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Every solution provided in the RD Sharma Solutions books is updated frequently according to the change in the syllabus.

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Is RD Sharma Solutions available according to particular exercises in Class 12 mathematics?

The students can refer according to the specific exercises they require. For instance, the RD Sharma Class 12 Chapter 2 Exercise 2.2 part is enough to work out the sums in exercise 2.2.

Are the solutions provided in the RD Sharma books prepared from the exam point of view?

The solutions provided in the RD Sharma book are prepared by experts and are constructed from the exam point of view.

Is the RD Sharma Solutions book enough for the CBSE board students to prepare for their public exams?

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RD Sharma Class 12 Exercise 2.2 Function Solutions Maths - Download PDF Free Online

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RD Sharma Class 12 Exercise 2.2 Function Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 2.2 Function Solutions Maths - Download PDF Free Online

Team Careers360Updated on 20 Jan 2022, 10:40 AM IST

The RD Sharma Solutions for Class 12 Mathematics help the students to understand and work out the problems efficiently. It is essential to possess the right guide while working on homework or assignments, especially on complex topics like the RD Sharma Class 12th Exercise 2.2. This single compilation is adequate for the CBSE students who are preparing for their board exams.

RD Sharma Class 12 Solutions Chapter 2 Functions - Other Exercise

Functions Exercise 2.2 Question 1 (i) .

Answer :$g\; o\; f \left ( x \right )=4x^{2}+12x+14 \; \text {and}\; f\; o\; g\; =2x^{2}+13$
Hint :$g\; o\; f$ means $f(x)$ is in $g(x)$ function
$f\; o\; g$ means $g(x)$ function is in $f(x)$ function
Given : $f : R \rightarrow R\; \text {and }\; g:R \rightarrow R$
$f \; o\; g: R \rightarrow R\; \text {and }\; g\; o\; f:R \rightarrow R$
$f(x)=2x+3,g(x)=x^{2}+5$
Solution :
Now first we find $g\; o\; f$
$g\; o\; f (x)= g\left ( 2x+3 \right )$
$=\left ( 2x+3\right )^{2}+5$
$g\; o\; f (x)= 4x^{2}+12x+9+5$ $\because \left [ \left ( a+b \right )^{2}=a^{2}+b^{2}+2ab \right ]$
$g\; o\; f (x)= 4x^{2}+12x+14$
Now we find $f\; o\; g$
$f\; o\; g (x)=f\left ( g\left ( x \right ) \right )$
$f\left ( x^{2}+5 \right )=2\left ( x^{2}+5 \right )+3$
$f\; o\; g(x)=2x^{2}+13$
Hence, $f\; o\; g(x)=2x^{2}+13\; \text {and}\; g\; o\; f(x)=4x^{2}+12x+14$

Then we find $f\; o\; g$
$f\; o\; g (x)=f(g(x))$
$=f\left ( x^{3} \right )$
$f\; o\; g\left ( x \right )= 2x^{3}+x^{6}$
Hence, $f\; o\; g = 2x^{3}+x^{6}$ and $g\; o\; f =\left ( 2x+x^{2} \right )^{3}$

Functions Exercise 2.2 Question 1 (ii) .

Answer :$f\; o\; g=2x^{3}+x^{6}\; \text {and}\; g\; o\; f=\left ( 2x+x^{2} \right )^{3}$
Hint :$g\; o\; f$ means $f(x)$ function is in $g(x)$ function
$f\; o\; g$ means $g(x)$ function is in $f(x)$ funcion
Given :$f\left ( x \right )=2x+x^{2}$
$g\left ( x \right )=x^{3}$
$f:R\rightarrow R\; \text {and}\; g:R\rightarrow R$
$f\; o\; g:R\rightarrow R\; \text {and}\; g\; o\; f:R\rightarrow R$
Solution :
First, we find $g\; o\; f$
$g\; o\; f\left ( x \right )=g\left ( f\left ( x \right ) \right )$
$=g \left ( 2x+x^{2} \right )$
$g\; o\; f\left ( x \right )= \left ( 2x+x^{2} \right )^{3}$
Then we find $f\; o\; g$
$f\; o\; g \left ( x \right )=f\left ( g\left ( x \right ) \right )$
$=f(x^{3})$
$f\; o\; g \left ( x \right )=2x^{3}+x^{6}$
Hence, $f\; o\; g =2x^{3}+x^{6}$ and $g\; o\; f = \left ( 2x+x^{2} \right )^{3}$

Functions Exercise 2.2 Question 1 (iii) .

Answer :$g\; o\; f = 3\left ( x^{2}+8 \right )^{3}+1\; \text {and}\; f\; o\; g=9x^{6}+6x^{3}+9$
Hint :$g\; o\; f$ means $f(x)$ function is in $g(x)$ function
$f\; o\; g$ means $g(x)$ function is in $f(x)$ function
Given :$f(x)=x^{2}+8$
$g(x)=3x^{3}+1$
$f:R\rightarrow R\; \text {and}\; g:R\rightarrow R$
$f\; o\; g:R\rightarrow R\; \text {and}\; g\; o\; f:R\rightarrow R$
Solution :
First, we find $g\; o\; f$
Thus, $g\; o\; f\left ( x \right )=g\left [ f\left ( x \right ) \right ]$
$g\; o\; f\left ( x \right )=g\left [ x^{2}+8 \right ]$
$g\; o\; f\left ( x \right )=3\left [ x^{2}+8 \right ]^{3}+1$
Similarly,
$f\; o\; g\left ( x \right )=f\left [ g(x) \right ]$
$f\; o\; g\left ( x \right )=f\left ( 3x^{3}+1 \right )$
$f\; o\; g\left ( x \right )= \left [ 3x^{3}+1 \right ]^{2}+8$
$f\; o\; g\left ( x \right )= \left [ 9x^{6}+1+6x^{3} \right ]^{2}+8$ $\because \left [ \left ( a+b \right )^{2} =a^{2}+b^{2}+2ab\right ]$
$f\; o\; g\left ( x \right )= 9x^{6}+1+6x^{3}+9$
Hence, $f\; o\; g = 9x^{6}+1+6x^{3}+9 \; \text {and}\; g\; o\; f=3\left ( x^{2}+8 \right )^{3}+1$

Functions Exercise 2.2 Question 1 (iv) .

Answer :$f\; o\; g=\left | x \right | \; \text {and} \; g\; o\; f=\left | x \right |$
Hint :$g\; o\; f$ means $f(x)$ function is in $g(x)$ function
$f\; o\; g$ means $g(x)$ function is in $f(x)$ function
Given :$f(x)=x$
$g(x)=\left | x \right |$
Solution :
Since $f:R\rightarrow R\; \text {and}\; g:R\rightarrow R$
$f\; o\; g:R\rightarrow R\; \text {and}\; g\; o\; f:R\rightarrow R$
Now, $g\; o\; f (x)=g(f(x))$
$=g(x)$
$g\; o\; f \left ( x \right )= \left | x \right |$
Similarly ,
$f\; o\; f (x)=f(g(x))$
$=f \left | x \right |$
$f\; o\; g(x)= \left | x \right |$
Hence, $f\; o\; g= \left | x \right |\; \text {and}\; g\; o\; f=\left | x \right |$

Functions Exercise 2.2 Question 1 (v) .

Answer :$f\; o\; g = 9x^{2}-18x+5$$f\; o\; g = 9x^{2}-18x+5\; \text {and}\; g\; o\; f=3x^{2}+6x-13$
Hint :$g\; o\; f$ means $f(x)$ function is in $g(x)$ function
$f\; o\; g$ means $g(x)$ function is in $f(x)$ function
Given :$f(x)=x^{2}+2x-3$
$g(x)= 3x-4$
Solution :
Since $f:R\rightarrow R \text {and}g:R\rightarrow R$
$f\; o\; g:R\rightarrow R \; \text {and}\; g\; o\; f:R\rightarrow R$
Now, $g\; o\; f (x)=g(f(x))=g(x)$
$g\; o\; f (x)=g(f(x))=g\left ( x^{2}+2x -3\right )$
$g\; o\; f (x)= 3\left ( x^{2}+2x -3\right )-4$
$g\; o\; f (x)= 3x^{2}+6x -9-4$
$g\; o\; f (x)= 3x^{2}+6x -13$
$f\; o\; g (x)= f(g(x))=f(x)$
$f\; o\; g (x)= f(g(x))=f \left ( 3x-4 \right )$
$f\; o\; g (x)= \left ( 3x-4 \right )^{2}+2\left ( 3x-4 \right )-3$
$=9x^{2}+16-24x+6x-8-3$ $\because \left [ \left ( a-b \right )^{2}=a^{2}+b^{2} -2ab\right ]$
$f\; o\; g (x)=9x^{2}-18x+5$
Hence, $f\; o\; g = 9x^{2}-18x+5\; \text {and}\; g\; o\; f=3x^{2}+6x-13$

Functions Exercise 2.2 Question 1 (vi).
Hint :$g\; o\; f$ means $f(x)$ function is in $g(x)$ function
$f\; o\; g$ means $g(x)$ function is in $f(x)$ function
Given :$f(x)=8x^{3}$
$g(x)=x^{\frac{1}{3}}$
Solution :
Since $f:R\rightarrow R \; \text {and} \; g:R\rightarrow R$
$f\; o\; g:R\rightarrow R \; \text {and} \; g\; o\; f:R\rightarrow R$
Now, $g\; o\; f(x)=g(f(x))=g(8x^{3} )$
$g\; o\; f(x)=(8x^{3} )^{\frac{1}{3}}$
$=\left ( 8 \right )^{\frac{1}{3}}=\left ( 2^{3}\right )^{\frac{1}{3}}$ $\because \left [ \text {as we know},\left ( a^{m} \right )^{\frac{1}{n}}=\left ( a \right )^{\frac{m}{n}} \right ]$
$g\; o\; f (x)=(8)^{\frac{1}{3}}(x^{3})^{\frac{1}{3}}$
$g\; o\; f (x)= 2x$
Similarly,
$f\; o\; g (x)=f(g(x))=f(x)^{\frac{1}{3}}$
$f\; o\; g (x)=8 (x^{\frac{1}{3}})^{3}$
$f\; o\; g (x)=8 x$ $\because \left [ (a^{m})^{\frac{1}{n}}=(a)^{\frac{m}{n}} \right ]$
Hence, $f\; o\; g=8x\; \text {and}\; g\; o\; f=2x$

Functions Exercise 2.2 Question 2

Answer :$f\; o\; g= \left \{ (1,1),(3,1),(4,3),(5,3) \right \}$
$g\; o\; f= \left \{ (3,3),(9,3),(12,9) \right \}$
Hint : The set which contains all the elements of all the ordered pairs of relation R is known as the domain of the relation. The set which contains all the second element, is known as the range of the relation.
Given :$f=\left \{ (3,1),(9,3),(12,4) \right \}$

$g=\left \{ (-1,-2),(-2,-4),(-3,-6),(4,8) \right \}$
Prove :$g\; o\; f$ and $f\; o\; g$ are both defined
Solution :$f:\left \{ 3,9,12 \right \}\rightarrow \left \{ 1,3,4 \right \}\; \text {and}\; g:\left \{ 1,3,4,5 \right \}\rightarrow \left \{ 3,9 \right \}$
Co-domain of f is a subset of the domain g
So, $g\; o\; f$ exist and $g\; o\; f:\left \{ 3,9,12 \right \}\rightarrow \left \{ 3,9 \right \}$
$(g\; o\; f)(3)=g(f(3))=g(1)=3$
$(g\; o\; f)(9)=g(f(9))=g(3)=3$
$(g\; o\; f)(12)=g(f(12))=g(4)=9$
$g\; o\; f =\left \{ \left ( 3,3 \right ),\left ( 9,3 \right ),\left ( 12,9 \right ) \right \}$
Co-domain of g is subset of the domain of f.
So, $f\; o\; g$ exist and $f\; o\; g:\left \{ 1,3,4,5 \right \}\rightarrow \left \{ 3,9,12 \right \}$
$(f\; o\; g)(1)=f(g(1))=f(3)=1$
$(f\; o\; g)(3)=f(g(3))=f(3)=1$
$(f\; o\; g)(4)=f(g(4))=f(9)=3$
$(f\; o\; g)(5)=f(g(5))=f(9)=3$
$f\; o\; g=\left \{ (1,1),(3,1),(4,3),(5,3) \right \}$
Hence proved, $g\; o\; f$ and $f\; o\; g$ are both defined.
$f\; o\; g=\left \{ (1,1),(3,1),(4,3),(5,3) \right \}$ and
$g\; o\; f=\left \{ (3,3),(9,3),(12,9) \right \}$

Functions Exercise 2.2 Question 3.

Answer :$g\; o\; f = \left \{ \left ( 1,2 \right ),\left ( 4,-4 \right ),\left ( 9,-6 \right ),\left ( 16,8 \right ) \right \}$
Hint : The set which contains all the elements of all the ordered pairs of relation R is known as the domain of the relation. The set which contains all the second element is known as the range of the relation.
Given :$f=\left \{ (1,-1),(4,-2),(9,-3),(16,4) \right \}$
$g=\left \{ (-1,-2),(-2,-4),(-3,-6),(4,8) \right \}$
Prove :$g\; o\; f$ is defined while $f\; o\; g$ is not defined.
Solution
Now, Domain of $f=\left \{ 1,4,9,16 \right \}$
Range of $f=\left \{ -1,-2,-3,4 \right \}$
Domain of $g=\left \{ -1,-2,-3,4 \right \}$
Range of $g=\left \{ -2,-4,-6,8 \right \}$
Clearly, Range of f = domain of g
$\therefore \; g\; o\; f \; \text {is defind}.$
But Range of $g\neq$ domain of f
$\therefore \; f\; o\; g \; \text {is not defind}.$
Now, $g\; o\; f(1)=g(-1)=-2$
$g\; o\; f(4)=g(-2)=-4$
$g\; o\; f(9)=g(-3)=-6$
$g\; o\; f(16)=g(4)=8$
$g\; o\; f= \left \{ (1,-2),(4,-4),(9,-6),(16,8) \right \}$
Hence proved, $g\; o\; f$ is defined but $f\; o\; g$ is mot defined.
$g\; o\; f = \left \{ \left ( 1,-2 \right ),\left ( 4,-4 \right ),\left ( 9,-6 \right ),\left ( 16,8 \right ) \right \}$

Functions Exercise 2.2 Question 4.

Answer : $g\; o\; f=\left \{ (a,a),(b,b),(c,c) \right \}\; \text {and}$
$f\; o\; g=\left \{ (u,u),(v,v),(w,w) \right \}$
Hint : For any function to be a bijective, the given function should be one-one and onto.\
Given :$A=\left \{ a,b,c \right \}$
$B=\left \{ u,v,w \right \}$
Solution :
f and g be two functions from A to B and from B to A ; $A\rightarrow B\; \text {and}\; g:B\rightarrow A$
$f=\left \{ (a,v),(b,u),(c,w) \right \}$$f=\left \{ (a,v),(b,u),(c,w)\right \}$
$g=\left \{ (u,b),(v,a),(w,c)\right \}$
For both f and g, Different elements of the domain have different images.
$\therefore$ f and f are onto
Again, for each element in co-domain of f and g, there in a preimage in domain
$\therefore$ f and f are onto
Thus, f and g are bijectives.
Now,
$g\; o\; f=\left \{ (a,a),(b,b),(c,c) \right \} \text {and}$
$f\; o\; g=\left \{ (u,u),(v,v),(w,w) \right \}$
Hence prove, f and g both are bijections
$g\; o\; f=\left \{ (a,a),(b,b),(c,c) \right \}\; \text {and}$ $f\; o\; g=\left \{ (u,u),(v,v),(w,w) \right \}$

Functions Exercise 2.2 Question 5.

Answer :$f\; o\; g (2)=633$
$g\; o\; f (1)=2188$
Hint :$g\; o\; f$ means $f(x)$ function is in $g(x)$ function
$f\; o\; g$ means $g(x)$ function is in $f(x)$ function
Given : $f:R\rightarrow R ; f(x)=x^{2}+8$
$g:R\rightarrow R ; g(x)=3x^{3}+1$
Solution :
First, we find $f\; o\; g$
$f\; o\; g(x)=f(g(x))=f(3x^{3}+1)$
$f\; o\; g(x)=(3x^{3}+1)^{2}+8$ $\because [(a+b)^{2}=a^{2}+b^{2}+2ab]$
$f\; o\; g(2)=(3\times(2)^{3}+1)^{2}+8$
$=\left ( 3 \times 8+1 \right )^{2}+8$
$=(25)^{2}+8$
$=625+8$
$f\; o\; g(2)=633$
Similarly,
$g\; o\; f(x)=g(f(x))=g(x^{2}+8)=3(x^{2}+8)^{3}+1$
$g\; o\; f(1)=3(1+8)^{3}+1\Rightarrow 3(9)^{3}+1$
$=3\times 729+1$
$=2188$
Hence, $f\; o\; g(2)=633 \; \text {and} \; g\; o\; f(1)=2188$

Function Exercise 2.2 Question 6.

Answer :$f\; o\; g=x \; \text {and}\; g\; o\; f=x$
Hint :$g\; o\; f$ means $f(x)$ function is in $g(x)$ function
$f\; o\; g$ means $g(x)$ function is in $f(x)$ function
Given :$f:R^{+}\rightarrow R^{+}$ defined by
$f(x)=x^{2}$
$g:R^{+}\rightarrow R^{+}$ defined by
$g(x)=\sqrt{x}$
Solution :
$f\; o\; g(x)=f(g(x))=f(\sqrt{x})$
$f\; o\; g(x)=(\sqrt{x})^{2}=x$
Similarly,
$g\; o\; f(x)=g(f(x))=g(x^{2})$
$g\; o\; f(x)=(\sqrt{x})^{2}=x$
Thus
$f\; o\; g(x)=g\; o\; f(x)$
$x=x$
Hence proved functions are equal.
$f\; o\; g=x\; \text {and}\; g\; o\; f=x$

Function Exercise 2.2 Question 7.

Answer : $f\; o\; g \neq g\; o\; f$
Hint :$g\; o\; f$ means $f(x)$ function is in $g(x)$ function
$f\; o\; g$ means $g(x)$ function is in $f(x)$ function
Given :$f:R \rightarrow R$ defined by
$f(x)=x^{2}$
$g:R\rightarrow R$ defined by
$g(x)= x+1$
Solution :
$f\; o\; g(x)=f(g(x))=f(x+1)$
$f\; o\; g(x)=(x+1)^{2}$ $\therefore [(a+b)^2=a^2+b^2+2ab]$
$f\; o\; g(x)=x^{2}+1+2x$ ......(i)
Similarly,
$g\; o\; f(x)=g(f(x))=g(x^{2})$
$g\; o\; f(x)= x^{2}+1$ .....(ii)
Now we have to make a comparison between eq. (i) and (ii)
$f\; o\; g(x) \neq g\; o\; f(x)$
$x^{2}+1+2x \neq x^{2}+1$
Hence proved $f\; o\; g \neq g\; o\; f$

Function Exercise 2.2 Question 8.

Answer :$f\; o\; g=g\; o\; f=I_{R}$
Hint :$g\; o\; f$ means $f(x)$ function is in $g(x)$ function
$f\; o\; g$ means $g(x)$ function is in $f(x)$ function
Given :$f:R \rightarrow R$ and $g:R \rightarrow R$ are defined by
$\\f(x)=x+1\\g(x)=x-1$
Solution :
$\begin{aligned} &f \circ g(x)=f(g(x))=f(x-1) \\ &f \circ g(x)=x-1+1 \\ & f \circ g(x)=x=I_{R} \end{aligned}$ .....(i)
Similarly,
$\begin{aligned} &g \circ f(x)=g(f(x))=g(x+1) \\ &g \circ f(x)=x+1-1 \\ &g \circ f(x)=x=I_{R} \end{aligned}$ ..... (ii)
From equation (i) and (ii)
$f\; o\; g=g\; o\; f=I_{R}$
Hence proved : $f\; o\; g=g\; o\; f=I_{R}$

Function Exercise 2.2 Question 9.

Answer :$h\; o(g\; o\; f)=(h\; o\; g)o\; f$
Hint :$g\; o\; f$ means $f(x)$ function is in $g(x)$ function
$f\; o\; g$ means $g(x)$ function is in $f(x)$ function
For associative property $h\; o(g\; o\; f)=(h\; o\; g)o\; f$
Given :
$\begin{aligned} &f: N \rightarrow Z_{0},: Z_{0} \rightarrow Q, R: Q \rightarrow R \\ &f(x)=2 x \\ &g(x)=\frac{1}{x} \\ &h(x)=e^{x} \end{aligned}$
Solution :
$\begin{aligned} &g \circ f: N \rightarrow Q \text { and } h \circ g: Z_{0} \rightarrow R \\ &h \circ(g \circ f): N \rightarrow R \text { and }(h \circ g) \circ f: N \rightarrow R \end{aligned}$
So, both have same domains $\left [ \text {since}f(x)=2x \right ]$
$\begin{gathered} (g \circ f)(x)=g[f(x)] \\ =g(2 x) \\ =\frac{1}{2 x} \\ (h \circ g)(x)=h[g(x)] \\ =h\left(\frac{1}{x}\right) \Rightarrow e^{\frac{1}{x}} \end{gathered}$ .....(ii)
Now,
$\begin{gathered} {[h \circ(g \circ f)](x)=h[(g \circ f)(x)]} \\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; =h\left(\frac{1}{2 x}\right) \end{gathered}$ ....[From (i)]
$=e^{\frac{1}{2x}}$ ....[From (ii)]
$[(h \circ g) \circ f(x)]=(h \circ g)[f(x)]$
$=(h \circ g)(2 x)$ ...[since $f(x)=2x$]
$=e^{\frac{1}{2x}}$ ....[From (ii)]
$[h \circ(g \circ f)](x)=[(h \circ g) \circ f(x)], \quad \in N$
So, $h \circ(g \circ f)=(h \circ g) \circ f$
Hence, the associative property has benn verified.

Function Exercise 2.2 Question 10.

Answer : $h\; o(g\; o\; f)=(h\; o\; g)o\; f$
Hint :$g\; o\; f$ means $f(x)$ function is in $g(x)$ function
$f\; o\; g$ means $g(x)$ function is in $f(x)$ function
Given : $f: N \rightarrow N, g: N \rightarrow N, h: N \rightarrow R$ defined as
$\begin{array}{r} f(x)=2 x \\ g(y)=3 y+4 \\ h(z)=\sin z \\ \text { All } x, y, z \in N \end{array}$
Solution :
$\begin{aligned} h \circ(g \circ f)(x)=& h(g \circ f(x))=h(g(f(x))) \\ &=h(g(2 x))=h(3(2 x)+4) \\ &=h(6 x+4) \\ &=\sin (6 x+4) \quad \forall x \in N \end{aligned}$
Similarly,
$\begin{aligned} ((h \circ g) \circ f)(x) &=(h \circ g)(f(x)) \\ &=(h \circ g)(2 x) \\ &=h(g(2 x))=h(3(2 x)+4) \end{aligned}$
$=h(6x+4)=\sin(6x+4) \; \; \; \; \; \; \; \forall x\; \epsilon \; N$
This shows, $h\; o(g\; o\; f)=(h\; o\; g)o\; f$

Functions Exercise 2.2 Question 11.

Answer :
$\begin{aligned} &f: N \rightarrow N \text { by } f(x)=x+1 \\ &g: N \rightarrow N \text { by } \\ &g(x)=\{x-1 \text { if } x>1,1 \text { if } x=1 \end{aligned}$
Hint : Onto functions means every range has some preimage in the domain of function.
Given :$f:N\rightarrow N \; \text {and}\; g:N\rightarrow N$
Solution :
Let $f:N\rightarrow N \; \text {be}\; f(x)=x+1$
and $g:N\rightarrow N \; \text {be}\; g(x)= \left \{ x-1,x>11,x=1 \right.$
We will first show that f is not onto
Checking f is not onto
$\begin{aligned} &f: N \rightarrow N \text { be } f(x)=x+1 \\ &y=f(x), \text { where } y \in N \\ &y=x+1 ; x=y-1 \end{aligned}$
for $y=1;x=1-1=0$
But 0 is not a natural number
$\therefore$ f is not onto.
Finding $g\; o\; f :$
$\begin{aligned} &f(x)=x+1 \\ &g(x)=\{x-1 \quad x>1 \quad 1 \quad x=1 \\ &f(x)=x+1 ; g(x)=1 \end{aligned}$
Since $g(x)=1,g(f(x))=1$
So, $g\; o\; f =1$
For $x>1$
Since,
$\begin{aligned} &g(x)=x-1, g(f(x))=f(x)-1 \\ &g \circ f=(x+1)-1 \\ &g \circ f=x \end{aligned}$
So, $g \circ f=\{x, x>11, x=1\}$
Let $g \circ f=y, \text { where } y \in N$
So, $y=\{x, x>11, x=1\}$
Here, y is a natural number. As $y=x$
So, x is also a natural number.
Hence, $g\; o\; f$ is onto.

Functions Exercise 2.2 Question 12.

Answer :$f(x)=x \; \text {and}\; g(x)=\left | x \right |$
where $f:N\rightarrow Z \; \text {and}\; g:Z\rightarrow Z$
Hint : An injective function is a function that maps distinct elements of its domain to the distinct elements of its co-domain.
Given : $f:N\rightarrow Z \; \text {and}\; g:Z\rightarrow Z$
Solution :
Let $f(x)=x \; \text {and}\; g(x)=\left | x \right |$
$g(x)=|x|=\{x \quad x \geq 0-x \quad x<0$
Checking $g(x)$ injective (one-one)
$g(1)=\left | 1 \right |=1$
$g(-1)=\left | -1 \right |=1$
Since, different elements 1, -1 have the same image 1
$\therefore$ g is not injective (one -one).
Checking for injective (one-one)
$\begin{aligned} &f: N \rightarrow Z \quad \& g: Z \rightarrow Z \\ &f(x)=x \text { and } g(x)=|x| \\ &g \circ f(x)=g(f(x))=|f(x)| \\ &=|x|=\{x, x \geq 0-x, \quad x<0\} \\ &\text { Here } g \circ f(x): N \rightarrow Z \end{aligned}$
So, x is always a natural number.
Here $\left | x \right |$ will always be a natural number.
So, $g\; o\; f (x)$ has a unique image
$\therefore$ $g\; o\; f (x)$ is injective.
Hence, $g\; o\; f$ is injective but g is not injective

Functions Exercise 2.2 Question 13.

Answer :$g\; o\; f$ is a one -one function.
Hint : one-one functions means every domain has a distinct image. If one-one is given for any function
$f(x) \text { as if } F\left(x_{1}\right)=F\left(x_{2}\right) \text { then } x_{1}=x_{2} \text { where } x_{1}, x_{2} \in \text { domains of } f(x)$
Given : $f:A\rightarrow B\; \text {and}\; g:B\rightarrow C$ one-one function
Solution :
$\begin{aligned} &\text { Let, } x, y \in A \text { such that } \\ &g \circ f(x)=g \circ f(y) \\ &g(f(x))=g(f(y)) \end{aligned}$
$f(x)=f(y)$ [ $\because$ g is one - one ]
$x=y$ [$\because$ f is one - one]
$g\; o\; f$ is a one - one function

Functions Exercise 2.2 Question 14.

Answer :$g\; o\; f$ is an onto function
Hint : Onto function means every range has some pre-image in the domain of function.
Given :$f:A\rightarrow B\; \text {and}\; g:B\rightarrow C$ are onto function
Prove :$g\; o\; f:A\rightarrow C$ is an onto function
Solution :
Let $y\; \epsilon \; C$ then
$\begin{aligned} &g \circ f(x)=y \\ &g(f(x))=y \end{aligned}$ ....(i)
Since, g is onto, for each element in C, then exists a pre-image in B.
$g(x)=y$ .....(ii)
From (i) and (ii)
$f(x)=x$
Since f is onto, for each element in B, there exist a pre image in A
$f(x)=x$ ....(iii)
From (ii) and (iii) we can conclude that for each $y\; \epsilon \; C$, there exist a preimage in A
Such that $g\; o\; f(x)=y$
$g\; o\; f$ is onto.

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The 2nd chapter in RD Sharma Class 12th Exercise 2.2 is regarding Functions in mathematics. This chapter focuses on the various kinds of functions, compositions of functions, and inverse of functions. Again, this is a specific topic to be grasped by students compared to the other chapters in the book. The first exercise, 2.1, is a bit essential in the mathematical functions. The second exercise, 2.2, dives deeply into the concepts of fog and gof, injective, surjective, and bijective.

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Chapter-wise RD Sharma Class 12 Solutions