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

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# 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

RD Sharma books are easily the most preferred books in the country. They are widely used by CBSE schools and a majority of teachers to set up question papers. RD Sharma's books are comprehensive, content-rich, and informative. They are a one-stop shop for CBSE . There are many solved examples for every chapter, followed by Level 1 and Level 2 questions

## RD Sharma Class 12 Solutions Chapter 2 Functions - Other Exercise

Functions Excercise: 2.4

functions exercise 2.4 question 1 (I)

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)
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)

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)

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)

$f^{-1}$ 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

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

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

$f^{-1}\left ( x \right )=\frac{x-5}{3}$
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 $f^{-1}\left ( x \right )=y$ â€¦(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

$f^{-1}\left ( x \right )=\frac{x-3}{4}$
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.
$f^{-1}\left ( x \right )=y$ â€¦ (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

$f^{-1}\left ( x \right )=\sqrt{x-4}$
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.
$\! \! \! \! \! \! \! \! f^{-1}\left ( x \right )=y\\\\x=f\left ( y \right )\\\\x=y^{2}+4\\\\x-4=y^{2}y=\sqrt{x-4}\\\\f^{-1}\left ( x \right )=\sqrt{x-4}$ â€¦ (i)

Function Exercise 2.4 Question 8

$f$ 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

$f$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.
$f^{-1}\left ( x \right )=y$ â€¦ (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

$f^{-1}\left ( x \right )=\sqrt[3]{x+3},f^{-1}\left ( 24 \right )=3,f^{-1}\left ( 5 \right )=2$
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 itâ€™s convertible.
$\! \! \! \! \! \! \! \! \!Let, f^{-1}\left (x \right )=y\\\\x=f\left ( y \right )\\\\x=y^{3}-3\\\\x+3=y^{3}\\\\y=\sqrt[3]{x+3}=f^{-1}\left ( x \right )$ â€¦(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

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.
$f^{-1}\left ( x \right )=y$ â€¦(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

$\left ( gof \right )^{-1}=f^{-1}og^{-1}$
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, itâ€™s invertible
Let $f^{-1}\left ( x \right )=y$ â€¦(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 itâ€™s invertible.
Let $g^{-1}\left ( x \right )=y$
$x=g\left ( y \right )$ â€¦(ii)
$x=y+2$
$y=x-2$
So, $g^{-1}\left ( x \right )=x-2$ [from (ii)]
$\left ( gof \right )^{-1}=f^{-1}og^{-1}$.
$f\left ( x \right )=2x,g\left ( x \right )=x+2,f^{-1}\left ( x \right )=\frac{x}{2},g^{-1}\left ( x \right )=x-2$
$\left ( f^{-1} og^{-1}\right )\left ( x \right )=f^{-1}\left ( x-2 \right )$ $\left [ \because \left ( f^{-1} og^{-1}\right )\left ( x \right )=f^{-1}\left ( g^{-1}\left ( x \right ) \right ) \right ]$
$\left ( f^{-1}og^{-1} \right )\left ( x \right )=f^{-1}\left ( \frac{x-2}{2} \right )$ â€¦(iii)
$\left ( gof \right )\left ( x \right )=g\left ( f\left ( x \right ) \right )$
$=g\left (2 x \right )$
$=2x+1$
Let $\left ( gof \right )\left ( x \right )=y$ â€¦(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

$f^{-1}\left ( x \right )=\frac{3x-2}{x-1}$
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

$f^{-1}\left ( y \right )g\left ( y \right )=\frac{\sqrt{25y+5 4}-3}{5}$
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

$f^{-1}\left ( 43 \right )2,f^{-1}\left ( 163 \right )=4$
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 andare identity functions.
Thus is invertible.
So,
Now,

Function Exercise 2.4 Question 16

$f^{-1}\left ( x \right )=\frac{4}{4-3x}$
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

$f^{-1}\left ( x \right )=\frac{1}{2}\log\left ( \frac{1+x}{1-x} \right )$
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 $f^{-1}\left ( x \right )=y$ â€¦ (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

$f^{-1}\left ( x \right )= \frac{1}{2}\log _{e}\left ( \frac{x}{2-x} \right )$
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

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

$g^{-1}\left ( x \right )= \frac{2}{\pi }\sin ^{-1}x$
Given:
$A= \left \{x\, \epsilon\, R-1\leq x\leq 1 \right \},f:A\rightarrow A,g:A\rightarrow A$
and
Hint:
or exists only if or is a bijection.

Solution:

is not one-one, sinceand 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

$f$ 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 canâ€™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

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

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)

$fog$ 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 â€˜Functions.â€™ 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 - â€˜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.

• RD Sharma Class 12th Exercise 2.4 solutions also aid in the clear understanding of ideas and preparing for not only 12th boards but also upcoming competitive exams such as BITSAT, JEE, and NEET.

• The questions and answers assist you in revising the material and scoring well on your tests. In addition, solved questions save a lot of time and effort, which is why students can breeze through this chapter after understanding the concepts.

• The material is updated to the latest version, which means you will be up to date with the bookâ€™s standards.

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.

JEE Main Highest Scoring Chapters & Topics
Just Study 40% Syllabus and Score upto 100%

## FAQ for Class 12 Book

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

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

Chapter-wise RD Sharma Class 12 Solutions

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

2. 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.

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.

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.

5. 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

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##### Orthotist and Prosthetist

Orthotists and Prosthetists are professionals who provide aid to patients with disabilities. They fix them to artificial limbs (prosthetics) and help them to regain stability. There are times when people lose their limbs in an accident. In some other occasions, they are born without a limb or orthopaedic impairment. Orthotists and prosthetists play a crucial role in their lives with fixing them to assistive devices and provide mobility.

6 Jobs Available
##### Veterinary Doctor

A veterinary doctor is a medical professional with a degree in veterinary science. The veterinary science qualification is the minimum requirement to become a veterinary doctor. There are numerous veterinary science courses offered by various institutes. He or she is employed at zoos to ensure they are provided with good health facilities and medical care to improve their life expectancy.

5 Jobs Available
##### Pathologist

A career in pathology in India is filled with several responsibilities as it is a medical branch and affects human lives. The demand for pathologists has been increasing over the past few years as people are getting more aware of different diseases. Not only that, but an increase in population and lifestyle changes have also contributed to the increase in a pathologist’s demand. The pathology careers provide an extremely huge number of opportunities and if you want to be a part of the medical field you can consider being a pathologist. If you want to know more about a career in pathology in India then continue reading this article.

5 Jobs Available
##### Gynaecologist

Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth.

4 Jobs Available
##### Surgical Technologist

When it comes to an operation theatre, there are several tasks that are to be carried out before as well as after the operation or surgery has taken place. Such tasks are not possible without surgical tech and surgical tech tools. A single surgeon cannot do it all alone. It’s like for a footballer he needs his team’s support to score a goal the same goes for a surgeon. It is here, when a surgical technologist comes into the picture. It is the job of a surgical technologist to prepare the operation theatre with all the required equipment before the surgery. Not only that, once an operation is done it is the job of the surgical technologist to clean all the equipment. One has to fulfil the minimum requirements of surgical tech qualifications.

3 Jobs Available
##### Oncologist

An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.

3 Jobs Available
##### Chemical Pathologist

Are you searching for a chemical pathologist job description? A chemical pathologist is a skilled professional in healthcare who utilises biochemical laboratory tests to diagnose disease by analysing the levels of various components or constituents in the patient’s body fluid.

2 Jobs Available
##### Biochemical Engineer

A Biochemical Engineer is a professional involved in the study of proteins, viruses, cells and other biological substances. He or she utilises his or her scientific knowledge to develop products, medicines or ways to improve quality and refine processes. A Biochemical Engineer studies chemical functions occurring in a living organism’s body. He or she utilises the observed knowledge to alter the composition of products and develop new processes. A Biochemical Engineer may develop biofuels or environmentally friendly methods to dispose of waste generated by industries.

2 Jobs Available
##### Actor

For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs.

4 Jobs Available
##### Acrobat

Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

3 Jobs Available
##### Video Game Designer

Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages. Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

3 Jobs Available
##### Talent Agent

The career as a Talent Agent is filled with responsibilities. A Talent Agent is someone who is involved in the pre-production process of the film. It is a very busy job for a Talent Agent but as and when an individual gains experience and progresses in the career he or she can have people assisting him or her in work. Depending on one’s responsibilities, number of clients and experience he or she may also have to lead a team and work with juniors under him or her in a talent agency. In order to know more about the job of a talent agent continue reading the article.

If you want to know more about talent agent meaning, how to become a Talent Agent, or Talent Agent job description then continue reading this article.

3 Jobs Available

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

3 Jobs Available
##### Videographer

Careers in videography are art that can be defined as a creative and interpretive process that culminates in the authorship of an original work of art rather than a simple recording of a simple event. It would be wrong to portrait it as a subcategory of photography, rather photography is one of the crafts used in videographer jobs in addition to technical skills like organization, management, interpretation, and image-manipulation techniques. Students pursue Visual Media, Film, Television, Digital Video Production to opt for a videographer career path. The visual impacts of a film are driven by the creative decisions taken in videography jobs. Individuals who opt for a career as a videographer are involved in the entire lifecycle of a film and production.

2 Jobs Available
##### Multimedia Specialist

A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications.

2 Jobs Available
##### Visual Communication Designer

Individuals who want to opt for a career as a Visual Communication Designer will work in the graphic design and arts industry. Every sector in the modern age is using visuals to connect with people, clients, or customers. This career involves art and technology and candidates who want to pursue their career as visual communication designer has a great scope of career opportunity.

2 Jobs Available
##### Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook.

5 Jobs Available
##### Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available
##### Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available
##### Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. Ever since internet cost got reduced the viewership for these types of content has increased on a large scale. Therefore, the career as vlogger has a lot to offer. If you want to know more about the career as vlogger, how to become a vlogger, so on and so forth then continue reading the article. Students can visit Jamia Millia Islamia, Asian College of Journalism, Indian Institute of Mass Communication to pursue journalism degrees.

3 Jobs Available
##### Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available
##### Content Writer

Content writing is meant to speak directly with a particular audience, such as customers, potential customers, investors, employees, or other stakeholders. The main aim of professional content writers is to speak to their targeted audience and if it is not then it is not doing its job. There are numerous kinds of the content present on the website and each is different based on the service or the product it is used for.

2 Jobs Available
##### Reporter

Individuals who opt for a career as a reporter may often be at work on national holidays and festivities. He or she pitches various story ideas and covers news stories in risky situations. Students can pursue a BMC (Bachelor of Mass Communication), B.M.M. (Bachelor of Mass Media), or MAJMC (MA in Journalism and Mass Communication) to become a reporter. While we sit at home reporters travel to locations to collect information that carries a news value.

2 Jobs Available
##### Linguist

Linguistic meaning is related to language or Linguistics which is the study of languages. A career as a linguistic meaning, a profession that is based on the scientific study of language, and it's a very broad field with many specialities. Famous linguists work in academia, researching and teaching different areas of language, such as phonetics (sounds), syntax (word order) and semantics (meaning).

Other researchers focus on specialities like computational linguistics, which seeks to better match human and computer language capacities, or applied linguistics, which is concerned with improving language education. Still, others work as language experts for the government, advertising companies, dictionary publishers and various other private enterprises. Some might work from home as freelance linguists. Philologist, phonologist, and dialectician are some of Linguist synonym. Linguists can study French, German, Italian

2 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product.

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
##### Production Manager

Production Manager Job Description: A Production Manager is responsible for ensuring smooth running of manufacturing processes in an efficient manner. He or she plans and organises production schedules. The role of Production Manager involves estimation, negotiation on budget and timescales with the clients and managers.

3 Jobs Available

A Team Leader is a professional responsible for guiding, monitoring and leading the entire group. He or she is responsible for motivating team members by providing a pleasant work environment to them and inspiring positive communication. A Team Leader contributes to the achievement of the organisation’s goals. He or she improves the confidence, product knowledge and communication skills of the team members and empowers them.

2 Jobs Available
##### Quality Systems Manager

A Quality Systems Manager is a professional responsible for developing strategies, processes, policies, standards and systems concerning the company as well as operations of its supply chain. It includes auditing to ensure compliance. It could also be carried out by a third party.

2 Jobs Available
##### Merchandiser

A career as a merchandiser requires one to promote specific products and services of one or different brands, to increase the in-house sales of the store. Merchandising job focuses on enticing the customers to enter the store and hence increasing their chances of buying a product. Although the buyer is the one who selects the lines, it all depends on the merchandiser on how much money a buyer will spend, how many lines will be purchased, and what will be the quantity of those lines. In a career as merchandiser, one is required to closely work with the display staff in order to decide in what way a product would be displayed so that sales can be maximised. In small brands or local retail stores, a merchandiser is responsible for both merchandising and buying.

2 Jobs Available
##### Procurement Manager

The procurement Manager is also known as  Purchasing Manager. The role of the Procurement Manager is to source products and services for a company. A Procurement Manager is involved in developing a purchasing strategy, including the company's budget and the supplies as well as the vendors who can provide goods and services to the company. His or her ultimate goal is to bring the right products or services at the right time with cost-effectiveness.

2 Jobs Available
##### Production Planner

Individuals who opt for a career as a production planner are professionals who are responsible for ensuring goods manufactured by the employing company are cost-effective and meets quality specifications including ensuring the availability of ready to distribute stock in a timely fashion manner.

2 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### ITSM Manager

ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks.

3 Jobs Available
##### Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack

3 Jobs Available
##### Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
##### Computer System Analyst

Individuals in the computer systems analyst career path study the hardware and applications that are part of an organization's computer systems, as well as how they are used. They collaborate closely with managers and end-users to identify system specifications and business priorities, as well as to assess the efficiency of computer systems and create techniques to boost IT efficiency. Individuals who opt for a career as a computer system analyst support the implementation, modification, and debugging of new systems after they have been installed.

2 Jobs Available
##### Test Manager

A Test Manager is a professional responsible for planning, coordinating and controlling test activities. He or she develops test processes and strategies to analyse and determine test methods and tools for test activities. The test manager jobs involve documenting tests that have been carried out, analysing and evaluating software quality to determine further recommended procedures.

2 Jobs Available
##### Azure Developer

A career as Azure Developer comes with the responsibility of designing and developing cloud-based applications and maintaining software components. He or she possesses an in-depth knowledge of cloud computing and Azure app service.

2 Jobs Available
##### Deep Learning Engineer

A Deep Learning Engineer is an IT professional who is responsible for developing and managing data pipelines. He or she is knowledgeable about analyzing and storing data collected from various sources.  A Career as a Deep Learning Engineer needs to help the  data scientists and analysts to create effective data sets.

2 Jobs Available