RD Sharma Class 12 Exercise 2.4 Functions Solutions Maths

RD Sharma Class 12 Exercise 2.4 Function Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 20, 2022 10:52 AM IST

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RD Sharma Class 12 Solutions Chapter 2 Functions - Other Exercise

Functions Excercise: 2.4

functions exercise 2.4 question 1 (I)

Answer:
No, $f$ is many one.
Given:
$f:\left \{ 1,2,3,4 \right \}\rightarrow \left \{ 10 \right \}\: with\, f=\left \{ \left ( 1,0 \right ),\left ( 2,10 \right ),\left ( 3,10 \right )\left ( 4,10 \right ) \right \}$
Hint:
A function has an inverse if it is one-one and onto.
Solution:
Let us check one-one
$f=\left \{ \left ( 1,10 \right )\left ( 2,10 \right ) , \left ( 3,10 \right ),\left ( 4,10 \right )\right \}$

Since all elements have an image, they don't have a unique image.
$\therefore \: \: \: \: \: \: f$ is not one-one
Since, $f$ is not one-one, it doesn't have an inverse
Functions excercise 2.4 question 1 (II)

Answer :
No, $g$ is many one.
Given :
$g=\left \{ 5,6,7,8 \right \}\rightarrow \left \{ 1,2,3,4 \right \}\: \: With\: \: g=\left \{ \left ( 5,4 \right ),\left ( 6,3 \right ),\left ( 7,4 \right ) \left ( 8,2 \right )\right \}$
Hint:
A function has an inverse if it is one-one and onto.
Solution:
Let us check one-one
$g=\left \{ \left ( 5,4 \right ),\left ( 6,3 \right ),\left ( 7,4 \right ) \left ( 8,2 \right )\right \}$

Since $5$ and $7$ have same image $4$ , $g$ is not one-one.
Since $g$ is not one-one, it doesn't have an inverse.

functions exercise 2.4 question 1 (iii)

Answer:
Yes, $h$ is a bijection.
Given:
$h:\left \{ 2,3,4,5 \right \}\rightarrow \left \{ 7,9,11,13\right \}\: with\: h=\left \{ \left ( 2,7 \right ),\left ( 3,9 \right ),\left ( 4,11 \right ),\left ( 5,13 \right ) \right \}$
Hint:
A function has an inverse if it is one-one and onto.
Solution:
Let us check one-one
$h=\left \{ \left ( 2,7 \right ),\left ( 3,9 \right ),\left ( 4,11 \right ),\left ( 5,13 \right ) \right \}$

Since all element has unique image $h$ is one-one.
Check onto,

Since each and every image there is a corresponding element.
$\therefore h$ is onto.
Since the function is both one-one and onto, it will have an inverse.
$h=\left \{ \left ( 2,7 \right ),\left ( 3,9 \right ),\left ( 4,11 \right ),\left ( 5,13 \right ) \right \}$
$h^{-1}=\left \{ \left ( 7,2 \right ),\left ( 9,3 \right ),\left ( 11,4 \right ),\left ( 13,5 \right ) \right \}$

Functions exercise 2.4 question 2 (I)

Answer:

Given:

Hint:

A function has an inverse if it is one-one and onto.

Solution:

Let us check one-one.

Here the different elements of the domain have different images in the co-domain.

Thus, this is one-one.

Range of

Therefore is a bijection and clearlyexists.

Hence .

Functions exercise 2.4 question 2 (ii)

Answer:
$f^{-1}$ doesn't exist.
Given:
$A=\left \{ 1,3,5,7,9 \right \},\: B=\left \{ 0,1,9,25,49,81 \right \},f\left ( x \right )=x^{2}$
Solution:
Let us check one-one
$f\left ( x \right )=x^{2},f\left ( 1 \right )=1,f\left ( 3 \right )=3^{2}=9,f\left ( 5 \right )=5^{2}=25,f\left ( 7 \right )=7^{2}=49,f\left ( 9 \right )=9^{2}=81$
$\therefore f=\left \{ \left ( 1,1 \right ),\left ( 3,9 \right ),\left ( 5,25 \right ),\left ( 7,49 \right ),\left ( 9,81 \right ) \right \}$
The different elements of the domain have different images in the codomain.
Thus, $f$ is one-one.
This is not onto because the element $0$ in the codomain $\left ( B \right )$ has no preimage in the domain $\left ( A \right )$ .
Therefore, $f^{-1}$ doesn't exist.

Functions exercise 2.4 question 3

Answer:
$\left ( gof \right )^{-1}=f^{-1}og^{-1}$
Given:
$f=\left \{ \left ( 1,a \right ),\left ( 2,b \right ),\left ( c,3 \right ) \right \},g=\left \{ \left ( a,apple \right ),\left ( b,ball \right ),\left ( c,cat \right ) \right \}$ , clearly $f$ and $g$ are bijections.
Hint:
Here, $f$ , $g$ and $fog$ are invertible.
Solution:
Now, $f^{-1}=\left \{ \left ( a,1 \right ),\left ( b,2 \right ),\left ( 3,c \right ) \right \},g^{-1}=\left \{ \left ( apple,a \right ),\left ( ball,b \right ),\left ( cat,c \right ) \right \}$
So, $f^{-1}og^{-1}=\left \{ \left ( apple,1 \right ),\left ( ball,2 \right ),\left ( cat,3 \right ) \right \}$
$f:\left \{ 1,2,3 \right \}\rightarrow \left \{ a,b,c \right \}\: and\: g:\left \{ a,b,c \right \}\rightarrow \left \{ apple,ball,cat \right \}$
So, $gof:\left \{ 1,2,3 \right \}\rightarrow \left \{ apple,ball,cat \right \}$
$\Rightarrow \left ( gof \right )\left ( 1 \right )=g\left [ f\left ( 1 \right ) \right ]=g\left ( a \right )=apple$
$\left ( gof \right )\left ( 2 \right )=g\left [ f\left ( 2 \right ) \right ]=g\left ( b \right )=ball$
$\left ( gof \right )\left ( 3 \right )=g\left [ f\left ( 3 \right ) \right ]=g\left ( c \right )=cat$
$\therefore$ $\left ( gof \right )=\left \{ \left ( 1,apple \right ),\left ( 2,ball \right ),\left ( 3,cat \right ) \right \}$
Clearly, $gof$ is a bijection.
So, $gof$ is invertible.
$\left ( gof \right )^{-1}=\left \{ \left ( apple,1 \right ),\left ( ball,2 \right ),\left ( cat,3 \right ) \right \}$
$From\: \left ( i \right ),\: \left ( iii \right ),$ we get,
$\left ( gof \right )^{-1}=f^{-1}og^{-1}$

Functions exercise 2.4 question 4 (0)

Answer:
$\left ( gof \right )^{-1}=f^{-1}og^{-1}$
Given:
$A=\left \{ 1,2,3,4 \right \},B=\left \{ 3,5,7,9 \right \},C=\left \{ 7,23,47,79 \right \}$
$f:A\rightarrow B,g:B\rightarrow Cdefined\: by\: f\left ( x \right )=2x+1,g\left ( x \right )=x^{2}-2.$
Hint:
Clearly $f$ and $g$ are bijections.
Solution:
$f\left ( x \right )=2x+1$
$f=\left \{ \left ( 1,2\left ( 1 \right )+1 \right ),\left ( 2,2\left ( 2 \right ) +1 \right ),\left ( 3,2\left ( 3 \right ) +1\right ),\left ( 4,2\left ( 4 \right ) +1\right )\right \}$
$=\left \{ \left ( 1,3 \right ),\left ( 2,5 \right ),\left ( 3,7 \right ),\left ( 4,9 \right ) \right \}$
$g\left ( x \right )=x^{2}-2$
$g=\left \{ \left ( 3,3^{2} -2\right ),\left ( 5,5^{2}-2 \right ),\left ( 7,7^{2} -2\right ),\left ( 9,9^{2}-2 \right ) \right \}$
$=\left \{ \left ( 3,7 \right ),\left ( 5,23 \right ),\left ( 7,47 \right ),\left ( 9,79 \right ) \right \}$
So, $f^{-1}=\left \{ \left ( 3,1 \right ),\left ( 5,2 \right ),\left ( 7,3 \right ),\left ( 9,4 \right ) \right \}$
$g^{-1}=\left \{ \left ( 7,3 \right ),\left ( 23,5 \right ),\left ( 47,7 \right ),\left ( 79,9 \right ) \right \}$
Now, $\left ( f^{-1}0g^{-1} \right ):C\rightarrow A$
$\left ( f^{-1}og^{-1} \right )=\left \{ \left ( 7,1 \right ),\left ( 23,2 \right ),\left ( 47,3 \right ),\left ( 79,4 \right ) \right \}$
Also $f:A\rightarrow B,g:B\rightarrow C$
$gof:A\rightarrow C,\left ( gof \right )^{-1}:C\rightarrow A$
So, $f^{-1}og^{-1}\: \: and\: \: \left ( gof \right )^{-1}$ have the same domains.
$\left ( gof \right )\left ( x \right )=g\left ( x \right )=g\left ( 2x+1 \right )=\left ( 2x+1 \right )^{2}-2$
$\left ( gof \right )\left ( x \right )=g\left [ f\left ( x \right ) \right ]$
$\left ( gof \right )\left ( x \right )=4x^{2}+4x+1-2$
$\left ( gof \right )\left ( x \right )=4x^{2}+4x-1$
Then, $\left ( gof \right )\left ( 1 \right )=g\left ( f\left ( 1 \right ) \right )=4+4-1=7$
$\left ( gof \right )\left ( 2 \right )=g\left ( f\left ( 2 \right ) \right )=16+8-1=23$
$\left ( gof \right )\left ( 3 \right )=g\left ( f\left ( 3 \right ) \right )=36+12-1=47$
$\left ( gof \right )\left ( 4 \right )=g\left ( f\left ( 4 \right ) \right )=64+16-1=79$
So, $\left ( gof \right )=\left \{ \left ( 1,7 \right ),\left ( 2,23 \right ),\left ( 3,47 \right ),\left ( 4,79 \right ) \right \}$
$\left ( gof \right )^{-1}=\left \{ \left ( 7,1 \right ),\left ( 23,2 \right ),\left ( 47,3 \right ),\left ( 79,4 \right ) \right \}$
From $\left ( i \right ),\left ( ii \right )$
$\left ( i \right ),\left ( ii\left ( gof \right ) \right )^{-1}=f^{-1}og^{-1}$
$\left \{ \left ( 7,1 \right ),\left ( 23,2 \right ),\left ( 47,3 \right ),\left ( 79,4 \right ) \right \}=\left \{ \left ( 7,1 \right ) ,\left ( 23,2 \right ),\left ( 47,3 \right ),\left ( 79,4 \right )\right \}$

Function Exercise 2.4 Question 5

Answer:
$f^{-1}\left ( x \right )=\frac{x-5}{3}$
Given:
$f:Q\rightarrow Q$ , defined by $f\left ( x \right )= 3x+5$
Hint:
Solution:
Injectivity of $f$ :
Let $x$ and $y$ be two elements of the domain $\left ( Q \right )$ , such that,
$f\left ( x \right )= f\left ( y \right )$
$3x+5=3y+5$
$3x=3y$
$x=y$
So, $f$ is one-one.
Subjectivity of $f$
Let $y$ be n the codomain $\left ( Q \right )$ , $f\left ( x \right )=y$ .
$3x+5=y$
$3x=y-5$
$x=\frac{y-5}{3}$
$f$ is onto.
So, $f$ is bijection.
Let $f^{-1}\left ( x \right )=y$ …(i)
$x =f\left ( y \right )$
$x=3y+5$
$x-5=3y$
$y=\frac{x-5}{3}$
$f^{-1}\left ( x \right )= \frac{x-5}{3}$ [From (i)]

Function Exercise 2.4 Question 6

Answer:
$f^{-1}\left ( x \right )=\frac{x-3}{4}$
Given:
$f\left ( x \right )=4x+3,f:R\rightarrow R$ .
Hint:
Bijection function should be fulfil the injectivity, surjectivity.
Solution:
Let $x,y$ be two elements of domain $\left ( R \right )$ , such that
$4x+3=4y+3$
$4x=4y$
$x=y$
So, $f$ is one-one.
Let $y$ be in the co-domain $\left ( R \right )$ , such that $f\left ( x \right )=y$ .
$4x+3=y$
$4x=y-3$
$x=\frac{y-3}{4}\: \epsilon\: R$
$f$ is onto.
So, $f$ is a bijection.
$f^{-1}\left ( x \right )=y$ … (i)
$x=f\left ( y \right )$
$x=4y+3$
$x-3=4y$
$y=\frac{x-3}{4}$
$f^{-1}\left ( x \right )=\frac{x-3}{4}$

Function Exercise 2.4 Question 7

Answer:
$f^{-1}\left ( x \right )=\sqrt{x-4}$
Given:
$f\left ( x \right )=x^{2}+4,f:R\rightarrow R_{+}\rightarrow \left [ 4,\infty \right ]$ .
Hint:
Bijection function should be fulfil the injectivity and surjectivity.
Solution:
Let $x,y$ be two elements of the domain $\left ( a \right )$ .
$\! \! \! \! \! \! \! \! \! \! f\left ( x \right )=f\left ( y \right )\\\\x^{2+4}=y^{2}+4\\\\x^{2}=y^{2}\\\\x=y$
So $f$ is one-one.
Let $y$ be in the co-domain $\left ( a \right )$ such that
$\! \! \! \! \! \! \! \! f\left ( x \right )=y\\\\x^{2}+4=y\\\\x^{2}=y-4\\\\x=\sqrt{y-4}\epsilon \: R$
$f$ is onto, so $f$ is bijection.
$\! \! \! \! \! \! \! \! f^{-1}\left ( x \right )=y\\\\x=f\left ( y \right )\\\\x=y^{2}+4\\\\x-4=y^{2}y=\sqrt{x-4}\\\\f^{-1}\left ( x \right )=\sqrt{x-4}$ … (i)

Function Exercise 2.4 Question 8

Answer:
$f$ is invertible.
Given:
$f\left ( x \right )=\frac{4x+3}{6x-4},x\neq \frac{2}{3}$
Hint:
If the function is invertible, $f\! o\! f=1$ .
Solution:
$f\left ( x \right )=\frac{4x+3}{6x-4},x\neq \frac{2}{3}$
$\left (f\! o\! f \right )\left ( x \right )=f\left ( f\left ( x \right ) \right )=f\left ( \frac{4x+3}{6x-4} \right )$
$\frac{4\left ( \frac{4x+3}{6x-4} \right )+3}{6\left ( \frac{4x+3}{6x-4}\right )-4}=\frac{16x+12+18x-12}{24x+18-24x+16}=\frac{34x}{34}=x$
$\therefore f\! o\! f\left ( x \right )=x$ for all $x\neq \frac{2}{3}$
$f\! o\! f=1$
Hence, the given function $f$ is invertible and inverse of $f$ is $f$ itself

Function Exercise 2.4 Question 9

Answer:
$f$ is invertible with the function.
$f^{-1}\left ( x \right )=\frac{\sqrt{x-6}-1}{3}$
Given:
$f:R_{+}\rightarrow \left [ 5,\infty \right ]$ given by $f\left ( x \right )=9x^{2}+6x-5$
Hint:
Bijection function should fulfil the injection and surjection condition.
Solution:
Let $x,y$ be two elements of domain $\left ( R^{+} \right )$ , such that $f\left ( x \right )=f\left ( y \right )$ .
$\! \! \! \! \! \! \! \! 9x^{2}+6x-5=9y^{2}+6y-5\\\\9x^{2}+6x=9y^{2}+6y\\\\x=y\left ( as,x,y\: \epsilon \: R^{+} \right )$
$f$ is one-one.
Subjectivity of $f$ :
Let $y$ is in the co-domain $\left ( Q \right )$ such that $f\left ( x \right )=y$
$\! \! \! \! \! \! \! \! \! 9x^{2}+6x-5=y\\\\9x^{2}+6x=y+5$
$9x^{2}+6x+1=y+6$ [Adding $1$ on both sides]
$\left ( 3x+1 \right )^{2}=6+y\\\\3x+1=\sqrt{y+6}\\\\3x=\sqrt{y+6}-1\\\\x=\frac{\sqrt{y+6}1}{3}\: \epsilon \: R^{+}\left ( domain \right )$
$f$ is onto.
So, $f$ is a bijection and hence, it is invertible.
$f^{-1}\left ( x \right )=y$ … (i)
$\! \! \! \! \! \! \! \! x=f\left ( y \right )\\\\x=9y^{2}+6y-5\\\\x+5=9y^{2}+6y$
$x+6=\left ( 3y+1 \right )^{2}$ [Adding

on both sides]

$\! \! \! \! \! \! \! \! \! 3y+1=\sqrt{x+6}\\\\3y=\sqrt{x+6}-1\\\\y=\frac{\sqrt{x+6}-1}{3}$
$f^{-1}\left ( x \right )=\frac{\sqrt{x+6}-1}{3}$ [From (i)]

Function Exercise 2.4 Question 10

Answer:
$f^{-1}\left ( x \right )=\sqrt[3]{x+3},f^{-1}\left ( 24 \right )=3,f^{-1}\left ( 5 \right )=2$
Given:
If $f:R\rightarrow R$ be defined by $f\left ( x \right )=x^{3}-3$ .
Hint:
Bijection function should be fulfil the injectivity, surjectivity condition.
Solution:
Let $x,y$ be two elements in domain $\left ( R \right )$ , such that $x^{3}-3=y^{3}-3$ .
$\! \! \! \! \! \! \! \! x^{3}=y^{3}\\\\x=y$
So, $f$ is one-one.
Surjectivity of $f$ .
Let $y$ be in the co-domain $\left ( R \right )$ such that $f\left ( x \right )=y$
$\! \! \! \! \! \! \! \! \! x^{3}-3=y\\\\x=\sqrt[3]{y+3}\: \epsilon \: R$
$f$ is onto.
So, $f$ is a bijection and hence it’s convertible.
$\! \! \! \! \! \! \! \! \!Let, f^{-1}\left (x \right )=y\\\\x=f\left ( y \right )\\\\x=y^{3}-3\\\\x+3=y^{3}\\\\y=\sqrt[3]{x+3}=f^{-1}\left ( x \right )$ …(i)
From (i)
$\! \! \! \! \! \! \! \! f^{-1}\left ( x \right )=\sqrt[3]{x+3}\\\\f^{-1}\left ( 24+3 \right )=\sqrt[3]{27}=\sqrt[3]{3^{3}}=3\\\\f^{-1}\left ( 5 \right )=\sqrt[3]{5+3}=\sqrt[3]{8}=2$

Function Exercise 2.4 Question 11

Answer:
Bijection, $f^{-1}\left ( 3 \right )=-1$ .
Given:
$f:R\rightarrow R$ is defined as $f\left ( x \right )=x^{3}+4$ .
Hint:
Bijection function should be fulfil the injection and surjection condition.
Solution:
Let $x,y$ be two elements of domain $\left ( R \right )$ .
$\! \! \! \! \! \! \! f\left ( x \right )=f\left ( y \right )\\\\x^{3}+4=y^{3}+4\\\\x^{3}=y^{3}\\\\x=y$
$f$ is one-one.

Let $y$ be in the co-domain $\left ( R \right )$ , such that $f\left ( x \right )=y$

$x^{2}+4=y\\\\x^{3}=y-4\\\\x=\sqrt[3]{y-4}\: \epsilon \: R \: domain$
$f$ is onto.
So, $f$ is a bijection, hence is invertible.
$f^{-1}\left ( x \right )=y$ …(i)
$x=f\left ( y \right )$
$\! \! \! \! \! \! \! x=y^{3}+4\\\\x-4=y^{3}\\\\y=\sqrt[3]{x-4}$
$\! \! \! \! So\: \: \: \: \: \: f^{-1}\left ( x \right )=\sqrt[3]{x-4}\\\\f^{-1}\left ( 3 \right )=\sqrt[3]{3-4}=\sqrt[3]{-1}=-1$ [from (i

Function Exercise 2.4 Question 12

Answer:
$\left ( gof \right )^{-1}=f^{-1}og^{-1}$
Given:
$f:Q\rightarrow Q,g:Q\rightarrow Q$ defined by $f\left ( x \right )=2x,g\left ( x \right )=x+2$ .
Hint:
Bijective function should be fulfil the injective and surjective function.
Solution:
$x,y$ be two elements of domain $\left ( Q \right )$ , such
$\! \! \! \! \! \! \! \! \! f\left (x \right )=f\left ( y \right )\\\\2x=2y\\\\x=y$
$f$ is one-one.
Let $y$ be in the co-domain $\left ( Q \right )$ , such that
$f\left ( x \right )=y$
$2x =y$
$x=\frac{y}{2}\epsilon \: Q$ .
$f$ is onto.
So, $f$ is a bijection and hence, it’s invertible
Let …(i)
$\\\\x=f\left ( y \right )\\\\y=\frac{x}{2}\\\\f^{-1}\left ( x \right )=\frac{x}{2}$ Injectivity of $g$
Let $x,y$ be two elements of domain $\left ( Q \right )$ .
$\! \! \! \! \! \! \! \! g\left ( x \right )=g\left ( y \right )\\\\x+2=y+2\\\\x=y$
So, $g$ is one-one.
Subjectivity of $g$ :
Let $y$ be in the co-domain $\left ( Q \right )$ , such that $g\left ( x \right )=y$ .
$x+2=y$
$x=2-y\: \epsilon \:$ (domain)
$g$ is onto.
So, $g$ is a bijection, hence it’s invertible.
Let $g^{-1}\left ( x \right )=y$
$x=g\left ( y \right )$ …(ii)
$x=y+2$
$y=x-2$
So, $g^{-1}\left ( x \right )=x-2$ [from (ii)]
$\left ( gof \right )^{-1}=f^{-1}og^{-1}$ .
$f\left ( x \right )=2x,g\left ( x \right )=x+2,f^{-1}\left ( x \right )=\frac{x}{2},g^{-1}\left ( x \right )=x-2$
$\left ( f^{-1} og^{-1}\right )\left ( x \right )=f^{-1}\left ( x-2 \right )$ $\left [ \because \left ( f^{-1} og^{-1}\right )\left ( x \right )=f^{-1}\left ( g^{-1}\left ( x \right ) \right ) \right ]$
$\left ( f^{-1}og^{-1} \right )\left ( x \right )=f^{-1}\left ( \frac{x-2}{2} \right )$ …(iii)
$\left ( gof \right )\left ( x \right )=g\left ( f\left ( x \right ) \right )$
$=g\left (2 x \right )$
$=2x+1$
Let $\left ( gof \right )\left ( x \right )=y$ …(vi)
$x=\left ( gof \right )\left ( y \right )$
$x=2y+2$
$y=\frac{x-2}{2}$
$\left ( gof \right )^{-1}=f^{-1}og^{-1}$ .

Function Exercise 2.4 Question 13

Answer:
$f^{-1}\left ( x \right )=\frac{3x-2}{x-1}$
Given:
$A=R-\left \{ 3 \right \},B=R-\left \{ 1 \right \},f:A\rightarrow B$ defined by $f\left ( x \right )=\frac{x-2}{x-3}$ .
Hint:
One-one means every domain has a distinct range. One means every image has some preimage in the domain function.
Solution:
We have, $A=R-\left \{ 3 \right \},B=R-\left \{ 1 \right \}$ .
The function $f:A\rightarrow B$ defined by $f\left ( x \right )=\frac{x-2}{x-3}$ .
Let $x,y\epsilon A$ such that $f\left ( x \right )=f\left ( y \right )$ then
$\frac{x-2}{x-3}=\frac{y-2}{y-3}$
$xy-3x-2y+6=xy-2x-3y+6$
$-x=-y$
$x=y$
$f$ is one-one.
Let $y\epsilon B$ , then $y\neq 1$ .
The function $f$ is onto if there exists $x\epsilon A$ such that $f\left ( x \right )=y$ .
$\Rightarrow \frac{x-2}{x-3}=y$
$x-2=xy-3y$
$x-xy=2-3y$
$x=\frac{2-3y}{1-y}\epsilon A\left ( y\neq 1 \right )$
Thus, for any $y\epsilon B$ , there exists $\frac{2-3y}{1-y}\epsilon A$ such that
$f\left ( \frac{2-3y}{1-y} \right )=\frac{\left ( \frac{2-3y}{1-y} \right )-2}{\left ( \frac{2-3y}{1-y} \right )-3}=\frac{2-3y-2+2y}{2-3y-3+3y}=\frac{-y}{-1}=y$
$f$ is onto.
So, $f$ is one-one and onto function.
$x=\frac{2-3y}{1-y}$
$f^{-1}\left ( x \right )=\frac{3x-2}{x-1}$

Function Exercise 2.4 Question 14

Answer:
$f^{-1}\left ( y \right )g\left ( y \right )=\frac{\sqrt{25y+5 4}-3}{5}$
Given:
$f:R^{+}\rightarrow \left [ -9,\infty \right ]$ given by $f\left ( x \right )=5x^{2}+6x-9$ .
We need to prove that $f^{-1}\left ( y \right )=\frac{\sqrt{54+25y}-3}{5}$
Solution:
$f\left ( x \right )=5x^{2}+6x-9$
Let $y=5x^{2}+6x-9$
Dividing by $5$ ,
$= x^{2}+\frac{6}{5}x-\frac{9}{5}$
$= x^{2}+2x\frac{3}{5}+\frac{9}{25}-\frac{9}{25}-\frac{9}{5}$
$= x^{2}+\left ( 2x\times \frac{3}{5} \right )+\left ( \frac{3}{5} \right )^{2}-\frac{9}{25}-\frac{9}{5}$ $\left [ \because \left ( a+b \right )^{2}=a^{2}+2ab+b^{2} \right ]$
$= \left ( x+\frac{3}{5} \right )^{2}-\frac{9-45}{25}$
$y= \left ( x+\frac{3}{5} \right )^{2}-\frac{54}{25}$
$\sqrt{y+\frac{54}{25}}=\left ( x+\frac{3}{5} \right )$
$\sqrt{\frac{25y+54}{25}}= x+\frac{3}{5}$
$x=\frac{\sqrt{25y+54}}{5}-\frac{3}{5}$
$x=\frac{\sqrt{25y+54}-3}{5}$
Let $g\left ( y \right )=\frac{\sqrt{25y+54}-3}{5}$
Now,
$fog\left ( y \right )=f\left ( g\left ( y \right ) \right )$
$=f\left (\frac{\sqrt{25y+54}-3}{5} \right )$
$=5\left (\frac{\sqrt{25y+54}-3}{5} \right )^{2}+6\left (\frac{\sqrt{25y+54}-3}{5} \right )-9$
$=5\left (\frac{25y+54+9-6\sqrt{25y+54-3}}{25} \right )+\left (\frac{6\sqrt{25y+54}-18}{5} \right )-9$
$=\frac{25y+63-6\sqrt{25y+54}}{5} +\frac{6\sqrt{25y+54}-18}{5} -9$
$=\frac{25y+63-18-45}{5}$
$=\frac{25y}{5}$
$=5y$
Identity function,
Also, $gof\left ( x \right )=g\left ( f\left ( x \right ) \right )$
$=g\left ( 5x^{2}+6x-9 \right )$
$= \frac{\sqrt{5\left ( 5x^{2}+6x-9 \right )+54}-3}{5}$
$= \frac{\sqrt{25x^{2}+30x-45+54}-3}{5}$
$= \frac{\sqrt{25x^{2}+30x+9}-3}{5}$
$=\frac{\sqrt{\left ( 5x+3 \right )^{2}}-3}{5}$
$=\frac{5x+3-3}{5}$
$= x$
So, $f$ is invertible, an identity function.
$f\left ( y \right )=g\left ( y \right )=\frac{\sqrt{25y+54}-3}{5}$

Function Exercise 2.4 Question 15

Answer:
$f^{-1}\left ( 43 \right )2,f^{-1}\left ( 163 \right )=4$
Given:
$f:N\rightarrow N$ is a function defined as $f\left ( x \right )=9x^{2}+6x-5$ .
Hint:
The function should be an identity function.
Solution:
Let $y=f\left ( x \right )=9x^{2}+6x-5$
$y=9x^{2}+6x+1-1-5$
$y=\left (9x^{2}+6x+1 \right )-6$
$y=\left ( 3x+1 \right )^{2}-6$
$y+6=\left ( 3x+1 \right )^{2}$
$\sqrt{y+6}=\left ( 3x+1 \right )$

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

(Let

)
Now,

=y+6-6

Identity function.

Identity function
Since

and

are identity functions.
Thus

is invertible.
So,

Now,

Function Exercise 2.4 Question 16

Answer:
$f^{-1}\left ( x \right )=\frac{4}{4-3x}$
Given:
$f:R-\left \{ \frac{-4}{3} \right \}\rightarrow R-\left \{ \frac{4}{3} \right \}\:$ is given by $\: f\left ( x \right )=\frac{4x}{3x+4}$
Injectivity,
Let $x,y\: \epsilon \: R-\left \{ \frac{-4}{3} \right \}$ be such that $f\left ( x \right )=f\left ( y \right )$
$\frac{4x}{3x+4}=\frac{4y}{3y+4}$
$4x\left ( 3y+4 \right )=4y\left ( 3x+4 \right )$
$12xy+16x= 12xy+16y$
$16x=16y$
$x=y$
$f$ is a one-one function.
Let $y\:$ be an arbitrary element of $R-\left \{ \frac{4}{3} \right \}$ .
Then $f\left ( x \right )= y$
$\frac{4x}{3x+4}=y$
$4x=3xy+4y$
$4x-3xy=4y$
$x=\frac{4y}{4-3y}$
As $y\: \epsilon \: R-\left \{ \frac{4}{3} \right \},\frac{4y}{4-3y}=\frac{-4}{3}$
Also, $\frac{4y}{4-3y}\neq \frac{-4}{3}$ because, $\frac{4y}{4-3y}= \frac{-4}{3}$
$12y=-16+12y\Rightarrow 0=-16$
Which is not possible.
Thus,
$x=\frac{4y}{4-3y}\: \epsilon \: R-\left \{ \frac{-4}{3} \right \}$
$f\left ( x \right )=f\left ( \frac{4x}{3x+4} \right )=\frac{4\left ( \frac{4y}{4-3y} \right )}{3\left ( \frac{4y}{4-3y} \right )+4}$
$=\frac{16y}{12y+16-12y}=\frac{16y}{16}$
$y$
So every element in $R-\left \{ \frac{4}{3} \right \}$ has pre image in $R-\left \{ \frac{-4}{3} \right \}$
Hence $f$ is onto.
Now, $x=\frac{4y}{4-3y}$ , replacing $x$ by $f^{-1}$ and $y$ by $x$ ,
$f^{-1}\left ( x \right )=\frac{4}{4-3x}$

Function Exercise 2.4 Question 17

Answer:
$f^{-1}\left ( x \right )=\frac{1}{2}\log\left ( \frac{1+x}{1-x} \right )$
Given:
$f\left ( x \right )=\frac{10^{x}-10^{-x}}{10^{x}+10^{-x}}$ is invertible
Hint:
$f^{-1}\left ( x \right )=y\Rightarrow f\left ( y \right )=x$
Solution:
Let $f^{-1}\left ( x \right )=y$ … (i)
$f\left ( y\right )=x$
$\frac{10^{y}-10^{-y}}{10^{y}+10^{-y}}=x$
$\frac{10^{-y}\left ( 10^{2y}-1 \right )}{10^{-y}\left ( 10^{2y}+1 \right )}=x$
$10^{2y}-1=x10^{2y}+x$
$10^{2y}\left ( 1-x \right )=1+x$
$10^{2y}=\frac{1+x}{1-x}$
$2y=\log\left ( \frac{1+x}{1-x} \right )$
$y=\frac{1}{2}\log\left ( \frac{1+x}{1-x} \right )=f^{-1}\left ( x \right )$
So, $f^{-1}\left ( x \right )=\frac{1}{2}\log\left ( \frac{1+x}{1-x} \right )$

Function Exercise 2.4 Question 18

Answer:
$f^{-1}\left ( x \right )= \frac{1}{2}\log _{e}\left ( \frac{x}{2-x} \right )$
Given:
$f\left ( x \right )= \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}+1$ is invertible.
Hint:
$f^{-1}\left ( x \right )=y\Rightarrow f\left ( y \right )=x$
Solution:
Let $f^{-1}\left ( x \right )=y$
$f\left ( y \right )=x$
$\frac{e^{y}-e^{-y}}{e^{y}+e^{-y}}+1=x$
$\frac{e^{-y}\left ( e^{2y}-1 \right )}{e^{-y}\left ( e^{2y}+1 \right )}=x-1$
$e^{2y}-1=xe^{2y}-e^{2y}+x-1$
$e^{2y}\left ( 2-x \right )=x$
$e^{2y}=\frac{x}{2-x}$
$2y=\log_{e}\frac{x}{2-x}$
$y=\frac{1}{2}\log_{e}\left (\frac{x}{2-x} \right )= f^{-1}\left ( x \right )$
So, $f^{-1}\left ( x \right )=\frac{1}{2}\log_{e}\left (\frac{x}{2-x} \right )$

Function Exercise 2.4 Question 19

Answer:
$s=\left \{ 0,1 \right \}$
Given:
$f\left ( x \right )=\left ( x+1 \right )^{2}-1,x\geq -1$
Hint:
Bijection function should fulfil the injection and surjection condition.
Solution:
Injectivity: let $x$ and $y\: \epsilon \: \left [ -1,\infty \right )$ , such that
$f\left ( x \right )=f\left ( y \right )$
$\left ( x+1 \right )^{2}-1=\left ( y+1 \right )^{2}-1$
$\left ( x+1 \right )^{2}=\left ( y+1 \right )^{2}$
$x+1 = y+1$
$x = y$
So, $f$ is an injection.
Surjectivity: let $y\: \epsilon \: \left [ -1,\infty \right )$ , then
$f\left ( x \right )=y$
$\left ( x+1 \right )^{2}-1=y$
$x+1 =\sqrt{y+1}$
$x =\sqrt{y+1}-1$
$x =\sqrt{y+1}-1$ is real for all $y\geq -1$
$\Rightarrow f$ is a surjection.
$f$ is a bijection. Hence, $f$ is invertible.
Let
$f^{-1}\left ( x \right )= y$
$f\left ( y \right )= x$
$\left ( y+1 \right )^{2}-1= x$
$y\pm \sqrt{x+1}-1= f^{-1}\left ( x \right )$
$f^{-1}\left ( x \right )=\pm \sqrt{x+1}-1$
$f\left ( x \right )=f^{-1}\left ( x \right )$
$\left (x +1 \right )^{2}-1\pm \sqrt{x+1}-1$
$\left (x +1 \right )^{2}\pm \sqrt{x+1}$
Squaring on both sides
$\left ( x+1 \right )^{4}=x+1$
$\left ( x+1 \right )\left [ \left ( x+1 \right ) ^{3}-1\right ]=0$
$\left ( x+1 \right )=0\: or\: \left ( x+1 \right ) ^{3}-1=0$
$x=-1\: or\: \left ( x+1 \right ) ^{3}=1$
$\left ( x+1 \right ) ^{3}=1$
$x+1 =1$
$x=0$
$x=-1 \: or\: 0$ or

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

Function Exercise 2.4 Question 20

Answer:
$g^{-1}\left ( x \right )= \frac{2}{\pi }\sin ^{-1}x$
Given:
$A= \left \{x\, \epsilon\, R-1\leq x\leq 1 \right \},f:A\rightarrow A,g:A\rightarrow A$

and

Hint:

exists only if

is a bijection.

Solution:

is not one-one, since

and

have the same image under

is not a bijection. So,

does not exist.
Injectivity of

So,

is one-one.
Surjectivity of

:
Range of

(Codomain of

)

is onto.

is a bijection. So,

exists.
Finding

Function Exercise 2.4 Question 21

Answer:
$f$ is not invertible
Given:
$f\left ( x \right )= \cos \left ( x+2 \right )$
Hint:
If the function is invertible, that should fulfil the one-one, onto condition.
Solution:
Let $x,y$ be two elements in the domain $\left ( R \right )$
$f\left ( x \right )= f\left ( y \right )$
$\cos \left ( x+2 \right )=\cos \left ( y+2 \right )$
$\left ( x+2 \right )=2\pi - \left ( y+2 \right )$
$x=2\pi -y-4$
So, we can’t say $x=y$
For example,
$\cos \frac{\pi }{2}= \cos \frac{3\pi }{2}=0$
So, $\frac{\pi }{2}$ and $\frac{3\pi }{2}=0$
So, $\frac{\pi }{2}$ and $\frac{3\pi }{2}$ have the same image $0$
$f$ is not one-one, bijection, surjection.
Hence, $f$ is not invertible

Function Exercise 2.4 Question 22

Answer:
Bijection
Given:
$A=\left \{ 1,2,3,4 \right \},B=\left \{ a,b,c,d \right \}$
Hint:
Bijection should fulfil the one-one, onto condition.
Solution:
$f_{1}=\left \{ \left ( 1,a \right ),\left (2,b \right ),\left ( 3,c\right ) ,\left ( 4,d \right )\right \}$
$f_{1}^{-1}=\left \{ \left ( a,1 \right ),\left (b,2 \right ),\left ( c,3\right ) ,\left ( d,4\right )\right \}$
$f_{2}=\left \{ \left ( 1,b \right ),\left (2,a \right ),\left ( 3,c\right ) ,\left ( 4,d \right )\right \}$
$f_{2}^{-1}=\left \{ \left ( b,1 \right ),\left (a,2 \right ),\left ( c,3\right ) ,\left ( d,4\right )\right \}$
$f_{3}=\left \{ \left ( 1,a \right ),\left (2,b \right ),\left ( 4,c\right ) ,\left ( 3,d \right )\right \}$
$f_{3}^{-1}=\left \{ \left ( a,1 \right ),\left (b,2 \right ),\left ( c,4\right ) ,\left ( d,3 \right )\right \}$
$f_{4}=\left \{ \left ( 1,b \right ),\left (2,a \right ),\left ( 4,c\right ) ,\left ( 3,d\right )\right \}$
$f_{4}^{-1}=\left \{ \left ( b,1 \right ),\left (a,2 \right ),\left ( c,4\right ) ,\left ( d,3\right )\right \}$
Clearly, all these are bijections because they are one-one and onto

Function Exercise 2.4 Question 23

Answer:
Bijective
Given:
$A,B$ be two sets, each with a finite number of elements. Prove that $A,B$ bijection.
Hint:
Bijection should fulfil the surjective injective condition.
Solution:
$A \: and\: B$ are two non-empty sets.
Let $f$ be a function from $A$ to $B$ .
It is given that there is injective map from $A$ to $B$ .
That means $f$ is one-one.
It also given that there is injective map from $B$ to $A$ .
That means $f$ is onto
$f$ is bijective.

Function Exercise 2.4 Question 24 (i)

Answer:
$fog$ is an injection.
Given:
$f:A\rightarrow A,g:A\rightarrow A$
Hint:
Injection means every domain has a distinct range.
Solution:
Let $x,y$ be two elements of the domain $\left ( A \right )$ such that
$\left ( fog \right )\left ( x \right )=\left ( fog \right )\left ( y \right )$
$f\left ( g\left ( x \right ) \right )=f\left ( g\left ( y \right ) \right )$
$g\left ( x \right )= g\left ( y \right )$ [As, $f$ is one-one]
$x=y$ [As, $g$ is one-one]
So, $fog$ is an injection.

RD Sharma Class 12th Exercise 2.4 deals with the chapter ‘Functions.’ First, you will learn about different types and properties of functions followed by examples. There are a total of 24 - level one questions in this exercise. They have relatively low complexity and can be solved by learning the fundamentals.

The following section contains 16 - level one questions and 8 - level two questions. Career360 has provided the best solutions on their website. The RD Sharma Class 12th Exercise 2.4 material has been developed with the specific needs of Class 12 students in mind. These solutions will give you a solid conceptual foundation for dealing with any queries.

The chapter has four exercises totaling more than 100 questions. For a better comprehension of our solutions, the formulae and questions have been broken down into steps. For students to cover the entire syllabus, they have to practice every question from the book. This is not possible as there are hundreds of questions per chapter. As there is a significant time constraint for students, they should refer to RD Sharma Class 12th Exercise 2.4 by Career360 to reduce their preparation time and quickly solve all the problems.

Benefits of RD Sharma Solutions for Class 12

Highly qualified specialists with years of experience have developed RD Sharma Class 12 Chapter 2 Exercise 2.4 - ‘Functions.'
Experts have provided these solutions in a simple and easy-to-understand manner to enable students to learn properly.
RD Sharma Class 12th Exercise 2.4 solutions are beneficial for completing homework and studying for CBSE examinations as they comply with the CBSE syllabus.
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The following are the fundamentals of RD Sharma Class 12 Chapter 2 Exercise 2.4

This exercise focuses on questions that are a little more advanced and complex.
Apart from that, read the chapter summary to gain a better understanding. It will assist you in quickly reviewing all of the major themes and formulas discussed in this chapter.
This chapter focuses on the different types of functions and their properties.
Class 12 RD Sharma Chapter 2 Exercise 2.4 Solution is presented clearly to allow students to complete the exercises without difficulty.

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FAQ for Class 12 Book

Q1. How do I get the solution of RD Sharma chapter 2 exercise 2.4 for class 12?

Downloading the Class 12 RD Sharma Chapter 2 Exercise 2.4 Solution is fairly straightforward. In addition, you can go to Career360's website and get the book in PDF format for free.

Q2. Do the RD Sharma Relations Ex 2.4 class 12 solutions follow the most recent syllabus?

The RD Sharma class 12 solution Chapter 2 Exercise 2.4 Solution is always updated with the latest syllabus. With each new edition of the NCERT book, the pdf of the RD Sharma Class 12 Solutions Chapter 2 Ex 2.4 is updated so that students can get answers to all of the problems in the book.

Q3. Is the content of the RD Sharma solutions up to date with the most recent syllabus?

RD Sharma book is updated each year to reflect the most up-to-date CBSE class 12 syllabus. As a result, you can be confident that the book will provide you with all of the knowledge you need.

Q4. Define a Function.

A function is the criteria that define the relationship between two or more variables. For example, if A is a variable, its function is defined as f(A). To learn more about functions, check RD Sharma Class 12 Solutions Chapter 2 Ex 2.4.

Q5. What are the different types of functions?

The different types of functions are

Injective function
Surjective function
Many to one function
Polynomial function
Identical function
Quadratic function etc.

To know more about this, check out RD Sharma Class 12 Solutions Functions Ex 2.4.
Chapter-wise RD Sharma Class 12 Solutions

Frequently Asked Questions (FAQs)

1. How do I get the solution of RD Sharma chapter 2 exercise 2.4 for class 12?

Downloading the Class 12 RD Sharma Chapter 2 Exercise 2.4 Solution is fairly straightforward. In addition, you can go to Career360's website and get the book in PDF format for free.

2. Do the RD Sharma Relations Ex 2.4 class 12 solutions follow the most recent syllabus?

3. . Is the content of the RD Sharma solutions up to date with the most recent syllabus?

RD Sharma book is updated each year to reflect the most up-to-date CBSE class 12 syllabus. As a result, you can be confident that the book will provide you with all of the knowledge you need.

4. Define a Function.

5. What are the different types of functions?

The different types of functions are