RD Sharma Class 12 Exercise 2.3 Function Solutions Maths - Download PDF Free Online

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RD Sharma Class 12 Exercise 2.3 Function Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 2.3 Function Solutions Maths - Download PDF Free Online

Kuldeep MauryaUpdated on 20 Jan 2022, 10:43 AM IST

RD Sharma books are no less than a Math bible for students as they contain concepts and chapters that help students understand the subject. Similarly, RD Sharma Class 12th Exercise 2.3 is specifically designed for Class 12 students to prepare for their exams. Therefore, students will find this material very helpful for their preparation as they are preparing for their Class 12 exams and the entrance exams that will be conducted in the future.

RD Sharma Class 12 Solutions Chapter 2 Functions - Other Exercise

Functions Excercise: 2.3

Functions Exercise 2.3 Question 1 (i).

Answer :$f\; o\; g(x)=x\; \text {and}\; g\; o\; f(x)=x$
Hint : If $f:A\rightarrow B$ and $g:B\rightarrow C$ be two given functions then the composite of f and g denoted by $g\; o\; f$
$\therefore(g \circ f): A \rightarrow C(g \circ f)(x)=g\{f(x)\} \forall x \in A$
And dom $(g\; o\; f)=$ dom f
Given : Here given that $f(x)=e^{x} \; \text {and}\; g(x)=log_{e}x$
Here we have find out $f\; o\; g$ and $g\; o\; f$
Solution :
Here first we will find out $f\; o\; g$
Here we have,
$\begin{aligned} &f(x)=e^{x} \text { and } g(x)=\log _{e} x \\ &\begin{aligned} \therefore f \circ g(x) &=f\{g(x)\} \\ &=f\left\{\log _{e} x\right\} \\ &=\log _{e}\left(e^{x}\right) \\ &=x \\ \therefore f \circ g(x) &=x \end{aligned} \end{aligned}$
Now, we find out $g\; o\; f$, we have ,
$\begin{aligned} &f(x)=e^{x} \& g(x)=\log _{e} x \\ &\therefore g \circ f(x)=g\{f(x)\} \\ &\qquad \begin{aligned} &=g\left\{e^{x}\right\} \\ &=\log _{e}\left(e^{x}\right) \\ &=x \end{aligned} \\ &\therefore g \circ f=x \end{aligned}$

Functions Exercise 2.3 Question 1 (ii).

Answer :
$\begin{aligned} &g \circ f(x)=\cos x^{2} \\ &f \circ g(x)=\cos ^{2} x \end{aligned}$
Hint : Domain of f and domain g = R
Range of $f=(0,\infty )$
Range of g $g=(-1,1 )$
$\therefore$ Range of f $\subset$ domain of g
$\Rightarrow \; \; \; \; \; \; \; \; g\; o\; f$ exists
and Range of g $\subset$ domain of f
$\Rightarrow \; \; \; \; \; \; \; \; f\; o\; g$ exists
Given : Here given that
$f(x)=x^{2}\; \text {and}\; g(x)=\cos x$
Solution :
Here first we will find out $g\; o\; f,$
We have,
$\begin{array}{r} f(x)=x^{2} \text { and } g(x)=\cos x \\ \therefore g \circ f(x)=g\{f(x)\} \\ =g\left\{x^{2}\right\} \\ =\cos x^{2} \end{array}$
and again for $f\; o\; g$
$\begin{aligned} &\therefore \operatorname{fog}(x)=f\{g(x)\} \\ &\qquad \begin{aligned} &=f\{\cos x\} \\ &=\cos ^{2} x \end{aligned} \end{aligned}$
NOTE :$f\; o\; g$ is defined only when range $(g)\subseteq \text {dom}\; (f) \; \text {and dom}\;f\; o\; g=\text {dom}(g) .$

Functions Exercise 2.3 Question 1 (iii).

Answer :
$\begin{aligned} &f \circ g(x)=|\sin x| \\ &g \circ f(x)=\sin |x| \end{aligned}$
Hint : The Range of $f=(0,\infty )$$f=(0,\infty )\; \subset$ domain $g(R)$
$\Rightarrow \; \; \; \; \; g\; o\; f$ exists
And Range of $g=[-1,1]\subset$ domain R
$\Rightarrow \; \; \; \; \; \; f\; o\; g$ exists
Given : Here given that
$f(x)=\left | x \right |\; \text {and}\; g(x)=\sin x$
Here we have to find out $f\; o\; g$ and $g\; o\; f$
Solution :
First , we will find out $f\; o\; g(x)$
$\begin{aligned} f \circ g(x) &=f\{g(x)\} \\ &=f\{\sin x\} \\ &=|\sin x| \end{aligned}$
Again, we find out $g\; o\; f(x)$
$\begin{aligned} g \circ f(x) &=g\{f(x)\} \\ &=g\{|x|\} \\ &=\sin |x| \end{aligned}$

Functions Exercise 2.3 Question 1 (iv).

Answer :
$\begin{aligned} &g \circ f(x)=e^{x+1} \\ &f \circ g(x)=e^{x}+1 \end{aligned}$
Hint : The Range of $f = R \; \subset \text {Domain of } f=R$
$\Rightarrow \; \; \; \; \; \; f\; o\; g$ exists
Given : Here given that
$f(x)=x+1 \text { and } g(x)=e^{x}$
Here we have to find out $g \circ f(x)\; \text {and}\; f \circ g(x)$
Solution :
First, we will find out $g\; o\; f (x)$
We have,
$\begin{aligned} g \circ f(x) &=g\{f(x)\} \\ &=g\{(x+1)\} \\ &=e^{x+1} \end{aligned}$
And again, we will find out $f\; o\; g (x)$
$\begin{gathered} f \circ g(x)=f\{g(x)\} \mid \\ \; \; \; \; \;\; \; \; =f\left\{e^{x}\right\} \\ \; \; \; \; \; \; \; \; =e^{x}+1 \\ \text { So, } g \circ f(x)=e^{x+1} \text { and } f \circ g(x)=e^{x}+1 \end{gathered}$

Functions Exercise 2.3 Question 1 (v).

Answer :
$\begin{aligned} &f \circ g(x)=\sin ^{-1}\left(x^{2}\right) \\ &g \circ f(x)=\left(\sin ^{-1} x\right)^{2} \end{aligned}$
Hint :
$\\\text {Domain} (f \circ g)=\{x: x \in \text {domain of g and} \; g(x) \in \text {domain of f}\}\\ \text {Domain of}\; (f \circ g)=\left\{x: x \in R\right. and \left.x^{2} \in[-1,1]\right\}\\ \text {Domain of}\; f \circ g=[-1,1]$
Given : Here given that
$f(x)=\sin ^{-1} x \; \text {and}\; g(x)=x^{2}$
Here, we have to compute $f\; o\; g(x) \; \text {and}\; g\; o\; f(x)$
Solution :
Here we have to compute $f\; o\; g$ :
$\begin{aligned} &\because f \circ g:[-1,1] \in R \\ &f \circ g=f\{g(x)\} \\ &\quad=f\left\{x^{2}\right\} \\ &\quad=\sin ^{-1}\left(x^{2}\right) \end{aligned}$
Now, we have to find out $g\; o\; f$:
Clearly, the range of f is subset of the domain of g.
$\begin{aligned} &f \circ f:[-1,1] \rightarrow R \\ &\qquad \begin{aligned} g \circ f(x) &=g\{f(x)\} \\ =g\left \{ \sin^{-1}x \right \} &=\left(\sin ^{-1} x\right)^{2} \end{aligned} \end{aligned}$

Functions Exercise 2.3 Question 1 (vi).

Answer :
$\begin{aligned} &g \circ f(x)=\sin (x+1) \\ &f \circ g(x)=\sin x+1 \end{aligned}$
Hint :$f: R \rightarrow R ; g: R \rightarrow[-1,1]$
Given : Here given that
$f(x)=x+1\; \text {and}\; g(x)=\sin x$
we have to compute $f\; o\; g (x)$ and $g\; o\; f (x)$
Solution :
First, we have to compute $f\; o\; g$
Clearly, the range of g is a subset of the domain f.
Set of the domain of f.
$\Rightarrow \; \; \; \; \; \; f\; o\; g:R\rightarrow R$
$\\ f\; o\; g (x)=f\left \{ g(x) \right \}$
$=f(\sin x)$
$= \sin (x+1)$
Now, we will compute $g\; o\; f$
$g\; o\; f (n)=g\left \{ f(n) \right \}$
$=g\left \{ x+1 \right \}$
$=\sin (x+1)$

Functions Exercise 2.3 Question 1 (vii).

Answer :
$\begin{aligned} &f \circ g(x)=2 x+4 \\ &g \circ f(x)=2 x+5 \end{aligned}$
Hint : Domain $f \circ g=\{x: x \in R \text { and } 2 x+3 \in R\}$
Now, we have to compute $f\; o\; g$ and $g\; o\; f$
Given : Here given that
$f(x)=x+1 \text { and } g(x)=2 x+3$
Solution :
First, we will compute $f\; o\; g$
Here, $f: R \rightarrow R, g: R \rightarrow R$
Clearly the range of g is a subset of the domain of f.
$\begin{aligned} \Rightarrow \quad f \circ g: R \rightarrow R & \\ f \circ g(x) &=f\{g(x)\} \\ &=f(2 x+3) \\ &=2 x+3+1 \\ &=2 x+4 \end{aligned}$
Now, we will compute $g\; o\; f$
Clearly the range of f is subset of the domain of g
$\begin{aligned} \Rightarrow \quad f \circ g: R \rightarrow R \\ g \circ f(x) &=g\{f(x)\} \\ &=g(x+1) \\ &=2(x+1)+3 \\ &=2 x+5 . \end{aligned}$
Hence, $f \circ g(x)=2 x+4 \text { and } g \circ f(x)=2 x+5$

Functions Exercise 2.3 Question 1 (viii).

Answer :
$\begin{aligned} &f \circ g(x)=C \\ &g \circ f(x)=\sin C^{2} \end{aligned}$
Hint : The range of g is a subset of the domain of f .
Given: Here given that
$f(x)=C \text { and } g(x)=\sin x^{2}$
Here we have to compute $f\; o\; g$and $g\; o\; f$.
Solution :
First , we will compute the $f\; o\; g$
We have, $f \circ g: R \rightarrow R$
$\begin{aligned} f \circ g(x) &=f\{g(x)\} \\ &=f\left(\sin x^{2}\right) \\ &=C \end{aligned}$
Now we will compute $g\; o\; f$
Clearly the range of f is a subset of the domain of g .
$\begin{aligned} \Rightarrow \quad f \circ g: R & \rightarrow R \\ g \circ f(x) &=g\{f(x)\} \\ &=g(C) \\ &=\sin C^{2} \end{aligned}$

Functions Exercise 2.3 Question 1 (ix).

Answer :
$\begin{aligned} &f \circ g(x)=\frac{3 x^{2}-4 x+2}{(1-x)^{2}} \\ &g \circ f(x)=\frac{\left(x^{2}+2\right)}{\left(x^{2}+1\right)} \end{aligned}$
Hint :$f: R\rightarrow [2,\infty ]$
Given : Here given that $f(x)=x^{2}+1 \text { and } g(x)=1-\frac{1}{1-x}$
For domain of $g=1-x\neq 0$
$\Rightarrow \; \; \; \; \; \; \; x\neq 1.$
Domain of $g=R-\{1\}$
$\\g(x)=1-\frac{1}{x} \\ =\frac{1-x-1}{1-x} \\ =\frac{-x}{1-x}$
For range of g
$\begin{array}{ll} & y=\frac{-x}{1-x} \\ \Rightarrow & y-x y=-x \\ \Rightarrow & y=x y-x \\ \Rightarrow & y=x(y-1) \\ \Rightarrow & x=\frac{y}{y-1} \end{array}$
Range of $g=R-\{1\}$
$\therefore g: R-\{1\} \rightarrow R-\{1\}$
Solution :
First, we compute $f\; o\; g$
Here clearly the range of g is subset of domain of f.
$\begin{aligned} &\Rightarrow \quad f \circ g: R-\{1\} \rightarrow R \\ &\Rightarrow \quad f \circ g(x)=f(g(x)) \\ &\qquad \begin{aligned} \Rightarrow & f\left(\frac{-x}{1-x}\right) \\ &=\left(\frac{x}{(1-x)}\right)^{2}+2 \\ &=\frac{x^{2}+2 x^{2}+2-4 x}{(1-x)^{2}} \\ & f \circ g(x)=\frac{3 x^{2}-4 x+2}{(1-x)^{2}} \end{aligned} \end{aligned}$
Now, we will compute $g\; o\; f$

Clearly the range of f is a subset of the domain of g

$\begin{aligned} \Rightarrow \quad g \circ f: R \rightarrow R \\ g \circ f(x) &=g(f(x)) \\ &=g\left(x^{2}+2\right) \\ &=1-\frac{1}{1-\left(x^{2}+2\right)} \\ &=\frac{-1}{1-\left(x^{2}+2\right)} \\ &=\frac{x^{2}+2}{x^{2}+1} \\ g \circ f(x) &=\frac{x^{2}+2}{x^{2}+1} \end{aligned}$

Functions Exercise 2.3 Question 2.

Given : Here given that
$f(x)=x^{2}+x+1 \text { and } g(x)=\sin x$
To prove : We have to prove that $f \circ g \neq g \circ f$
Solution :
First, we have to compute $f\; o\; g$ and $g\; o\; f$
For $f\; o\; g$
$\begin{aligned} f \circ g(x)=& f(g(x)) \\ &=f(\sin x) \\ &=\sin ^{2} x+\sin x+1 \end{aligned}$ ...(i)
And now for $g\; o\; f$
$\begin{aligned} g \circ f(x)=& g(f(x)) \\ &=g\left(x^{2}+x+1\right) \\ &=\sin \left(x^{2}+x+1\right) \end{aligned}$ ... (ii)
From enq. (1) and (2) we proved that
$f \circ g \neq g \circ f$

Functions Exercise 2.3 Question 3.

Given : Here given that $f(x)=\left | x \right |$
To prove : Here we have to prove that
$f\; o\; f=f$
Solution :
Here we have to compute
$f\; o\; f (x)$
We have ,
$\begin{aligned} &(f \circ f)(x)=f(f(x)) \\ &=f(|x|) \\ &=|| x|| \\ &=|x| \\ &=f(x) \end{aligned}$
So, $(f \circ f)(x)=f(x) \forall x \in R$
Hence, $f\; o\; f=f$

Functions Exercise 2.3 Question 4 (i).

Answer :$f \circ g(x)=2 x^{2}+7$
Given : Here given that
$f(x)=2 x+5 \text { and } g(x)=x^{2}+1$
Here we have to compute $f\; o\; g$
Solution :
Since $f(x)$ and $g(x)$ are polynomials.
$\begin{array}{ll} \Rightarrow \quad & f: R \rightarrow R \text { and } g: R \rightarrow R \\ \Rightarrow \quad & f \circ g(x)=f(g(x)) \\ & =f\left(x^{2}+1\right) \\ & =2\left(x^{2}+1\right)+5 \\ & =2 x^{2}+2+5 \\ & =2 x^{2}+7 \end{array}$
Hence, $f\; o\; g(x)=2x^{2}+7$

Functions Exercise 2.3 Question 4 (ii).

Answer : $g \circ f(x)=4 x^{2}+20 x+26$
Given : Here given that $f(x)=2 x+5 \text { and } g(x)=x^{2}+1$
Here we have to compute $g\; o\; f$
Solution :
Since $f(x)$ and $g(x)$ are polynomials.
$\Rightarrow \quad f: R \rightarrow R \text { and } g: R \rightarrow R$
$\begin{aligned} \Rightarrow \quad g \circ f(x) &=g(f(x)) \\ &=g(2 x+5) \\ &=(2 x+5)^{2}+1 \\ &=4 x^{2}+20 x+26 \end{aligned}$
Hence, $g \circ f(x)=4 x^{2}+20 x+26$

Functions Exercise 2.3 Question 4 (iii).

Answer :$f \circ f(x)=4 x+15$
Given : Here given that $f(x)=2 x+5 \text { and } g(x)=x^{2}+1$
Here we have to compute $f\; o\; f$
Solution :
Since $f(x)$ and $g(x)$ are polynomials
$\Rightarrow \quad f: R \rightarrow R \text { and } g: R \rightarrow R$
$\begin{aligned} \Rightarrow \quad f \circ f(x) &=f(f(x)) \\ &=f(2 x+5) \\ &=2(2 x+5)+5 \\ &=4 x+10+5 \\ &=4 x+15 \end{aligned}$
Hence, $f \circ f(x)=4 x+15$

Functions Exercise 2.3 Question 4 (iv).

Answer :$f^{2}(x)=4 x^{2}+20 x+25$
Given : $f(x)=2 x+5 \text { and } g(x)=x^{2}+1$
Here we have to find out $f^{2}$
Hint : If we want to $f^{2}$ then first we compute $f (x)\times f(x).$
Solution :
Since $f (x)=2x-5$ is a polynomial.
$f:R\rightarrow R$
$\begin{aligned} f^{2}(x) &=f(x) \times f(x) \\ &=(2 x+5)(2 x+5) \\ &=(2 x+5)^{2} \\ f^{2}(x) &=4 x^{2}+20 x+25 \end{aligned}$
In this question we have to prove that
$f\; o\; f=f^{2}$
Then first we have to find out $f\; o\; f$
For $f\; o\; f$
$\begin{aligned} &\operatorname{fof}(x)=f(f(x)) \\ &=f(2 x+5) \\ &=2(2 x+5)+5 \\ &=4 x+10+5 \\ &\operatorname{fof}(x)=4 x+15 \end{aligned}$ ...(1)
Again we will compute $f^{2}$
Since
$\begin{aligned} &f^{2}(x)=f(x) \times f(x) \\ &=(2 x+5)(2 x+5) \\ &=4 x^{2}+20 x+25 \end{aligned}$ ...(2)
From (1) and (2) we have
$(f\; o\; f)(x)=f^{2}(x)$
Hence proved

Function Exercise 2.3 Question 5.

Answer : $f \circ g(x)=\sin 2 x ; g \circ f(x)=2 \sin x$ and they are not equal functions.
Given : Here given that $f(x)=\sin x \; \text {and} \; g(x)=2 x$
Here we have to compute $g\; o\; f$ and $f\; o\; g$
Hint : First we will compute $g\; o\; f$
we know that $f: R \rightarrow[-1,1] \text { and } g: R \rightarrow R$
Clearly the range of f is a subset of the domain of g.
Second, we will compute $f\; o\; g$
Clearly the range of g is a subset of the domain of f.
Solution :
$\begin{aligned} &g \circ f: R \rightarrow R \\ &(g \circ f)(x)=g(f(x)) \\ &\qquad \begin{aligned} &=g(\sin x) \\ &=2 \sin x \end{aligned} \end{aligned}$ ....(1)
$\begin{aligned} &f \circ g: R \rightarrow R \\ &\text { So, } f \circ g(x)=f(g(x)) \\ &\qquad \begin{aligned} &=f(2 x) \\ &=\sin 2 x \end{aligned} \end{aligned}$ ....(2)
From (1) and (2) clearly $f\; o\; g\neq g\; o\; f$
Hence, they are not equal functions.

Function Exercise 2.3 Question 6.

Given : Here given that
$f(x)=\sin x ; g(x)=2 x \& h(x)=\cos x$
To prove : Here we have to prove that $f \circ g=g \circ(f h)$.
Solution :
We know that
$f: R \rightarrow[-1,1] \text { and } g: R \rightarrow R$
Clearly, the range of g is a subset of the domain of f .
$f \circ g: R \rightarrow R$
Now,
$\begin{aligned} (f h)(x)=f(x) h(x) &=(\sin x)(\cos x) \\ &=\frac{1}{2} \sin (2 x) \end{aligned}$
Domain of h is R.
Since range of $\sin x \text { is }[-1,1] ;-1 \leq \sin \sin 2 x \leq 1$
$\Rightarrow \quad-\frac{1}{2} \leq \sin \sin \frac{x}{2} \leq \frac{1}{2}$
Range of $f h=\left[-\frac{1}{2}, \frac{1}{2}\right]$
So, $(f h): R \rightarrow\left[-\frac{1}{2}, \frac{1}{2}\right]$
Clearly the range of fh is a subset of g
$g \circ(f h): R \rightarrow R$
Domain of $f\; o\; g\; \text {and}\; g\; o\; (fh)$ are the same gain
So,
$\begin{aligned} f \circ g(x) &=f(g(x)) \\ (f \circ g)(x) &=f(2 x) \\ &=\sin (2 x) \end{aligned}$
And again ,
$\begin{aligned} &g \circ(f h)(x)=g(f(x) \cdot h(x)) \\ &=g(\sin x \cos x) \\ &=2 \sin x \cos x \\ &=\sin 2 x \end{aligned}$
$\Rightarrow \quad f \circ g(x)=g o(f h)(x) ; \forall x \in R$
Hence, $f \circ g=g \circ(f h)$

Functions Exercise 2.3 Question 7.

Given : Here given that f is a real function and g is a function given by $g(x)=2x$.
To prove : Here we have to prove that $g\circ f=f+f$
Solution : Now, we will take L.H.S. if L.H.S. = R.H.S. then it will be proved.
Take, L.H.S.
$\begin{aligned} g \circ & f(x)=g(f(x)) \\ &=g(f(x)) \\ &=2(f(x)) \\ &=f(x)+f(x) \\ &=\text { R.H.S. } \end{aligned}$
Hence, $g\circ f=f+f$

Functions Exercise 2.3 Question 8.

Answer :
$\begin{aligned} &(g o f)(x)=\log _{e} \sqrt{1-x} \\ &(f \circ g)(x)=\log _{e} \sqrt{1-x} \end{aligned}$
Hint : Domain of f and g are R
Range of $f=(-\infty ,1)$
Range of $f=(0,e)$
Given : Here given that $f(x)=\sqrt{1-x} \text { and } g(x)=\log _{e} x$
Solution :
Clearly Range $f\; \; \subset$ domain of g
$\Rightarrow \quad g \circ f$ exists
And Range $g\; \; \subset$ domain of f
$\Rightarrow \quad f \circ g$
$\begin{aligned} \therefore(g \circ f)(x) &=g(f(x)) \\ &=g(\sqrt{1-x}) \\ &=\log _{e} \sqrt{1-x} \end{aligned}$
Again, we will compute $f \circ g$
$\begin{aligned} \therefore(f \circ g)(x) &=f(g(x)) \\ &=f\left(\log _{e} x\right) \\ (f \circ g)(x)=& \log _{e} \sqrt{1-x} \end{aligned}$

Functions Exercise 2.3 Question 9.

Answer :
$\begin{aligned} &(f \circ g)(x)=\tan \left(\sqrt{1-x^{2}}\right) \\ &(g \circ f)(x)=\tan \left(\sqrt{1-x^{2}}\right) \end{aligned}$
Given : Here given that
$\begin{aligned} &f:\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \rightarrow R \text { and }\\ &g:[-1,1] \rightarrow R \text { defined as } f(x)=\tan x \text { and } g(x)=\sqrt{1-x^{2}} \end{aligned}$
Hint : Since $\in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right), y \in(-\infty, \infty)$
Range of $f\; \; \subset$ domain of $g = \left [ -1,1 \right ]$
Solution :
First, we compute $(f\; \circ \; g)(x)$
We know that
Let $y=f(x)$
$\begin{array}{ll} \Rightarrow \quad y=\tan \tan x \\ \Rightarrow \quad x=y ; x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \\ \Rightarrow \quad(f \circ g)(x)=f(g(x)) \\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;=f\left(\sqrt{1-x^{2}}\right) \\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; =\tan \left(\sqrt{1-x^{2}}\right) \end{array}$
Again we will compute $(g\; \circ \; f)(x)$
$\begin{aligned} (g \circ f)(x) &=g(f(x)) \\ &=g(\tan x) \\ g \circ f(x)=& \tan \left(\sqrt{1-x^{2}}\right) \end{aligned}$

Functions Exercise 2.3 Question 10.

Answer :
$\begin{aligned} &(f \circ g)(x)=\sqrt{x^{2}+4} \\ &(g \circ f)(x)=x+4 \end{aligned}$
Given : Here given that
$f(x)=\sqrt{x+3} \text { and } g(x)=x^{2}+1$
Here we have to find out $f\; \circ \; g$ and $g\; \circ \; f$
Solution :
First, we will compute $f\; \circ \; g$
$\begin{aligned} (f \circ g)(x) &=f(g(x)) \\ &=f\left(x^{2}+1\right) \\ &=\sqrt{x^{2}+1+3} \\ (f \circ g)(x) &=\sqrt{x^{2}+4} \end{aligned}$
Again, we will compute $g\; \circ \; f$
$\begin{aligned} (g \circ f)(x) &=g(f(x)) \\ &=g(\sqrt{x+3}) \\ &=(\sqrt{x+3})^{2}+1 \\ (g \circ f)(x) &=x+4 \end{aligned}$
Hence, $(f \circ g)(x)=\sqrt{x^{2}+4} \text { and }(g \circ f)(x)=x+4$

Functions Exercise 2.3 Question 11 (i) .

Answer :$(f \circ f)(x)=\sqrt{\sqrt{x-2}-2}$
Hint : Domain $f=(2,\infty )$ and Range $(f)=(0,\infty )$
Clearly, the range of $(f)$ is not a subset of domain of f.
Given : Here f be a real function given by $f(x)=\sqrt{x-2}$
Here we have to compute $f\; \circ \; f$
Solution :
We know that
$\because$ Domain of $(f \circ f)=\{x: x \in \text { Domain of } f \text { and } f(x) \in \text { Domain }(f)\}$
$\begin{aligned} &=\{x: x \in(2, \infty) \text { and } \sqrt{x-2} \in(2, \infty) \\ &=\{x: x \in(2, \infty) \text { and } \sqrt{x-2} \geq 2\} \\ &=\{x: x \in(2, \infty) \text { and } x-2 \geq 4\} \\ &=\{x: x \in(2, \infty) x \geq 6 \\ &=(6, \infty) \end{aligned}$
Now,
$\begin{aligned} (f \circ f)(x) &=f(f(x)) \\ &=f(\sqrt{x-2}) \\ &=\sqrt{\sqrt{x-2}-2} \\ \therefore f \circ f:(6, \infty) \rightarrow R \end{aligned}$
Hence, $(f \circ f)(x)=\sqrt{\sqrt{x-2}-2}$

Functions Exercise 2.3 Question 11 (ii) .

Answer :$(f \circ f \circ f)(x)=\sqrt{\sqrt{\sqrt{x-2}-2}-2}$
Hint : Domain $(f)=(2, \infty)$ and range of $(f)=(0, \infty)$
Given : Here f be a real function given by $f(x)=\sqrt{x-2}$
We have to compute $f \circ f \circ f$
Solution :
Clearly the range f is not subset of domain of f.
$\begin{aligned} &\therefore \text { Domain }(f \circ f)=\{x: x \in(2, \infty) \text { and } \sqrt{x-2} \in(2, \infty)\} \\ &=\{x: x \in(2, \infty) \text { and } \sqrt{x-2} \geq 2\} \\ &=\{x: x \in(2, \infty) \text { and } x-2 \geq 4\} \\ &=\{x: x \in(2, \infty) \text { and } x-2 \geq 6 \\ &=(6, \infty) \end{aligned}$
Clearly the range of $f=(0, \infty) \not \subset \text { Domain of }(f \circ f)$
$\begin{aligned} &\therefore \text { Domain of }(f \circ f \circ f)=\{x: x \in(2, \infty) \text { and } \sqrt{x-2} \in(6, \infty)\\ &\begin{aligned} &=\{x: x \in(2, \infty) \text { and } \sqrt{x-2} \geq 6\} \\ &=\{x: x \in(2, \infty) \text { and } x-2 \geq 36\} \\ &=\{x: x \in(2, \infty) \text { and } x \geq 38\} \end{aligned} \end{aligned}$
Now, we will compute $(f \circ f \circ f)(x)$
First we compute $(f \circ f )(x)$
$\begin{aligned} (f \circ f)(x) &=f(f(x)) \\ &=f(\sqrt{x-2}) \\ (f \circ f(x)) &=\sqrt{\sqrt{x-2}-2} \end{aligned}$
Now,
$\begin{aligned} (f \circ f \circ f)(x) &=(f \circ f)(f(x)) \\ &=(f \circ f)(\sqrt{x-2}) \end{aligned}$
$=\sqrt{\sqrt{\sqrt{x-2}-2}-2}$
Hence, $(f \circ f \circ f)(x)=\sqrt{\sqrt{\sqrt{x-2}-2}-2}$

Functions Exercise 2.3 Question 11 (iii) .

Answer :$(f \circ f \circ f)(38)=0$
Hint :$\text { Domain }(f)=(2, \infty) \text { and Range }(f)=(0, \infty)$
Given : Let f be a real function given by $f(x)=\sqrt{x-2}$
Now, we will compute $(f \circ f \circ f)(38)$
Solution :
Clearly the range $(f)$ is not a subset of domain of $(f)$
$\begin{aligned} &\therefore \text { Domain of }(f \circ f)=\{x: x \in \text { Domain of } f \text { and } f(x) \in \text { Domain }(f)\}\\ &=\{x: x \in(2, \infty) \text { and } \sqrt{x-2} \in(2, \infty)\}\\ &=\{x: x \in(2, \infty) \text { and } \sqrt{x-2} \geq 2\}\\ &=\{x: x \in(2, \infty) \text { and } x-2 \geq 4\}\\ &=\{x: x \in(2, \infty) \text { and } x \geq 6\}\\ &=(6, \infty) \end{aligned}$
Clearly range of $f=[0, \infty) \not \subset \text { Domain of }(f \circ f)$
$\text{Domain of}((f \circ f) \circ f)=\{x: x \in \text {Domain of } f \text { and } f(x) \in \text {Domain}(f \circ f)\}$
$\therefore \text { Domain of }((f \circ f) \circ f)=\{x: x \in[2, \infty) \text { and } \sqrt{x-2} \in[\not \subset, \infty)\}$
$\begin{aligned} &=\{x: x \in[2, \infty) \text { and } \sqrt{x-2} \geq 6\} \\ &=\{x: x \in[2, \infty) \text { and } x-2 \geq 36\} \\ &=\{x: x \in[2, \infty) \text { and } x \geq 38\} \\ &=[38, \infty) \end{aligned}$
Now, we will compute $(f \circ f \circ f)(x)$
$\begin{aligned} (f \circ f)(x) &=f(f(x)) \\ &=f(\sqrt{x-2}) \\ &=\sqrt{\sqrt{x-2}-2} \end{aligned}$
Again we will compute $(f \circ f \circ f)(x)$
$\begin{aligned} &(f \circ f \circ f)(x)=(f \circ f)(f(x)) \\ &\qquad \begin{aligned} &\; \; \; \; \; \; \; \; \; \; \; \; \; \;=(f \circ f)(\sqrt{x-2}) \\ &\; \; \; \; \; \; \; \; \; \; \; \; \; \; =(f \circ f)(\sqrt{x-2}) \end{aligned} \\ &\qquad\; \; \; \; \; \; \; \; \; \; \; \; \; \; =\sqrt{\sqrt{\sqrt{x-2}-2}-2} \end{aligned}$
Now, $(f \circ f \circ f)(38)=\sqrt{\sqrt{\sqrt{38-2}-2}-2}$
$\begin{aligned} &=\sqrt{\sqrt{\sqrt{36}-2}-2} \\ &=\sqrt{\sqrt{6-2}-2} \\ &=\sqrt{2-2} \\ &=0 \end{aligned}$
Hence, $(f \circ f \circ f)(38)=0$

Functions Exercise 2.3 Question 11 (iv) .

Given : Here given that
$f(x)=\sqrt{x-2}$
Now we have to compute $f^{2}$ and
To prove : We have to prove that $f \circ f \neq f^{2}$
Solution :
First, we will compute $f^{2}$
$\begin{aligned} \mathrm{f}^{2}(\mathrm{x}) &=\mathrm{f}(\mathrm{x}) \times \mathrm{f}(\mathrm{x}) \\ &=\sqrt{x-2} \times \sqrt{x-2} \\ &=\mathrm{x}-2 \end{aligned}$
Clearly, Domain $(f)=[2, \infty) \text { and Range }(f)=[0, \infty]$
We observe that the range $(f)$ is not subset of domain of f
$\begin{aligned} &\therefore \text { Domain of }(f \circ f)=\{x: x \in \text { Domain }(f) \text { and } f(x) \in \text { Domain }(f)\}\\ &=\{x: x \in[2, \infty) \text { and } \sqrt{x-2} \geq[2, \infty)\}\\ &=\{x: x \in[2, \infty) \text { and } x-2 \geq 4\}\\ &=\{x: x \in[2, \infty) \text { and } x \geq 6\}\\ &=[6, \infty) \end{aligned}$
Now, we will compute $(f \circ f)(x)$
$\begin{aligned} &(f \circ f(x))=f(f(x)) \\ &=f(\sqrt{x-2})=\sqrt{\sqrt{x-2}-2} \end{aligned}$
and again for $f^{2}$
$\begin{aligned} &f^{2}(x)=[f(x)]^{2} \\ &=[\sqrt{x-2}]^{2} \\ &=x-2 \end{aligned}$
Here we see that $f \circ f \neq f^{2}$
Hence, $f \circ f \neq f^{2}$

Functions Exercise 2.3 Question 12 .

Answer :
$f \circ f(x)=\left\{\begin{array}{ll} 2+x ; & 0 \leq x \leq 1 \\ 2-x ; & 1<x \leq 2 \\ 4-x ; & 2<x \leq 3 \end{array}\right.$
Hint : Range of $f=[0,3] \subseteq \text { domain of } f$.
Given : Given that $f(x)=\left\{\begin{array}{cc} 1+x ; & 0 \leq x \leq 2 \\ 3-x ; & 2 \leq x \leq 3 \end{array}\right.$
Here we have find out $f\; o\; f$
Solution :
We have,
Range of $\mathrm{f}=[0,3] \subseteq \text { domain of } \mathrm{f}$
$\begin{aligned} &\therefore f \circ f(x)=\mathrm{f}(\mathrm{f}(\mathrm{x})) \\ &=\mathrm{f}\left\{\begin{array}{cl} 1+\mathrm{x} ; & 0 \leq x \leq 2 \\ 3-\mathrm{x} ; & 2 \leq x \leq 3 \end{array}\right. \\ &\Rightarrow f \circ f(x)=\left\{\begin{array}{cl} 2+x ; & 0 \leq x \leq 1 \\ 2-x ; & 1 \leq x \leq 2 \\ 4-x ; & 2 \leq x \leq 3 \end{array}\right. \end{aligned}$

Functions Exercise 2.3 Question 13 .

Answer : $f \circ g(x)=\left\{\begin{array}{ll} 0, & x \geq 0 \\ -4 x, & x<0 \end{array}\right.$
$\begin{aligned} &g \circ f(x)=0 \text { for all } \mathrm{x} \\ &f \circ g(-3)=12, f \circ g(5)=0 \text { and } g \circ f(-2)=0 \end{aligned}$
Hint : We know about modulus function,
$\begin{aligned} &\mathrm{f}(\mathrm{x})=|x|+\mathrm{x} \\ &\Rightarrow \mathrm{f}(\mathrm{x})=\left\{\begin{array}{l} x+x, \text { if } x \geq 0 \\ -x+x, \text { if } x<0 \end{array}\right. \end{aligned}$
Given : Let f, $g:R\rightarrow R$ be two function defined as :
$\begin{aligned} &\mathrm{f}(\mathrm{x})=|x|+\mathrm{x} \\ &g(\mathrm{x})=|x|-\mathrm{x} \end{aligned}$
Here, we have to find out $f \circ g(x), g \circ f(x), f \circ g(-3), f \circ g(5) \text { and } g \circ f(-2)$
Solution :
Here,
$\begin{aligned} &\mathrm{f}(\mathrm{x})=|x|+\mathrm{x} \text { and } \mathrm{g}(\mathrm{x})=|x|-\mathrm{x} \\ &\mathrm{f}(\mathrm{x})=\left\{\begin{array}{l} x+x, \text { if } x \geq 0 \\ -x+x, \text { if } x<0 \end{array}\right. \\ &\Rightarrow \mathrm{f}(\mathrm{x})=\left\{\begin{array}{c} 2 x, \text { if } x \geq 0 \\ 0, \text { if } x<0 \end{array}\right. \\ &\mathrm{g}(\mathrm{x})=\left\{\begin{array}{l} x-x, \text { if } x \geq 0 \\ -x-x, \text { if } x<0 \end{array}\right. \\ &\Rightarrow g(\mathrm{x})=\left\{\begin{array}{c} 0, \text { if } x \geq 0 \\ -2 x, \text { if } x<0 \end{array}\right. \end{aligned}$
Thus for $x \geq 0, g \circ f(x)=g(f(x))$
$\begin{aligned} &=g(2 x) \\ &=0 \end{aligned}$
For $x<0, g \circ f(x)=g(f(x))$
$\begin{gathered} =\mathrm{g}(0) \\ =0 \\ \Rightarrow g \circ f(x)=0 \forall \mathrm{x} \in \mathrm{R} \end{gathered}$
Thus for $x \geq 0, f \circ g(x)=\mathrm{f}(\mathrm{g}(\mathrm{x}))$
$\begin{aligned} &=\mathrm{f}(0) \\ &=2(0) \\ &=0 \end{aligned}$
For $x<0, f \circ g(x)=\mathrm{f}(\mathrm{g}(\mathrm{x}))$
$\begin{aligned} &=f(-2 x) \\ &=-4 x \end{aligned}$
$\Rightarrow f \circ g(x)=\left\{\begin{array}{ll} 0, & x \geq 0 \\ -4 x, & x<0 \end{array}\right.$
Again,
$\begin{aligned} &\mathrm{f}(\mathrm{x})=|x|+\mathrm{x} \text { and } \mathrm{g}(\mathrm{x})=|x|-\mathrm{x} \\ &\therefore \begin{aligned} \therefore \circ g(x) &=\mathrm{f}(\mathrm{g}(\mathrm{x})) \\ &=\lg (\mathrm{x}) \mid+\mathrm{g}(\mathrm{x}) \\ &=|| x|-x|+|x|-\mathrm{x} \end{aligned} \\ &\begin{aligned} \therefore g \circ f(x) &=\mathrm{g}(\mathrm{f}(\mathrm{x})) \\ &=|\mathrm{f}(\mathrm{x})|-\mathrm{f}(\mathrm{x}) \\ &=|| x|+x|-[|x|+\mathrm{x}] \end{aligned} \end{aligned}$
Then,
$\begin{aligned} &f \circ g(-3)=||-3|-(-3)|+|-3|-(-3) \\ &=|3+3|+3+3 \\ &=6+6 \\ &\quad=12 \\ &\therefore f \circ g(-3)=12 \end{aligned}$
$\begin{aligned} &\begin{aligned} f \circ g(5) &=|| 5|-(5)|+|5|-(5) \\ &=5-5+5-5 \\ &=0 \\ \therefore f \circ g &(5)=0 \end{aligned} \end{aligned}$$\begin{aligned} &\begin{aligned} \therefore g \circ f(-2) &=||-2|+(-2)|-[|-2|+(-2)] \\ &=|2+2|-[2+2] \\ &=4-4 \\ &=0 \\ \therefore g \circ f(-2) &=0 \end{aligned} \end{aligned}$

RD Sharma Class 12th Exercise 2.3 covers many essential topics related to functions and will help students quickly understand many fun and exciting concepts. Here you will learn about different kinds of functions and solve fundamental problems involving them. As this is an important chapter, students can expect many questions from it in entrance exams.

In RD Sharma Class 12th Exercise 2.3, you will learn basic definitions of identity function, modulus function, fractional part function, signum function, etc. Then, you will learn some theories on relations between functions like 'one to one,' 'onto' function, etc., followed by problems. Next, the numerical are divided into different levels to gradually build their problem-solving skills and move on to complex problems.

RD Sharma Class 12th Exercise 2.3 has 20 - Level 1 and 6 - Level 2 example questions that students can use to build their basic problem-solving skills. Level 1 sums contain all the introductory concepts you have covered first in this chapter, followed by Level 2, which has more advanced questions. The Level 2 questions carry more marks but have more complexity, so students have to be perfect in Level 1 sums first.

There are many ways to solve a problem, you don't need to follow the textbook's steps. Students are free to use any method they like to solve a problem as long as it is valid. Getting a thorough knowledge of this material can help students find alternative ways to solve a problem. This saves time and allows students to reduce complexity for their sums.

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