineering Maths - Relations And Functions

Show that if $f:A\rightarrow B$ and $g:B\rightarrow C$ are one-one, then $g\circ f:A\rightarrow C$ is also one-one.

Given

$f:A\rightarrow B$ is one one

${ a }_{ i }\in A,{ b }_{ i }\in B$

$f({ a }_{ i })={ b }_{ i }$

$g:A\rightarrow B$ is one one

${ b }_{ i }\in B,{ c }_{ i }\in C$

$g({ b }_{ i })={ c }_{ i }$ for all

${ b }_{ i }$

$g(f({ a }_{ i }))={ c }_{ i }$ for all

${ a }_{ i }$ only one

${ c }_{ i }$

$g(f(x))$ is one one.

Let $f:R\rightarrow R$ be defined by $f(x)={x}^{3}+5$ then find ${f}^{-1}(x)$

let

$f(x)=x^3+5=y$

$y-5=x^3$

$x=(y-5)^{\dfrac{1}{3}}$

so therefore

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

If the function $f:[1,\infty)\rightarrow [1,\infty)$ is defined by $f(x)=2^{x(x-1)}$ ,then find $f^{-1}(4)$

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

$log_{2}(y)=x^{2}-x$

$log_{2}(y)=(x-\dfrac{1}{2})^{2}-\dfrac{1}{4}$

$\dfrac{4log_{2}(y)+1}{4}=(x-\dfrac{1}{2})^{2}$

$x-\dfrac{1}{2}=\dfrac{\pm\sqrt{4log_{2}(y)+1}}{2}$
Since the range of f(x) is

$R^{+}$
hence

$x-\dfrac{1}{2}=\dfrac{\sqrt{4log_{2}(y)+1}}{2}$

$x=\dfrac{1+\sqrt{4log_{2}(y)+1}}{2}$
Replacing x and y

$f^{-1}(x)=\dfrac{1+\sqrt{4log_{2}(x)+1}}{2}$
Therefore

$f^{-1}(4)=\dfrac{1+\sqrt{8+1}}{2}$

$=\dfrac{4}{2}$

$=2$

Let $f: R\rightarrow R$ be defined as $f(s) = 10x + 7$ . Find the function $g: R\rightarrow R$ such that $gof = fog = 1_{R}$

Step:1

$f\left( x \right)=10x+7\\ Let\quad f\left( x \right)=y\\ y=10x+7\\ y-7=10x\\ \therefore 10x=y-7\quad where\quad g:Y\rightarrow N\\ x=\cfrac { y-7 }{ 10 } \\ Let\quad g\left( y \right)=\cfrac { y-7 }{ 10 } \\$

Step:2

$gof=g\left( f\left( x \right) \right)=g\left( 10x+7 \right)\\ =\cfrac { (10x+7)-7 }{ 10 } =\cfrac { 10x }{ 10 } =x={ I }_{ R }$

Step:3

$fog=f\left( g\left( y \right) \right)=f[\cfrac { y-7 }{ 10 } ]=10(\cfrac { y-7 }{ 10 } )+y\\ =y+0=y={ I }_{ R }\\$

Since,

$gof=fog={ I }_{ R }$

$f$ is invertible.

and inverse of

$f=g\left( y \right)=\cfrac { y-3 }{ 4 }$

Let $f:R\rightarrow R$ be defined by $f(x)=2x+3$ , find ${f}^{-1}(4)$ .

Given that:

$f(x)=2x+3$

To find:

$f^{-1}(4)=?$

Solution:

A function is one-to=one for which every element of the range of the function corresponds to exactly one element of the domain.

Since,

$f$ is a linear function, we can say

$f$ is one-to-one and hence

$f^{-1}$ exits.

Let

$y=f(x)=2x+3$

or,

$y-3=2x$

or,

$\dfrac {y-3}2=x$

or,

$y=\dfrac{x-3}2$ (Interchange

$x$ and

$y$ )

or,

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

or,

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

or,

$f^{-1}(4)=\dfrac12$

Consider $f:R - \left\{ - \dfrac{4}{3} \right \} \rightarrow R - \left\{ \dfrac{4}{3} \right \}$ given by $f(x) = \dfrac{4x + 3}{3x + 4}$ . Show that $f$ is bijective. Find the inverse of $f$ and hence if the value of $f^{-1}(0)$ is $A$ and $x$ is $B$ such that $f^{-1} (x) = 2$ , then find $(A+B)\times100$ .

Given

$f(x)=\cfrac{4x+3}{3x+4}$

For one-one function, we have

$f(x_1) = f(x_2)\implies x_1=x_2$

Thus, consider

$f(x_1) = f(x_2)$

$\implies \cfrac{4x_1+3}{3x_1+4}=$

$\cfrac{4x_2+3}{3x_2+4}$

$\implies 12x_1x_2+16x_1+9x_2+12=12x_1x_2+9x_1+16x_2+12$

$\implies 7x_1=7x_2$

$\implies x_1=x_2$

$\therefore \ f(x)$ is one-one function.

For onto function:

Let

$f(x)=y$

$\implies \cfrac{4x+3}{3x+4} =y$

$\implies x =$

$\cfrac{3-4y}{3y-4}$

Domain of the above function is

$R-\{ \frac{4}{3}\}$ , which is range of function

$f(x)$ .

So, given function

$f(x)$ is onto.

Hence, given function

$f(x)$ is bijective.

Now, let us find the inverse of

$f(x)$

Let the given function

$f(x)$ be

$y$

$\Rightarrow f(x)=\cfrac{4x+3}{3x+4}=y$

So we get

$f^{-1}{(y)}=x=\cfrac{3-4y}{3y-4}$

So the inverse of function

$f$ is

$f^{-1}{(x)}=\cfrac{3-4x}{3x-4}$

Also,

$A = f^{-1}{(0)}$

$\Rightarrow A = f^{-1}{(0)}=-\cfrac{3}{4}$

Given that

$f^{-1}{(x)}=\cfrac{3-4x}{3x-4}=2$

$\Rightarrow 3-4x=6x-8$

$\Rightarrow B = x=\cfrac{11}{10}$

$\therefore 100\times (A+B) = 35$

If $f(x)=\dfrac{(4x+3)}{(6x-4)}$ , $x\neq \dfrac{2}{3}$ , show that $fof(x)=x$ , for all $x\neq\dfrac{2}{3}$ . What is the inverse of f?

$f(x)=\cfrac{4x+3}{6x-4}$

$fof(x)=\cfrac{4(\cfrac{4x+3}{6x-4})+3}{6(\cfrac{4x+3}{6x-4})-4}$

$=\cfrac{16x+12+18x-12}{24x+18-24+16}$

$=\cfrac{34x}{34}=x.$

$y=\cfrac{4x+3}{6x-4}$

$6xy-4y=4x+3$

$6xy-4x=4y+3$

$x=\cfrac{4y+3}{6y-4}$

$\therefore f(x)=\cfrac{4x+3}{6x-4}$

Let $f: N\rightarrow Y$ be a function defined as $f(x)=4x+3$ , where, $Y=\{y\in N : y=4x+3\ for\ some\ x\in N\}$ , Show that f is invertible. Find the inverse.

$f(x)=4x+3$

$\rightarrow$ let

$f(x_{1})=f(x_{2}), x_{1}x_{2}\in N$

$4x_{1}+3=4x_{2}+3$

$\therefore 4x_{1}=4x_{2}$

$\therefore x_{1}=x_{2}$

Thus

$f(x_{1})=f(x_{2})\Rightarrow x_{1}=x_{2}$ Hence the

$fx$ is one-one.

Let

$y\in y$ be a number of the form

$y=4k+3$

$y=f(x)$

$\therefore 4k+3=4x+3$

$\therefore k=x$

$\rightarrow$ This corresponding to any

$y\in y$ we have

$x\in N$

The function is onto

the function being both one one and onto is invertible

$y=4x+3$

$\therefore y-3=4x$

$\therefore x=\dfrac{y-3}{4}$

$\therefore f(x)=\dfrac{x-3}{4}$ or

$g(y)=\dfrac{4-3}{4}$ is inverse function.

Let $f:N\rightarrow N$ be a function defined as $f(x)=4{x}^{2}+12x+15$ . Show that $f:N\rightarrow S$ is invertible. Find the inverse of $f$ and hence find ${f}^{-1}(31)$ and ${f}^{-1}(87)$ .

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

$y-15={4x}^{2}+12x$

${4x}^{2}+12x+15-y=0$

$x=\cfrac { -12\pm \sqrt { 144-4(15-y) } \times 4 }{ 8 }$

$=\cfrac { -12\pm \sqrt { 144-(60-4y)\times 4 } }{ 8 }$

$=\cfrac { -12\pm \sqrt { 144-240-{16y}^{2}} }{ 8 }$

$=\cfrac { -12\pm \sqrt { {16y}^{2}-6 } }{ 8 }$

$=\cfrac { -3\pm \sqrt { {y}^{2}-6 } }{ 2 }$

${f}^{-1}(31)$

$=\cfrac { -3\pm \sqrt { {31}^{2}-6 } }{ 2 }\\=\cfrac{-3\pm30.90}{2}\\=\cfrac{27.9}{2} \text{ or } \cfrac{-33.90}{2}\\=13.95 \text{ or }-16.95$

${f}^{-1}(87)$

$=\cfrac{-3\pm \sqrt{{87}^{2}-6}}{2}\\ =\cfrac{-3\pm 86.96}{2}\\=41.98 \text{ or }-44.98$

$\therefore$

${f}^{-1}(31)=13.95$ or

$-16.95,$

${f}^{-1}(87)=41.98$ or

$-44.98$ .

Show that $f:\left[ -1,1 \right] \rightarrow R$ , given by $f\left( x \right) =\dfrac { x }{ \left( x+2 \right) }$ is one-one. Find the inverse of the function $f:\left[ -1,1 \right] \rightarrow Range\ f.$

Given

$f:[-1, 1]\to R$

$f(x)=\dfrac {x}{x+2}$

To show, one-one we have to show , if two different element of Domain gives same, then they are equal

Let

$x, y \in [-1, 1]$

Let

$f(x)=f(y)$

$\Rightarrow \ \dfrac {x}{x+2}=\dfrac {y}{y+2}$

$\Rightarrow \ xy+2x=xy+2y$

$\Rightarrow \ 2x=2y$

$\Rightarrow \ x=y$

Hence they function is one-one

Now let

$y=\dfrac {x}{x+2}\ \Rightarrow \ yx+2y=x$

$\Rightarrow \ x(y-1)=-2y$

$\Rightarrow \ x=\dfrac {2y}{1-y}$

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

Work out the inverse function for each mapping.

$x \rightarrow 5 x + 1$

$x \rightarrow 3 x - 7$

$x \rightarrow \dfrac { x } { 5 }$

$x \rightarrow \dfrac { x + 9 } { 4 }$

1099715_1175669_ans_96bbd1139f8b46ad981b48ed516d6527.jpg

Let $f$ be any real function and let $g$ be a function given by $g(x)=2x$ . Prove that $g\circ f=f+f$ .

$f:R\rightarrow R$ ......... [ Given ]

Since,

$g(x)=2x$ is a polynomial,

$g:R\rightarrow R$

Clearly,

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

$f+f:R\rightarrow R$

So, domains of

$g\circ f$ and

$f+f$ are the same.

$\Rightarrow$

$(g\circ f)(x)=g[f(x)]$

$=2f(x)$ ......... [ Since,

$g(x)=2x$ ]

$\Rightarrow$

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

$=2f(x)$

We can see that,

$\Rightarrow$

$(g\circ f)(x)=(f+f)(x),$

$\forall\,x\in R$

$\therefore$

$g\circ f=f+f$

If $f:\left (-\dfrac {\pi}{2},\dfrac {\pi}{2}\right)\rightarrow R$ and $g:[-1,1]\rightarrow R$ be defined as $f(x)=\tan x$ and $g(x)=\sqrt{1-{x}^{2}}$ respectively. Describe $f\circ g$ and $g\circ f$ .

$f:\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\rightarrow R$ and

$g:[-1,1]\rightarrow R$ defined as

$f(x)=\tan x$ and

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

Range of

$f:$

Let

$y=f(x)$

$\Rightarrow$

$y=\tan x$

$\Rightarrow$

$x=\tan^{-1}y$

Since,

$x\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right),\,y\in(-\infty,\infty)$

$\therefore$ Range of

$f$ is subset of domain

$g=[-1,1]$

$\therefore$

$g\circ f$ exists.

By similar argument

$f\circ g$ exists.

$\Rightarrow$

$f\circ g(x)=f[g(x)]$

$=f(\sqrt{1-x^2})$ ......... [ Since,

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

$=\tan\sqrt{1-x^2}$ ......... [ Since,

$f(x)=\tan x$ ]

$\Rightarrow$

$g\circ f(x)=g[f(x)]$

$=g(\tan x)$ ......... [ Since,

$f(x)=\tan x$ ]

$=\sqrt{1-\tan^2x}$ ......... [ Since,

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

If $f(x)=2x+5$ and $g(x)={x}^{2}+1$ be two real functions, then describe given functions:
(i) $f\circ g$
(ii) $g\circ f$
(iii) $f\circ f$
(iv) ${f}^{2}$
Also, show that $f\circ f\ne {f}^{2}$ .

$f\circ g$ means

$g(x)$ function is in

$f(x)$ function.

$g\circ f$ means

$f(x)$ function is in

$g(x)$ function.

$f(x)$ and

$g(x)$ are polynomials.

$\Rightarrow$

$f:R\rightarrow R$ and

$g:R\rightarrow R.$

So,

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

$g\circ f:R\rightarrow R$

$(i)$

$(f\circ g)(x)=f[g(x)]$

$=f(x^2+1)$ ......... [ Since,

$g(x)=x^2+1$ ]

$=2(x^2+1)+5$ ......... [ Since,

$f(x)=2x+5$ ]

$=2x^2+2+5$

$=2x^2+7$

$(ii)$

$(g\circ f)(x)=g[f(x)]$

$=g(2x+5)$ ......... [ Since,

$f(x)=2x+5$ ]

$=g(2x+5)^2+1$ ......... [ Since,

$g(x)=x^2+1$ ]

$=4x^2+20x+26$

$(iii)$

$(f\circ f)(x)=f[f(x)]$

$=f(2x+5)$ ......... [ Since,

$f(x)=2x+5$ ]

$=2(2x+5)+5$

$=4x+10+5$

$=4x+15$ ---- ( 1 )

$(iv)$

$f^2(x)=f(x)\times f(x)$

$=(2x+5)(2x+5)$

$=(2x+5)^2$

$=4x^2+20x+25$ ----- ( 2 )

From ( 1 ) and ( 2 ),

$\Rightarrow$

$(f\circ f)(x)\ne f^2(x)$

$\therefore$

$f\circ f\ne f^2$

If $f(x)=\sqrt{1-x}$ and $g(x)=\log _{ e }{ x }$ are two real functions, then describe functions $f\circ g$ and $g\circ f$ .

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

For domain,

$1-x\ge 0$

$\Rightarrow$

$x\le 1$

$\Rightarrow$ Domain of

$f=(-\infty , 1]$

$\Rightarrow$

$f:(\infty , 1]\rightarrow (0,\infty)$

$g(x)=\log_e x$

Clearly, the range of

$g$ is not a subset of the domain of

$f$ .

So, we need to compute the domain of

$f\circ g.$

$\Rightarrow$ Domain

$(f\circ g)=(0,e)\rightarrow R$

$\Rightarrow$

$(f\circ g)(x)=f[g(x)]$

$=f(\log_e x)$

$=\sqrt{1-\log_e x}$

The range of

$f$ is a subset of the domain of

$g.$

$\Rightarrow$

$g\circ f:(-\infty , 1]\rightarrow R$

$\Rightarrow$

$(g\circ f)(x)=g[f(x)]$

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

$=\log_e(1-x)^{\frac{1}{2}}$

$=\dfrac{1}{2}\log_e(1-x)$

If $f,g:R\rightarrow R$ be two functions defined as $f(x)=\left| x \right| +x$ and $g(x)=\left| x \right| -x$ for all $x\in R$ . Then, find $f\circ g$ and $g\circ f$ . Hence, find $f\circ g(-3)$ , $f\circ g (5)$ and $g\circ f(-2)$ .

$f(x)=|x|+x$ ......... [ Given ]

$g(x)=|x|-x,$

$\forall x\in R$

$\Rightarrow$

$f\circ g=f[g(x)]$

$=f(|x|-x)$ ......... [ Since,

$g(x)=|x|-x$ ]

$=||x|-x||+(|x|-x)$ ......... [ Since,

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

$\Rightarrow$

$f(g(x))=\begin{cases}0&x\ge 0\\4x& x<0\end{cases}$

$\Rightarrow$

$f(g(x))=\begin{cases}4x& x>0\\ 0& x\ge 0\end{cases}$

$\Rightarrow$

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

$=g(|x|+x)$ ......... [ Since,

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

$=||x|+x|-(|x|+x)$ ......... [ Since,

$g(x)=|x|-x$ ]

$\Rightarrow$

$g(f(x))=\begin{cases}0&x\ge 0\\0& x<0\end{cases}$

$\therefore$

$g(f(x))=g\circ f=0$

$\Rightarrow$

$f\circ g(-3)=4(-3)=-12$ ......... [ Since,

$f\circ g=4$ for

$x<0$ ]

$\Rightarrow$

$f\circ g(5)=0$ ......... [ Since,

$f\circ g=0$ for

$x\ge 0$ ]

$\Rightarrow$

$g\circ f(-2)=0$ ......... [ Since,

$g\circ f=0$ for

$x< 0$ ]

If $f(x)=\sqrt{x+3}$ and $g(x)={x}^{2}+1$ be two real functions, then find $f\circ g$ and $g\circ f$ .

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

For domain,

$x+3\ge 0$

$\Rightarrow$

$x\ge -3$

Domain of

$f=[-3,\infty)$

Range of

$f=(-3,\infty)$

Similarly, range of

$g=(1,\infty)$

Then, rangle of f is subset of domain

$g$ and range of

$g$ is subset of

$f$ .

$\therefore$

$f\circ g$ and

$g\circ f$ exist.

$\Rightarrow$

$f\circ g(x)=f[g(x)]$

$=f(x^2+1)$ ......... [ Since,

$g(x)^2+1$ ]

$=\sqrt{x^2+1+3}$ ......... [ Since,

$f(x)=\sqrt{x+3}$ ]

$=\sqrt{x^2+4}$

$\Rightarrow$

$g\circ f(x)=g[f(x)]$

$=g(\sqrt{x+3})$ ......... [ Since,

$f(x)=\sqrt{x+3}$ ]

$=(\sqrt{x+3})^2+1$ ......... [ Since,

$g(x)^2+1$ ]

$=x+3+1$

$=x+4$

Let $A=\left\{ x\in R|-1\le x\le 1 \right\}$ and let $f:A\rightarrow A,g:A\rightarrow A$ be two functions defined by $f(x)={x}^{2}$ and $g(x)=\sin \cfrac { \pi x }{ 2 }$ .Show that ${g}^{-1}$ exists but ${f}^{-1}$ does not exist. Also find ${g}^{-1}$ .

$f$ is not one-one because

$f\left(-1\right)={\left(-1\right)}^{2}=1$

and

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

$\Rightarrow\,-1$ and

$1$ have the same image under

$f$

$\Rightarrow\,f$ is not a bijection.

So,

${f}^{-1}$ does not exist.

Injectivity of

$g$

Let

$x$ and

$y$ be any two elements in the domain of

$A$ such that

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

$\Rightarrow\,\sin{\dfrac{\pi x}{2}}=\sin{\dfrac{\pi y}{2}}$

$\Rightarrow\,\dfrac{\pi x}{2}=\dfrac{\pi y}{2}$

$\Rightarrow\,x=y$

So,

$g$ is one-one.

Surjectivity of

$g:$

Range of

$g=\left[\sin{\left(-\dfrac{\pi}{2}\right),\sin{\dfrac{\pi}{2}}}\right]=\left[-1,1\right]=A$ (co-domain of

$g$ )

$\Rightarrow\,g$ is onto

$\Rightarrow\,g$ is a bijection.

So,

${g}^{-1}$ exists.

Also,

Let

${g}^{-1}{\left(x\right)}=y$ .......

$(1)$

$\Rightarrow\,g\left(y\right)=x$

$\Rightarrow\,\sin{\dfrac{\pi y}{2}}=x$

$\Rightarrow\,y=\dfrac{2}{\pi}{\sin}^{-1}{x}$

$\Rightarrow\,{g}^{-1}{\left(x\right)}=\dfrac{2}{\pi}{\sin}^{-1}{x}$ from

$(1)$

Let $f:\rightarrow R$ , $g:R\rightarrow R$ be two functions defined by $f(x)={x}^{2}+x+1$ and $g(x)=1-{x}^{2}$ . Write $f\circ g(-2)$ .

$f\circ g$ means

$g(x)$ function is in

$f(x)$ function.

$\Rightarrow$

$(f\circ g)(x)=f[g(x)]$

$=f(1-x^2)$

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

$\Rightarrow$

$(f\circ g)(x)=(1-x^2)^2+(1-x^2)+1$

$\Rightarrow$

$(f\circ g)(-2)=[1-(-2)^2]^2+[1-(-2)^2]+1$

$=[1-4]^2+[1-4]+1$

$=[-3]^2-3+1$

$=9-3+1$

$=7$

If $f:\left\{ 5,6 \right\} \rightarrow \left\{ 2,3 \right\}$ and $g:\left\{ 2,3 \right\} \rightarrow \left\{ 5,6 \right\}$ are given by $f=\left\{ \left( 5,2 \right) ,\left( 6,3 \right) \right\}$ and $g=\left\{ \left( 2,5 \right) ,\left( 3,6 \right) \right\}$ , find $f\circ g$ .

$f\circ g$ means

$g(x)$ function is in

$f(x)$ function.

$f:\{5,6\}\rightarrow \{2,3\}$ and

$g:\{2,3\}\rightarrow \{5,6\}$ given by

$f=\{(5,2),(6,3)\}$ and

$g=\{(2,5),(3,6)\}$

$\Rightarrow$

$f\circ g(2)=f[g(2)]$

$=f(5)$

$=2$

$\Rightarrow$

$f\circ g(3)=f[g(3)]$

$=f(6)$

$=3$

So,

$f\circ g:\{2,3\}\rightarrow \{2,3\}$ is defined as

$\Rightarrow$

$f\circ g=\{(2,2),(3,3)\}$

Let f(x) be a continuous and g(x) is a discontinuous function then prove that f(x)+g(x) is discontinuous at x=a

Given that f(x) is a continuous function and g(x) is a discontinuous function then for some arbitrary real number a we must have

$\underset{x\rightarrow a}{\lim}f(x)=f(a)$

and

$\underset{x\rightarrow a}{\lim}g(x)\neq g(a)$

Now,

$\underset{x\rightarrow a}{lim}[f(x)+g(x)]$

$=\underset{x\rightarrow a}{lim}f(x)+\underset{x\rightarrow a}{lim}g(x)\neq f(a)+g(a)$ [Using equations (1) and(2)]

$\Rightarrow$

$f(x)+g(x)$ is discontinuous

If $f$ is an invertible function, define as $f\left ( x \right )=\cfrac{3x-4}{5}$ , write $f^{-1}\left ( x \right )$ .

let,

$f(x)=y=\cfrac{3x-4}{5}$

So, to find inverse of

$f(x)$ we need

$x$ in terms of

$y$ which is

$f^{-1}(x)$

$\therefore y=\cfrac{3x-4}{5}$

$\Rightarrow 5y=3x-4$

$\Rightarrow 5y+4=3x$

$\Rightarrow x=\cfrac{5y+4}{3}$

$\therefore f^{-1}(y)=\cfrac{5y+4}{3}$

$\therefore f^{-1}(x)=\cfrac{5x+4}{3}$ ......

$\{$ replacing

$y$ with

$x\}$

If $A=\left\{a,b,c,d \right\}$ and $f=\left\{(a,b),(b,d),(c,a),(d,c)\right\}$ , show that $f$ is one-one from $A$ onto $A$ . Find $f^{-1}$

For

$f^{-1}$ to exist the function should be one-one and onto.

Here,

$f$ is one=one since each element of

$A$ is assigned to distinct element of the set

$A$ . Also,

$f$ is onto since

$f (A)=A$ .

Moreover,

$f^{-1}=\left\{(b,a),(d,b),(a,c),(c,d)\right\}$

Let $f:R\to R$ be the function defined by $f(x)=4x-3\forall x\in R$ . Then write $f^{-1}$ .

Given that

$f(x)=4x-3=y(say)$ , then

To find the

$f^{-1}$ we must represent

$x$ in terms of the variable

$y$ .

So,

$4x=y+3$

$\Rightarrow \ x=\dfrac {y+3}{4}$
Hence

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

Show that $f;[-1,1]\rightarrow R$ , given by $f(x)=\dfrac{x}{(x+2)}$ is one-one. Find the inverse of the function $f;[-1,1]\rightarrow$ Range $f$ .
(Hint: For $y\in$ Range, $f, y=f(x)=\dfrac{x}{x+2}$ for some $x$ in $[-1,1]$
$x=\dfrac{2y}{(1-y)}$

Let

$x_{1}, x_{2}\in D_{f}$

$f(x_{1})=f(x_{2})\rightarrow \dfrac{x_{1}}{x_1+2}=\dfrac{x_2}{x_2+2}$

$\Rightarrow x_1(x_2+2)=x_2(x_1+2)$

$\Rightarrow x_1x_2+2x_1=x_1x_2+2x_2$

$\Rightarrow x_1=x_2$
hence

$f$ is one-one
let

$y=\dfrac{x}{x+2}\Rightarrow xy+2y=x$

$\Rightarrow xy-x=-2y$

$\Rightarrow x(1-y(|)=2y$

$\Rightarrow x=\dfrac{2y}{1-y}$

$\forall\ y\ \in\ R, \exists\ x\ \in\ D_f$ such that

$f\left(\dfrac{2y}{1-y}\right)=y$ , hence

$f$ is onto

$\therefore$ Inverse exist

$\therefore f^{-1}(x)=\dfrac{2x}{1-x}\forall\ x\ \in\ D_{f-1}=R_f=R-\left\{1\right\}$

Consider $f:R\rightarrow R$ given by $f(x)=4x+3$ . Show that $f$ is invertible. Find the inverse of $f$ .

Let

$x_1, x_2\in R$

$f(x_1)=f(x_2)\Rightarrow 4x_1+3=4x_2+3$

$\Rightarrow x_1=x_2$
hence

$f$ is one-one
let

$y=4x+3\Rightarrow y-3=4x$

$\Rightarrow x=\dfrac{y-3}{4}$

$\forall\ y\ \in\ R\ \exists\ x\ \in\ R$ such that

$f\left(\dfrac{y-3}{4}\right)=\dfrac{4(y-3)}{4}+3=y$
hence inverse exist. And

$f^{-1}:R\rightarrow R$ such that

$f^{-1}(y)=\dfrac{y-3}{4}$ .

Consider $f : R - \left \{ -\dfrac{4}{3} \right \} \rightarrow R - \left \{ \dfrac{4}{3} \right \}$ given by $f(x) = \dfrac{4x + 3}{3x + 4}.$ Show that f is
bijective. Find the inverse of f and hence find $f^{-1} (0)$ and x such that $f^{-1}(x) = 2.$

Let

$x_1 , x_2 \in R - \left\{\dfrac{-4}{3}\right\}$ and

$f(x_1) = f(x_2)$

$\therefore \dfrac{4x_1 + 3}{3x_1 + 4} = \dfrac{4x_2 + 3}{3x_2 + 4} \Rightarrow (4x_1 + 3)( 3x_1 + 4) = (4x_2 + 3)(3x_1 + 4)$

$\Rightarrow 12x_1 x_2 + 16x_1 + 9x_2 + 12 = 12x_1 x_2 + 16x_2 + 9 x_2 + 12$

$\Rightarrow 16(x_1 - x_2) - 9 (x_1 - x_2) = 0 = 7(x_1 - x_2) = 0$

$\Rightarrow x_1 - x_2 = 0 \Rightarrow x_1 = x_2$

$\Rightarrow f$ is a one one function.

Let

$y = \dfrac{4x + 3}{3x + 4}$ for

$y \in R - \left\{\dfrac{4}{3}\right\}$

$\Rightarrow 3xy + 4y = 4x + 3 \Rightarrow 4x - 3xy = 4y - 3 \Rightarrow x(4 - 3y) = 4y - 3$

$\Rightarrow x = \dfrac{4y - 3}{4 - 3y} \Rightarrow \forall y \in R - \left\{\dfrac{4}{3}\right\}, x \ \in \ R - \left\{-\dfrac{4}{3} \right\}$

$\Rightarrow f$ is onto function and hence it is bijective.

Now,

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

for

$f^{-1} (x) = 2 , \dfrac{4x - 3}{4 - 3x} = 2 \Rightarrow 4x - 3 = 8 - 6x$

$\Rightarrow 10x = 11 \Rightarrow x = \dfrac{11}{10}$

Prove that the function $f : N \to N,$ defined by $f(x) = x^2 + x + 1$ is one-one but not onto. Find inverse of $f : N \to S$ , where $S$ is range of $f$ .

$f(x)=x^2+x+1;\,\,f:N\rightarrow N$ (Given)

Let

$x_1,x_2\,\epsilon \,D_f$ and

$f(x_1)=f(x_2)$

$\Rightarrow x^2_1+x_1+1=x^2_2+x_2+1\Rightarrow x^2_1-x^2_2+x_1-x_2=0$

$\Rightarrow (x_1-x_2)(x_1+x_2)+(x_1-x_2)=0\Rightarrow (x_1+x_2+1)(x_1-x_2)=0$

$\Rightarrow x_1-x_2=0\,\,[\because x_1+x_2+1\neq 0\,for\,any\,x\epsilon N]$

$\Rightarrow x_1=x_2\Rightarrow f$ is one-one function.

Now,

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

$\Rightarrow f(x)=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}$ ........increasing function.

$\because f(1)=3\Rightarrow f(x)\ge 3$ for energy

$x\epsilon N$

$\Rightarrow f(x)$ will not assume the values

$1$ and

$2$

$\Rightarrow f$ is not onto function.

Now, let

$y=x^2+x+1\Rightarrow x^2+x+1-y=0$

$\Rightarrow x=\dfrac{-1\pm \sqrt{1-4(1-y)}}{2}$ [Using Discriminant rule]

$x=\dfrac{-1\pm \sqrt{4y-3}}{2}\Rightarrow f^{-1}(x)=\dfrac{-1\pm \sqrt{4x-3}}{2}$

$\Rightarrow f^{-1}(x)=\dfrac{-1+\sqrt{4x-3}}{2}\,\,[\because f(1)=3\Rightarrow f^{-1}(3)=1]$

Let $g$ be a real valued differentiable function on $R$ such that $g(x) = 3e^{x-2}+4\displaystyle\int_{2}^{x}\sqrt{2t^{2}+6t+5}dt \quad\forall x \in R$ and let $g^{-1}$ be the inverse function of $g$ . If $(g^{-1})' (3)$ is equal to $\dfrac{p}{q}$ where $p$ and $q$ are relatively prime, then find $\dfrac{p+q}{6}$

$g\left( x \right) =3{ e }^{ x-2 }+4\int _{ 2 }^{ x }{ \sqrt { 2{ t }^{ 2 }+6t+5 } } dt\\ g\left( x \right) =y\\ \Rightarrow g^{ -1 }\left( y \right) =x$

Differentiating the above equation w.r.t

$x$ gives

$g^{ { -1 }^{ ' } }\left( g\left( x \right) \right) g^{ ' }\left( x \right) =1$

$\Rightarrow g^{ { -1 }^{ ' } }\left( g\left( x \right) \right) =\dfrac { 1 }{ g^{ ' }\left( x \right) }$

When

$\displaystyle x=2,\quad g\left( x \right) =3{ e }^{ 2-2 }+4\int _{ 2 }^{ 2 }{ \sqrt { 2{ t }^{ 2 }+6t+5 } } dt= 3+0=3$

$g'\left( x \right) =3{ e }^{ x-2 }+4\sqrt { 2{ x }^{ 2 }+6x+5 }$

$g'(2) = 23$

$x=2 g(x)=3$ implies

$(g)^{-1}(3)=2$

Now,

$g^{-1'}(g(x)) = g^{-1'}(3) = \dfrac1{g'(2)}$

$g^{ { -1 }^{ ' } }\left( 3 \right) =\dfrac { 1 }{ g^{ ' }\left( 2 \right) } =\dfrac { 1 }{ 23 } =\dfrac { p }{ q }$

$\therefore p+q=24\ (componendo) \therefore \dfrac { p+q }{ 6 } =4\\ \quad \quad \quad \quad$

Let $A = R - \left \{3\right \}$ and $B = R - \left \{1\right \}$ . Consider the function $f: A\rightarrow B$ defined by $f(x) = \left (\dfrac {x - 2}{x - 3}\right )$ Show that f is one-one and onto and hence find $f^{-1}$

Let $x_1, x_2 \epsilon A$ .
Now, $f(x_1)=f(x_2)\Rightarrow \dfrac{x_1-2}{x_1-3}=\dfrac{x_2-2}{x_2-3}$
$(x_1-2)(x_2-3)=(x_1-3)(x_2-2)$
$x_1x_2-3x_1-2x_2+6=x_1x_2-2x_1-3x_2+6$
$-3x_1-2x_2=-2x_1-3x_2$
$-x_1=-x_2\Rightarrow x_1=x_2$
Hence f is one-one function.
For Onto:
Let $y=\dfrac{x-2}{x-3}$
$xy-3y=x-2\Rightarrow xy-x=3y-2$
$x(y-1)=3y-2$
$x=\dfrac{3y-2}{y-1}$ --- (1)
From above it is obviously that $\forall$ y except 1, i.e., $\forall y \epsilon B=R-[1]\exists x\epsilon A$ .
Hence f is onto function.
Thus f is one-one function.
It $f^{-1}$ is inverse of f then $f^{-1}(y)=\dfrac{3y-2}{y-1}$ [from (1)].

Let $f$ be a real function given by $f(x)=\sqrt{x-2}$ . Find the following.
(i) $f\circ f$ (ii) $f\circ f\circ f$ (iii) $(f\circ f\circ f)(38)$ (iv) ${f}^{2}$
Also, show that $f\circ f\ne {f}^{2}$ .

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

For domain,

$\Rightarrow$

$x-2\ge 0$

$\Rightarrow$

$x\ge 2$

Domain of

$f=[2,\infty)$

Since,

$f$ is a square root function, range of

$f=(0,\infty)$

So,

$f:[2,\infty)\rightarrow (0,\infty)$

$(i)$

$f\circ f$

Range of

$f$ is not a subset of the domain of

$f$ .

$\Rightarrow$ Domain

$(f\circ f)=\{x:x:\in[2,\infty)\,and\,\sqrt{x-2}\in[2,\infty)\}$

$=\{x:x\in [2,\infty)\,and\,\sqrt{x-2}\ge 2\}$

$=\{x:x:\in[2\infty)\,and\,x-2\ge 4\}$

$=\{x:x\in[2,\infty)\,and\,x\ge 6\}$

$=[6,\infty)$

$\Rightarrow$

$(f\circ f)(x)=f(f(x))$

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

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

$(ii)$

$f\circ f\circ f=(f\circ f)\circ f$

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

We have

$f:[2,\infty)\rightarrow (0,\infty)$ and

$f\circ f:[6,\infty)\rightarrow R$

$\Rightarrow$ Range of

$f$ is not a subset of the domain of

$f\circ f.$

$\Rightarrow$ Domain

$((f\circ f)\circ f)=\{x:x\in$ Domain of

$f$ and

$f(x)\in$ Domain of

$f\circ f\}$

$=\{x:x\in [2,\infty)\,and\,\sqrt{x-2}\in[6,\infty)\}$

$=\{x:x\in[2,\infty)\,and\,\sqrt{x-2}\ge 6\}$

$=\{x:x\in[2,\infty)\,and\,x-2\ge 36\}$

$=\{x:x\in[2,\infty)\,and\,x\ge 38\}$

$=\{x:x\ge 38\}$

$=[38,\infty)$

$\Rightarrow$

$f\circ f:[38,\infty)\rightarrow R$

$\Rightarrow$

$f\circ f\circ f(x)=(f\circ f)\circ f(x)$

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

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

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

$(iii)$

$(f\circ f\circ f)(x)$

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

So,

$(f\circ f\circ f)(38)$

$=\sqrt{\sqrt{\sqrt{38-2}-2}-2}$

$=\sqrt{\sqrt{\sqrt{36}-2}-2}$

$=\sqrt{\sqrt{6-2}-2}$

$=\sqrt{2-2}$

$=0$

$(iv)$ We hae,

$f\circ f=\sqrt{\sqrt{x-2}-2}$

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

$\therefore$ from

$(i)$ and

$(iv)$ we can say that,

$\therefore$

$f\circ f\ne f^2$

If $f:R\rightarrow R$ be defined by $f(x)={x}^{3}-3$ , then prove that ${f}^{-1}$ exists and find a formula for ${f}^{-1}$ . Hence, find ${f}^{-1}(24)$ and ${f}^{-1}(5)$ .

$one-one$ test of

$f:$

Let

$x$ and

$y$ be two elements in domain

$(R),$

Such that,

$f(x)=f(y)$

$\Rightarrow$

$x^3-3=y^3-3$

$\Rightarrow$

$x^3=y^3$

$\Rightarrow$

$x=y$

$\therefore$

$f$ is one-one.

$onto$ test for

$f:$

Let

$y$ be in the co-domain

$(R)$ , such that

$f(x)=y$

$\Rightarrow$

$x^3-3=y$

$\Rightarrow$

$x^3=y+3$

$\Rightarrow$

$x=\sqrt[3]{y+3}\in R$

Since,

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

$\therefore$

$f$ is invertible.

Find

$f^{-1}:$

Let

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

$\Rightarrow$

$x=f(y)$

$\Rightarrow$

$x=y^3-3$

$\Rightarrow$

$x+3=y^3$

$y=\sqrt[3]{x+3}$

From ( 1 ),

$f^{-1}=\sqrt[3]{x+3}$

$\Rightarrow$ Now,

$f^{-1}(24)=\sqrt[3]{24+3}$

$=\sqrt[3]{27}$

$=3$

$\Rightarrow$ Now,

$f^{-1}(5)=\sqrt[3]{5+3}$

$=\sqrt[3]{8}$

$=2$

Let $f:N\rightarrow N$ be a function defined as $f(x)=9{x}^{2}+6x-5$ . Show that $f:N\rightarrow S$ , where $S$ is the range of $f$ , is invertible. Find the inverse of $f$ and hence find ${f}^{-1}(43)$ and ${f}^{-1}(163)$ .

$f:N\rightarrow N$ is a function defined as

$f(x)=9x^2+6x-5$

Let

$y=f(x)=9x^2+6x-5$

$\Rightarrow$

$y-9x^2+6x-5$

$\Rightarrow$

$y=9x^2+6x+1-1-5$

$\Rightarrow$

$y=(9x^2+6x+1)-6$

$\Rightarrow$

$y=(3x+1)^2-6$

$\Rightarrow$

$y+6=(3x+1)^2$

$\Rightarrow$

$\sqrt{y+6}=3x+1$

$(\because y\in N)$

$\Rightarrow$

$\sqrt{y+6}-1=3x$

$\Rightarrow$

$x=\dfrac{\sqrt{y+6}-1}{3}$

Let

$x=g(y)$

$\Rightarrow$

$g(y)=\dfrac{\sqrt{y+6}-1}{3}$

Now,

$f\circ g(y)=f[g(y)]$

$=f\left(\dfrac{\sqrt{y+6}-1}{3}\right)$

$=9\left(\dfrac{\sqrt{y+6}-1}{3}\right)^2+6\left(\dfrac{\sqrt{y+6}-1}{3}\right)-5$

$=9\left(\dfrac{y+6-2\sqrt{y+6}+1}{9}\right)-5$

$=9\left(\dfrac{y+6-2\sqrt{y+6}+1}{9}\right)+2(\sqrt{y+6}-1)-5$

$=y+6-2\sqrt{y+6}+1+2\sqrt{y+6}-2-5$

$=y$

$=I_y,$ Identify function

$\Rightarrow$

$g\circ f(x)=g[f(x)]$

$=g(9x^2+6x-5)$

$=\dfrac{\sqrt{(9x^2+6x-5)+6}-1}{3}$

$=\dfrac{\sqrt{(9x^2+6x+1)}-1}{3}$

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

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

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

$=x$

$=I_x$ , Indenfity function.

Since,

$f\circ g(y)$ and

$g(g\circ f(x)$ are identify function.

$\therefore$

$f$ is invertible.

So,

$f^{-1}(x)=g(x)=\dfrac{\sqrt{x+6}-1}{3}$

Now,

$\Rightarrow$

$f^{-1}(43)=\dfrac{\sqrt{43+6}-1}{3}$

$=\dfrac{\sqrt{46}-1}{3}$

$=\dfrac{7-1}{3}$

$=\dfrac{6}{3}$

$=2$ .

$\Rightarrow$

$f^{-1}(163)=\dfrac{\sqrt{163+6}-1}{3}$

$=\dfrac{\sqrt{169}-1}{3}$

$=\dfrac{13-1}{3}$

$=\dfrac{12}{3}$

$=4$

Let $f(x) =\left\{\begin{matrix} x+a & \text{if} \ x < 0 \\ |x-1| & \text{if} \ x \ge 0 \end{matrix}\right.$ and
$g(x) = \left\{\begin{matrix} x+1 & \text{if} \ x < 0\\ (x-1)^2+b & \text{if} \ x \ge 0, \end{matrix}\right.$
Where a and b are non-negative real number. Determine the composite function g o f. if is continuous for all real x, determine the Values of a and b. Further for these values of a and b is g o f differentiable at x=0? Justify your answer

1976015_1752592_ans_3a0fce8756e44abb997fb60f2f98f4df.jpeg

Relations And Functions - Class 12 Engineering Maths - Extra Questions

Show that if $f:A\rightarrow B$ and $g:B\rightarrow C$ are one-one, then $g\circ f:A\rightarrow C$ is also one-one.

Let $f:R\rightarrow R$ be defined by $f(x)={x}^{3}+5$ then find ${f}^{-1}(x)$

If the function $f:[1,\infty)\rightarrow [1,\infty)$ is defined by $f(x)=2^{x(x-1)}$ ,then find $f^{-1}(4)$

Let $f: R\rightarrow R$ be defined as $f(s) = 10x + 7$ . Find the function $g: R\rightarrow R$ such that $gof = fog = 1_{R}$

Let $f:R\rightarrow R$ be defined by $f(x)=2x+3$ , find ${f}^{-1}(4)$ .

If $f(x)=\dfrac{(4x+3)}{(6x-4)}$ , $x\neq \dfrac{2}{3}$ , show that $fof(x)=x$ , for all $x\neq\dfrac{2}{3}$ . What is the inverse of f?

Let $f: N\rightarrow Y$ be a function defined as $f(x)=4x+3$ , where, $Y=\{y\in N : y=4x+3\ for\ some\ x\in N\}$ , Show that f is invertible. Find the inverse.

Let $f:N\rightarrow N$ be a function defined as $f(x)=4{x}^{2}+12x+15$ . Show that $f:N\rightarrow S$ is invertible. Find the inverse of $f$ and hence find ${f}^{-1}(31)$ and ${f}^{-1}(87)$ .

Show that $f:\left[ -1,1 \right] \rightarrow R$ , given by $f\left( x \right) =\dfrac { x }{ \left( x+2 \right) }$ is one-one. Find the inverse of the function $f:\left[ -1,1 \right] \rightarrow Range\ f.$

Work out the inverse function for each mapping.

$x \rightarrow 5 x + 1$

$x \rightarrow 3 x - 7$

$x \rightarrow \dfrac { x } { 5 }$

$x \rightarrow \dfrac { x + 9 } { 4 }$

Let $f$ be any real function and let $g$ be a function given by $g(x)=2x$ . Prove that $g\circ f=f+f$ .

If $f:\left (-\dfrac {\pi}{2},\dfrac {\pi}{2}\right)\rightarrow R$ and $g:[-1,1]\rightarrow R$ be defined as $f(x)=\tan x$ and $g(x)=\sqrt{1-{x}^{2}}$ respectively. Describe $f\circ g$ and $g\circ f$ .

If $f(x)=2x+5$ and $g(x)={x}^{2}+1$ be two real functions, then describe given functions:
(i) $f\circ g$
(ii) $g\circ f$
(iii) $f\circ f$
(iv) ${f}^{2}$
Also, show that $f\circ f\ne {f}^{2}$ .

If $f(x)=\sqrt{1-x}$ and $g(x)=\log _{ e }{ x }$ are two real functions, then describe functions $f\circ g$ and $g\circ f$ .

If $f,g:R\rightarrow R$ be two functions defined as $f(x)=\left| x \right| +x$ and $g(x)=\left| x \right| -x$ for all $x\in R$ . Then, find $f\circ g$ and $g\circ f$ . Hence, find $f\circ g(-3)$ , $f\circ g (5)$ and $g\circ f(-2)$ .

If $f(x)=\sqrt{x+3}$ and $g(x)={x}^{2}+1$ be two real functions, then find $f\circ g$ and $g\circ f$ .

Let $A=\left\{ x\in R|-1\le x\le 1 \right\}$ and let $f:A\rightarrow A,g:A\rightarrow A$ be two functions defined by $f(x)={x}^{2}$ and $g(x)=\sin \cfrac { \pi x }{ 2 }$ .Show that ${g}^{-1}$ exists but ${f}^{-1}$ does not exist. Also find ${g}^{-1}$ .

Let $f:\rightarrow R$ , $g:R\rightarrow R$ be two functions defined by $f(x)={x}^{2}+x+1$ and $g(x)=1-{x}^{2}$ . Write $f\circ g(-2)$ .

If $f:\left\{ 5,6 \right\} \rightarrow \left\{ 2,3 \right\}$ and $g:\left\{ 2,3 \right\} \rightarrow \left\{ 5,6 \right\}$ are given by $f=\left\{ \left( 5,2 \right) ,\left( 6,3 \right) \right\}$ and $g=\left\{ \left( 2,5 \right) ,\left( 3,6 \right) \right\}$ , find $f\circ g$ .

Let f(x) be a continuous and g(x) is a discontinuous function then prove that f(x)+g(x) is discontinuous at x=a

If $f$ is an invertible function, define as $f\left ( x \right )=\cfrac{3x-4}{5}$ , write $f^{-1}\left ( x \right )$ .

If $A=\left\{a,b,c,d \right\}$ and $f=\left\{(a,b),(b,d),(c,a),(d,c)\right\}$ , show that $f$ is one-one from $A$ onto $A$ . Find $f^{-1}$

Let $f:R\to R$ be the function defined by $f(x)=4x-3\forall x\in R$ . Then write $f^{-1}$ .

Show that $f;[-1,1]\rightarrow R$ , given by $f(x)=\dfrac{x}{(x+2)}$ is one-one. Find the inverse of the function $f;[-1,1]\rightarrow$ Range $f$ .
(Hint: For $y\in$ Range, $f, y=f(x)=\dfrac{x}{x+2}$ for some $x$ in $[-1,1]$
$x=\dfrac{2y}{(1-y)}$

Consider $f:R\rightarrow R$ given by $f(x)=4x+3$ . Show that $f$ is invertible. Find the inverse of $f$ .

Consider $f : R - \left \{ -\dfrac{4}{3} \right \} \rightarrow R - \left \{ \dfrac{4}{3} \right \}$ given by $f(x) = \dfrac{4x + 3}{3x + 4}.$ Show that f is
bijective. Find the inverse of f and hence find $f^{-1} (0)$ and x such that $f^{-1}(x) = 2.$

Prove that the function $f : N \to N,$ defined by $f(x) = x^2 + x + 1$ is one-one but not onto. Find inverse of $f : N \to S$ , where $S$ is range of $f$ .

Let $A = R - \left \{3\right \}$ and $B = R - \left \{1\right \}$ . Consider the function $f: A\rightarrow B$ defined by $f(x) = \left (\dfrac {x - 2}{x - 3}\right )$ Show that f is one-one and onto and hence find $f^{-1}$

Let $f$ be a real function given by $f(x)=\sqrt{x-2}$ . Find the following.
(i) $f\circ f$ (ii) $f\circ f\circ f$ (iii) $(f\circ f\circ f)(38)$ (iv) ${f}^{2}$
Also, show that $f\circ f\ne {f}^{2}$ .

If $f:R\rightarrow R$ be defined by $f(x)={x}^{3}-3$ , then prove that ${f}^{-1}$ exists and find a formula for ${f}^{-1}$ . Hence, find ${f}^{-1}(24)$ and ${f}^{-1}(5)$ .

Let $f:N\rightarrow N$ be a function defined as $f(x)=9{x}^{2}+6x-5$ . Show that $f:N\rightarrow S$ , where $S$ is the range of $f$ , is invertible. Find the inverse of $f$ and hence find ${f}^{-1}(43)$ and ${f}^{-1}(163)$ .

Class 12 Engineering Maths Extra Questions

Relations And Functions - Class 12 Engineering Maths - Extra Questions

Show that if f:A→Bf:A\rightarrow B and g:B→Cg:B\rightarrow C are one-one, then g∘f:A→Cg\circ f:A\rightarrow C is also one-one.

Let f:R→Rf:R\rightarrow R be defined by f(x)=x3+5f(x)={x}^{3}+5 then find f−1(x){f}^{-1}(x)

If the function f:[1,∞)→[1,∞) f:[1,\infty)\rightarrow [1,\infty) is defined by f(x)=2x(x−1) f(x)=2^{x(x-1)},then find f−1(4) f^{-1}(4)

Let f:R→Rf: R\rightarrow R be defined as f(s)=10x+7f(s) = 10x + 7. Find the function g:R→Rg: R\rightarrow R such that gof=fog=1Rgof = fog = 1_{R}

Let f:R→Rf:R\rightarrow R be defined by f(x)=2x+3f(x)=2x+3, find f−1(4){f}^{-1}(4).

If f(x)=(4x+3)(6x−4)f(x)=\dfrac{(4x+3)}{(6x-4)}, x≠23x\neq \dfrac{2}{3}, show that fof(x)=xfof(x)=x, for all x≠23x\neq\dfrac{2}{3}. What is the inverse of f?

Let f:N→Yf: N\rightarrow Y be a function defined as f(x)=4x+3f(x)=4x+3, where, Y={y∈N:y=4x+3 for some x∈N}Y=\{y\in N : y=4x+3\ for\ some\ x\in N\}, Show that f is invertible. Find the inverse.

Let f:N→Nf:N\rightarrow N be a function defined as f(x)=4x2+12x+15f(x)=4{x}^{2}+12x+15. Show that f:N→Sf:N\rightarrow S is invertible. Find the inverse of ff and hence find f−1(31){f}^{-1}(31) and f−1(87){f}^{-1}(87) .

Show that f:[−1,1]→Rf:\left[ -1,1 \right] \rightarrow R, given by f(x)=x(x+2)f\left( x \right) =\dfrac { x }{ \left( x+2 \right) } is one-one. Find the inverse of the function f:[−1,1]→Range f.f:\left[ -1,1 \right] \rightarrow Range\ f.

Work out the inverse function for each mapping.x→5x+1x \rightarrow 5 x + 1x→3x−7x \rightarrow 3 x - 7x→x5x \rightarrow \dfrac { x } { 5 } x→x+94x \rightarrow \dfrac { x + 9 } { 4 }

Let ff be any real function and let gg be a function given by g(x)=2xg(x)=2x. Prove that g∘f=f+fg\circ f=f+f.

If f:(−π2,π2)→Rf:\left (-\dfrac {\pi}{2},\dfrac {\pi}{2}\right)\rightarrow R and g:[−1,1]→Rg:[-1,1]\rightarrow R be defined as f(x)=tanxf(x)=\tan x and g(x)=√1−x2g(x)=\sqrt{1-{x}^{2}} respectively. Describe f∘gf\circ g and g∘fg\circ f.

If f(x)=2x+5f(x)=2x+5 and g(x)=x2+1g(x)={x}^{2}+1 be two real functions, then describe given functions:(i) f∘gf\circ g(ii) g∘fg\circ f(iii) f∘ff\circ f(iv) f2{f}^{2}Also, show that f∘f≠f2f\circ f\ne {f}^{2}.

If f(x)=√1−xf(x)=\sqrt{1-x} and g(x)=logexg(x)=\log _{ e }{ x } are two real functions, then describe functions f∘gf\circ g and g∘fg\circ f.

If f,g:R→Rf,g:R\rightarrow R be two functions defined as f(x)=|x|+xf(x)=\left| x \right| +x and g(x)=|x|−xg(x)=\left| x \right| -x for all x∈Rx\in R. Then, find f∘gf\circ g and g∘fg\circ f. Hence, find f∘g(−3)f\circ g(-3), f∘g(5)f\circ g (5) and g∘f(−2)g\circ f(-2).

If f(x)=√x+3f(x)=\sqrt{x+3} and g(x)=x2+1g(x)={x}^{2}+1 be two real functions, then find f∘gf\circ g and g∘fg\circ f.

Let f:→Rf:\rightarrow R, g:R→Rg:R\rightarrow R be two functions defined by f(x)=x2+x+1f(x)={x}^{2}+x+1 and g(x)=1−x2g(x)=1-{x}^{2}. Write f∘g(−2)f\circ g(-2).

Let f(x) be a continuous and g(x) is a discontinuous function then prove that f(x)+g(x) is discontinuous at x=a

If ff is an invertible function, define as f\left ( x \right )=\cfrac{3x-4}{5}f\left ( x \right )=\cfrac{3x-4}{5}, write f^{-1}\left ( x \right )f^{-1}\left ( x \right ).

If A=\left\{a,b,c,d \right\}A=\left\{a,b,c,d \right\} and f=\left\{(a,b),(b,d),(c,a),(d,c)\right\}f=\left\{(a,b),(b,d),(c,a),(d,c)\right\}, show that ff is one-one from AA onto AA. Find f^{-1}f^{-1}

Let f:R\to Rf:R\to R be the function defined by f(x)=4x-3\forall x\in Rf(x)=4x-3\forall x\in R. Then write f^{-1}f^{-1}.

Consider f:R\rightarrow Rf:R\rightarrow R given by f(x)=4x+3f(x)=4x+3. Show that ff is invertible. Find the inverse of ff.

Prove that the function f : N \to N, f : N \to N, defined by f(x) = x^2 + x + 1f(x) = x^2 + x + 1 is one-one but not onto. Find inverse of f : N \to Sf : N \to S, where SS is range of ff.

Let ff be a real function given by f(x)=\sqrt{x-2}f(x)=\sqrt{x-2}. Find the following.(i) f\circ ff\circ f (ii) f\circ f\circ ff\circ f\circ f (iii) (f\circ f\circ f)(38)(f\circ f\circ f)(38) (iv) {f}^{2}{f}^{2}Also, show that f\circ f\ne {f}^{2}f\circ f\ne {f}^{2}.

If f:R\rightarrow Rf:R\rightarrow R be defined by f(x)={x}^{3}-3f(x)={x}^{3}-3, then prove that {f}^{-1}{f}^{-1} exists and find a formula for {f}^{-1}{f}^{-1}. Hence, find {f}^{-1}(24){f}^{-1}(24) and {f}^{-1}(5){f}^{-1}(5).

Let f:N\rightarrow Nf:N\rightarrow N be a function defined as f(x)=9{x}^{2}+6x-5f(x)=9{x}^{2}+6x-5. Show that f:N\rightarrow Sf:N\rightarrow S, where SS is the range of ff, is invertible. Find the inverse of ff and hence find {f}^{-1}(43){f}^{-1}(43) and {f}^{-1}(163){f}^{-1}(163).

Class 12 Engineering Maths Extra Questions

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Show that if $f:A\rightarrow B$ and $g:B\rightarrow C$ are one-one, then $g\circ f:A\rightarrow C$ is also one-one.

Let $f:R\rightarrow R$ be defined by $f(x)={x}^{3}+5$ then find ${f}^{-1}(x)$

If the function $f:[1,\infty)\rightarrow [1,\infty)$ is defined by $f(x)=2^{x(x-1)}$ ,then find $f^{-1}(4)$

Let $f: R\rightarrow R$ be defined as $f(s) = 10x + 7$ . Find the function $g: R\rightarrow R$ such that $gof = fog = 1_{R}$

Let $f:R\rightarrow R$ be defined by $f(x)=2x+3$ , find ${f}^{-1}(4)$ .

If $f(x)=\dfrac{(4x+3)}{(6x-4)}$ , $x\neq \dfrac{2}{3}$ , show that $fof(x)=x$ , for all $x\neq\dfrac{2}{3}$ . What is the inverse of f?

Let $f: N\rightarrow Y$ be a function defined as $f(x)=4x+3$ , where, $Y=\{y\in N : y=4x+3\ for\ some\ x\in N\}$ , Show that f is invertible. Find the inverse.

Let $f:N\rightarrow N$ be a function defined as $f(x)=4{x}^{2}+12x+15$ . Show that $f:N\rightarrow S$ is invertible. Find the inverse of $f$ and hence find ${f}^{-1}(31)$ and ${f}^{-1}(87)$ .

Show that $f:\left[ -1,1 \right] \rightarrow R$ , given by $f\left( x \right) =\dfrac { x }{ \left( x+2 \right) }$ is one-one. Find the inverse of the function $f:\left[ -1,1 \right] \rightarrow Range\ f.$

Work out the inverse function for each mapping.

$x \rightarrow 5 x + 1$

$x \rightarrow 3 x - 7$

$x \rightarrow \dfrac { x } { 5 }$

$x \rightarrow \dfrac { x + 9 } { 4 }$

Let $f$ be any real function and let $g$ be a function given by $g(x)=2x$ . Prove that $g\circ f=f+f$ .

If $f:\left (-\dfrac {\pi}{2},\dfrac {\pi}{2}\right)\rightarrow R$ and $g:[-1,1]\rightarrow R$ be defined as $f(x)=\tan x$ and $g(x)=\sqrt{1-{x}^{2}}$ respectively. Describe $f\circ g$ and $g\circ f$ .

If $f(x)=2x+5$ and $g(x)={x}^{2}+1$ be two real functions, then describe given functions:
(i) $f\circ g$
(ii) $g\circ f$
(iii) $f\circ f$
(iv) ${f}^{2}$
Also, show that $f\circ f\ne {f}^{2}$ .

If $f(x)=\sqrt{1-x}$ and $g(x)=\log _{ e }{ x }$ are two real functions, then describe functions $f\circ g$ and $g\circ f$ .

If $f,g:R\rightarrow R$ be two functions defined as $f(x)=\left| x \right| +x$ and $g(x)=\left| x \right| -x$ for all $x\in R$ . Then, find $f\circ g$ and $g\circ f$ . Hence, find $f\circ g(-3)$ , $f\circ g (5)$ and $g\circ f(-2)$ .

If $f(x)=\sqrt{x+3}$ and $g(x)={x}^{2}+1$ be two real functions, then find $f\circ g$ and $g\circ f$ .

Let $f:\rightarrow R$ , $g:R\rightarrow R$ be two functions defined by $f(x)={x}^{2}+x+1$ and $g(x)=1-{x}^{2}$ . Write $f\circ g(-2)$ .

If $f$ is an invertible function, define as $f\left ( x \right )=\cfrac{3x-4}{5}$ , write $f^{-1}\left ( x \right )$ .

If $A=\left\{a,b,c,d \right\}$ and $f=\left\{(a,b),(b,d),(c,a),(d,c)\right\}$ , show that $f$ is one-one from $A$ onto $A$ . Find $f^{-1}$

Let $f:R\to R$ be the function defined by $f(x)=4x-3\forall x\in R$ . Then write $f^{-1}$ .

Consider $f:R\rightarrow R$ given by $f(x)=4x+3$ . Show that $f$ is invertible. Find the inverse of $f$ .

Prove that the function $f : N \to N,$ defined by $f(x) = x^2 + x + 1$ is one-one but not onto. Find inverse of $f : N \to S$ , where $S$ is range of $f$ .

Let $f$ be a real function given by $f(x)=\sqrt{x-2}$ . Find the following.
(i) $f\circ f$ (ii) $f\circ f\circ f$ (iii) $(f\circ f\circ f)(38)$ (iv) ${f}^{2}$
Also, show that $f\circ f\ne {f}^{2}$ .

If $f:R\rightarrow R$ be defined by $f(x)={x}^{3}-3$ , then prove that ${f}^{-1}$ exists and find a formula for ${f}^{-1}$ . Hence, find ${f}^{-1}(24)$ and ${f}^{-1}(5)$ .

Let $f:N\rightarrow N$ be a function defined as $f(x)=9{x}^{2}+6x-5$ . Show that $f:N\rightarrow S$ , where $S$ is the range of $f$ , is invertible. Find the inverse of $f$ and hence find ${f}^{-1}(43)$ and ${f}^{-1}(163)$ .