RD Sharma Class 12 Exercise 9.1 Differentiability Solutions Maths - Download PDF Free Online

Differentiability Exercise 9.2 Question 2

Differentiability Exercise 9.2 Question 3

Differentiability Exercise 9.2 Question 4

Differentiability Exercise 9.2 Question 5

HInt : The function y = f(x) is said to be differentiable in the closed interval [a, b] if R f ` (a) and L f ` (b) exist and f `(x) exist for every point of (a, b).
If the left hand limit, right hand limit and the value of the function at x = c exist and are equal to each other, then f is said to be continuous at x = c.
Given: $f\left ( x \right )=|x|+|x-1|$
Solution:
The given function f(x) can be defined as:
$f\left ( x \right )=$ $\left\{\begin{array}{cc} x+x+1, & 1<x<0 \\ 1, & 0 \leq \mathrm{x} \leq 1 \\ -x-x+1, & 1<x<2 \end{array}\right.$
$\Rightarrow \mathrm{f}(\mathrm{x})=\left\{\begin{array}{c} 2 x+1,-1<x<0 \\ 1, \quad 0 \leq \mathrm{x} \leq 1 \\ -2 x+1,1<x<2 \end{array}\right.$
We know that a polynomial and a constant function is continuous and differentiable everywhere. So f(x) is continuous and differentiable for x $\epsilon$ (-1, 0) and x $\epsilon$ (0, 1) and (1, 2).
Continuity at $x=0:$
$\begin{aligned} &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}(2 x+1)=1 \\ &\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{-}} 1=1 \\ &\mathrm{f}(0)=2(0)+1=1 \end{aligned}$
Since, $\lim_{x\rightarrow 0^{-}}f\left ( x \right )=\lim_{x\rightarrow 0^{+}}f\left ( x \right )=f\left ( 0 \right ),f\left ( x \right )$is continuous at x = 0.
Continuity at x = 1:
$\begin{aligned} &\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}} 1=1 \\ &\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{-}} 1=1 \\ &\mathrm{f}(1)=1 \end{aligned}$
Since, $\lim_{x\rightarrow 1^{-}}f\left ( x \right )=\lim_{x\rightarrow 1^{+}}f\left ( x \right )=f\left ( 0 \right ),f\left ( x \right )$ is continuous at x = 1.
Differentiability at x = 0:
$\begin{aligned} &\text { LHD at } \mathrm{x}=0: \lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0} \\ &=\lim _{x \rightarrow 0^{-}} \frac{2 x+1-1}{x-0} \\ &=\lim _{x \rightarrow 0^{-}} \frac{2 x}{x} \\ &=2 \end{aligned}$
$\begin{aligned} &\text { RHD at } \mathrm{x}=0: \lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0} \\ &=\lim _{x \rightarrow 0^{+}} \frac{1-1}{x} \\ &=\lim _{x \rightarrow 0^{+}} \frac{0}{h} \\ &=0 \end{aligned}$
LHD at x = 0 $\neq$ RHD at x = 0
Hence, f(x) is continuous but not differentiable at x = 0.

Differentiability at x = 1:
$LHD\: at\: x=1:\lim_{x\rightarrow 1^{-}}\frac{f\left ( x \right )-f\left ( 1 \right )}{x-1}$
$\lim_{x\rightarrow 1^{-}}\frac{1-1}{x-1}$
$=0$
$RHD\: at\: x=1:\lim_{x\rightarrow 1^{+}}\frac{f\left ( x \right )-f\left ( 1 \right )}{x-1}$
$=\lim_{x\rightarrow 1^{+}}\frac{2x+1-1}{x-1}$
$=\lim_{x\rightarrow 1^{+}}\frac{2x}{x-1}$
$=\frac{1}{0}$
$=\infty$

LHD at x = 1 $\neq$ RHD at x = 1

Hence, f(x) is continuous but not differentiable at x = 1.

Differentiability Exercise 9.1 Question 7 Sub Question 2

Since f(x) is continuous at x = 0, we have to find its differentiability using the formula,

$f'\left ( x \right )=\lim_{x\rightarrow 0^{-}}\frac{f\left ( x \right )-f\left ( 0 \right )}{x-0}$

$LHD at x = 0: \lim_{x\rightarrow 0^{-}}\frac{f\left ( x \right )-f\left ( 0 \right )}{x-0}$

$\lim_{x\rightarrow 0}\frac{f\left ( 0-h \right )-f\left ( 0 \right )}{\left ( 0-h \right )-0}$

$=\lim_{h\rightarrow 0}\frac{\left ( -h \right )^{m}\sin\left ( \frac{-1}{h} \right )}{-h}$ $\left \{ \frac{a^{m}}{a^{n}} \Rightarrow a^{m-n}\right \}$

$=\lim_{h\rightarrow 0}-\left ( -h \right )^{m-1}\sin \frac{1}{h}$ $\sin \left ( \frac{1}{0} \right )=\sin (\infty),$

$\Rightarrow Not \: Defined$ $\left [ 0< m< 1 \right ]$

$\begin{aligned} &\mathrm{RHD} \text { at } \mathrm{x}=0: \lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0} \\ &=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{(0+h)-0} \\ &=\lim _{h \rightarrow 0} \frac{h^{m} \sin \left(\frac{1}{h}\right)}{h} \\ &=\lim _{h \rightarrow 0}(h)^{m-1} \sin \left(\frac{1}{h}\right) \end{aligned}$

$\Rightarrow$Not defined

(LHD at x = 0) $\neq$ (RHD at x = 0)

Hence, f(x) is continuous but not differentiable at x = 0.

Differentiability Exercise 9.1 Question 7 Sub Question 1

Hint: The function y = f(x) is said to be differentiable in the closed interval [a, b] if R f ` (a) and L f ` (b) exist and f `(x) exist for every point of (a, b).
If the left hand limit, right hand limit and the value of the function at x = c exist and are equal to each other, then f is said to be continuous at x = c.
Given: $f\left ( x \right )=$$\left\{\begin{array}{c} x^{m} \sin \left(\frac{1}{x}\right), x \neq 0 \\ 0, x=0 \end{array}\right.$
Solution:
Differentiability at x = 0:
$LHD\: at x=0: \lim_{x\rightarrow 0^{-}}\frac{f\left ( x \right )-f\left ( 0 \right )}{x-0}$
$= \lim_{h\rightarrow 0}\frac{f\left ( 0-h \right )-f\left ( 0 \right )}{\left ( 0-h \right )-0}$
$= \lim_{h\rightarrow 0} \frac{\left ( 0-h \right )^{m}\sin\left ( \frac{1}{-h} \right )}{-h}$
$= \lim_{h\rightarrow 0} \left ( -h \right )^{m-1}\sin \left ( \frac{1}{-h} \right )$ $\left \{ \frac{a^{m}}{a^{n}}\Rightarrow a^{m-n} \right \}$
$= \lim_{h\rightarrow 0} \left ( -h \right )^{m-1}\sin \left ( \frac{1}{h} \right )$ $\left \{ \sin \left ( -\theta \right )\Rightarrow -\sin \theta \right \}$
$\begin{aligned} &\text { RHD at } x=0: \lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0} \\ &=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{(0+h)-0} \\ &=\lim _{h \rightarrow 0} \frac{h^{m} \sin \left(\frac{1}{h}\right)}{h} \\ &\Rightarrow\{\text { when }-1 \leq k \leq 1\} \end{aligned}$ $\left\{\begin{array}{c} (0)^{m-1}=0 \\ \sin \left(\frac{1}{0}\right)=\sin (\infty) \end{array}\right.$
$\Rightarrow 0\; x\; k$
$\Rightarrow 0$
$(LHD \: at \: x = 0)= (RHD\: at\: x = 0)$

Hence, f(x) is differentiable at x = 0.

Differentiability Exercise 9.1 Question 7 Sub Question 3

Hint: The function y = f(x) is said to be differentiable in the closed interval [a, b] if R f ` (a) and L f ` (b) exist and f `(x) exist for every point of (a, b).
If the left hand limit, right hand limit and the value of the function at x = c exist and are equal to each other, then f is said to be continuous at x = c.
Given: $f\left ( x \right )=$$\left\{\begin{array}{c} x^{m} \sin \left(\frac{1}{x}\right), x \neq 0 \\ 0, x=0 \end{array}\right.$
Solution:
$\begin{aligned} &\mathrm{LHL}=\lim _{x \rightarrow 0^{-}} \mathrm{f}(\mathrm{x}) \\ &=\lim _{h \rightarrow 0} \mathrm{f}(0-\mathrm{h}) \\ &=\lim _{h \rightarrow 0}(-h)^{m} \sin \left(\frac{1}{-h}\right) \end{aligned}$
As we know $(0)^{1}=0 \text { and } \sin \left(\frac{1}{0}\right)=\sin (\infty) \text {, }$
$\Rightarrow not \: defined \: as\: \: m\leq 0$
$\begin{aligned} &\mathrm{RHL}=\lim _{x \rightarrow 0^{+}} \mathrm{f}(\mathrm{x}) \\ &=\lim _{h \rightarrow 0} \mathrm{f}(0+\mathrm{h}) \\ &=\lim _{h \rightarrow 0}(h)^{m} \sin \left(\frac{1}{h}\right) \end{aligned}$
$\Rightarrow not \: defined \: as\: \: m\leq 0$
Since RHL and LHL are not defined, f (x) is not continuous.
$\begin{aligned} &\mathrm{LHD} \text { at } \mathrm{x}=0: \lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}\\ &=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{(0-h)-0}\\ &=\lim _{h \rightarrow 0} \frac{(-h)^{m} \sin \left(\frac{-1}{h}\right)}{-h}\\ &=\lim _{h \rightarrow 0}-(-h)^{m-1} \sin \left(\frac{1}{h}\right)\\ &\text { Since }\left\{\frac{a^{m}}{a^{n}} \Rightarrow a^{m-n}\right\} \text { and }\{\sin (-\theta) \Rightarrow-\sin \theta\} \text {, } \end{aligned}$
$\Rightarrow not \: defined \: as\: \: m\leq 0$
$\begin{aligned} &\mathrm{RHD} \text { at } \mathrm{x}=0: \lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}\\ &=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{(0+h)-0}\\ &=\lim _{h \rightarrow 0} \frac{(h)^{m} \sin \left(\frac{1}{h}\right)}{h}\\ &=\lim _{h \rightarrow 0}(h)^{m-1} \sin \left(\frac{1}{h}\right)\\ &\text { Since }\left\{\frac{a^{m}}{a^{n}} \Rightarrow a^{m-n}\right\} \text { and }\{\sin (-\theta) \Rightarrow-\sin \theta\} \text {, } \end{aligned}$
$\Rightarrow not \: defined \: as\: \: m\leq 0$
LHD and RHD does not exist .
Hence, f(x) is neither continuous nor differentiable at x = 0.

Differentiability Exercise 9.1 Question 9

Now consider,

$\begin{aligned} &\mathrm{RHL}=\lim _{x \rightarrow 1^{+}} \mathrm{f}(\mathrm{x}) \\ &=\lim _{h \rightarrow 0} \mathrm{f}(1+\mathrm{h}) \\ &=\lim _{h \rightarrow 0}-(2(1+\mathrm{h})-3) \\ &=\lim _{h \rightarrow 0}-(2+2 \mathrm{~h}-3) \\ &=\lim _{h \rightarrow 0}-(-1+2 \mathrm{~h}) \\ &=1 \end{aligned}$

$LHL=RHL=f\left ( 1 \right )$

Since f(x) is continuous at x = 1, we have to find its differentiability using the formula,

$\begin{aligned} &f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ &\mathrm{LHD} \text { at } \mathrm{x}=1: \lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1} \\ &=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{1-h-1} \\ &=\lim _{h \rightarrow 0} \frac{\sin \left(n\left(\frac{1-h}{2}\right)\right)-1}{-h} \\ &=\lim _{h \rightarrow 0} \frac{\sin \left(\frac{n}{2}-\frac{n h}{2}\right)-1}{-h} \end{aligned}$ $\left [ \therefore \sin \left ( \frac{n}{2}-\theta \right )=\cos \theta \right ]$

$=-1$

$\begin{aligned} &\mathrm{RHD} \text { at } \mathrm{x}=1: \lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1} \\ &=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{1+h-1} \\ &=\lim _{h \rightarrow 0} \frac{-[2+2 h-3]-1}{h} \\ &=\lim _{h \rightarrow 0} \frac{-2 h}{h} \\ &=-2 \end{aligned}$
LHD at x = 1 $\neq$ RHD at x = 1. Hence, f(x) is continuous but not differentiable.

Differentiability Exercise 9.1 Question 10

Since f(x) is differentiable at x = 1, (LHD at x = 1) = (RHD at x = 1)

$2a=-1$

$a=\frac{-1}{2}$

Substituting ‘a’ in (1),

$a-b=1$

$\frac{-1}{2}-b=1$

$-b=1+\frac{1}{2}$

$b=\frac{-3}{2}$

Hence,$a=\frac{-1}{2}$ and $b=\frac{-3}{2}$

Differentiability Exercise 9 .1 Question 11

Since f(x) is differentiable, so (LHD at x = 1) = (RHD at x = 1)

$\Rightarrow b=5$

$\Rightarrow f\left ( 1 \right )=1+3+a$

$=4+a$

$\begin{aligned} &\mathrm{LHL} \Rightarrow \lim _{x \rightarrow 1^{-}} f(x) \\ &\Rightarrow \lim _{h \rightarrow 0} f(1-h) \\ &\Rightarrow \lim _{h \rightarrow 0}(1-h)^{2}+3(1-h)+a \\ &\therefore\left\{(a-b)^{2}=a^{2}+b^{2}-2 a b\right\} \\ &=\lim _{h \rightarrow 0} 1+h^{2}-2 h+3-3 h+a \\ &=4+a \end{aligned}$

$\begin{aligned} &\mathrm{RHL} \Rightarrow \lim _{x \rightarrow 1^{+}} f(x) \\ &\Rightarrow \lim _{h \rightarrow 0} f(1+h) \\ &\Rightarrow \lim _{h \rightarrow 0} b(1+h)+2 \\ &=\lim _{h \rightarrow 0} b+b h+2 \end{aligned}$

$=b+2$
Since f(x) is differentiable, f(x) is continuous.

LHL = RHL

$b+2=4+a$

we know that $b=5$

$5+2=4+a$

$7=4+a$

$a=3$

Hence, $a=3,$ and $b=5$