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

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

Edited By Kuldeep Maurya | Updated on Jan 20, 2022 04:17 PM IST

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## Differentiability Excercise: 9.1

Differentiability Exercise 9.1 Question 1

Answer: f(x) is continuous but not differentiable at x = 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 = 3 exist and are equal to each other, then f is said to be continuous at x = 3.
Given: $f\left ( x \right )=|x-3|$
Solution:
Therefore, we can write given function as,
$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{c} -(x-3), x<3 \\ x-3, x \geq 3 \end{array}\right.$
But $f\left ( 3 \right )=3-3=0$
$LHL=\lim_{x\rightarrow 3^{-}}f\left ( x \right )$
$=\lim_{h\rightarrow 0}f\left ( 3-h \right )$
$=\lim_{h\rightarrow 0}\: 3-\left ( 3-h \right )$
$=3-\left ( 3-0 \right )$
$=0$
Now Consider,
$RHL=\lim_{x\rightarrow 3^{+}}f\left ( x \right )$
$=\lim_{h\rightarrow 0} f\left ( 3+h \right )$
$=\lim_{h\rightarrow 0} 3+h-3$
$=3+0-3$
$=0$
$LHL=RHL=f \left ( 3 \right )$
Since f(x) is continuous at x = 3, we have to find its differentiability using the formula,
$f'\left ( x \right )=\lim_{h\rightarrow 0}\frac{f\left ( x+h \right )-f\left ( x \right )}{h}$
Left Hand Derivate (LHD) =$=\lim_{h\rightarrow 0}\frac{f\left ( c+h \right )-f\left ( c \right )}{h}$, denoted by L f (c)
Right Hand Derivate (RHD =$=\lim_{h\rightarrow 0}\frac{f\left ( c+h \right )-f\left ( c \right )}{h}$, denoted by R f (c)
\begin{aligned} &\text { LHD at } \mathrm{x}=3: \lim _{x \rightarrow 3^{3}} \frac{f(x)-f(3)}{x-3} \\ &=\lim _{h \rightarrow 0} \frac{f(3-h)-f(3)}{3-h-3} \\ &=\lim _{h \rightarrow 0} \frac{3-(3-h)-0}{-h} \\ &=\lim _{h \rightarrow 0} \frac{h}{-h} \\ &=-1 \end{aligned}
\begin{aligned} &\mathrm{RHD} \text { at } \mathrm{x}=3: \lim _{x \rightarrow 3^{+}} \frac{f(x)-f(3)}{x-3} \\ &=\lim _{h \rightarrow 0} \frac{f(3+h)-f(3)}{3+h-3} \\ &=\lim _{h \rightarrow 0} \frac{3+h-3-0}{h} \\ &=\lim _{h \rightarrow 0} \frac{h}{h} \\ &=1 \end{aligned}
LHD at x = 3 $\neq$ RHD at x = 3
Hence, f(x) is continuous but not differentiable at x = 3

Differentiability Exercise 9.2 Question 2

Answer: f(x) is not differentiable at x = 0.
Hint: For differentiability, LHD = RHD where LHD is left hand derivative and RHD is right hand derivative. Also, LHD and RHD should exist in given limit.
Given: $f\left ( x \right )=x^{\frac{1}{3}}$
Solution:
As We Know
\begin{aligned} &f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ &\text { LHD 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)^{\frac{1}{3}}}{-h} \\ &=\lim _{h \rightarrow 0} \frac{(-1)^{\frac{1}{3}} h^{\frac{1}{3}}}{(-1) h} \\ &=\lim _{h \rightarrow 0}(-1)^{\frac{-2}{3}} h^{\frac{-2}{3}} \end{aligned}
$\Rightarrow Not \: \: defined$

Differentiability Exercise 9.2 Question 3

Answer: f(x) is differentiable at x = 3 and f  (3) = 12.
Hint: For differentiability, LHD = RHD where LHD is left hand derivative and RHD is right hand derivative. Also, LHD and RHD should exist in given limit.
Given: $f\left ( x \right )=$$\left\{\begin{array}{c} 12 x-13, \text { if } x \leq 3 \\ 2 x^{2}+5, \text { if } x>3 \end{array}\right.$
Solution:
As we know
$f'\left ( x \right )=\lim_{h\rightarrow 0}\frac{f\left ( x+h \right )-f\left ( x \right )}{h}$
Now, we have to check differentiability of given function at x = 3:
$LHD at x=3:\lim_{x\rightarrow 3^{-}}\frac{f\left ( x \right )-f\left ( 3 \right )}{x-3}$
\begin{aligned} &=\lim _{h \rightarrow 0} \frac{f(3-h)-f(3)}{3-h-3} \\ &=\lim _{h \rightarrow 0} \frac{[12(3-h)-13]-[12(3)-13]}{-h} \\ &=\lim _{h \rightarrow 0} \frac{36-12 h-13-36+13}{-h} \\ &=\lim _{h \rightarrow 0} \frac{-12 h}{-h} \\ &=12 \end{aligned}
\begin{aligned} &\text { RHD at } \mathrm{x}=0: \lim _{x \rightarrow 3^{+}} \frac{f(x)-f(3)}{x-3} \\ &=\lim _{h \rightarrow 0} \frac{f(3+h)-f(3)}{3+h-3} \\ &=\lim _{h \rightarrow 0} \frac{\left[2(3+h)^{2}+5\right]-[12(3)-13]}{h} \\ &=\lim _{h \rightarrow 0} \frac{18+12 h+2 h^{2}+5-36+13}{h} \\ &=\lim _{h \rightarrow 0} \frac{2 h^{2}+12 h}{h} \\ &=\lim _{h \rightarrow 0} \frac{h(2 h+12)}{h} \\ &=\lim _{h \rightarrow 0} 2 h+12 \\ &=12 \end{aligned}
(LHD at x = 3) = (RHD at x = 3)
Hence, f(x) is differentiable at x = 3.

Differentiability Exercise 9.2 Question 4

Answer: f(x) is continuous but not differentiable at x = 2.
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(x)=$$\left\{\begin{array}{cc} 3 x-2, & 0 \leq \mathrm{x} \leq 1 \\ 2 x^{2}-x, & 12 \end{array}\right.$
Solution:
Now we have to check for continuity at x = 2.For continuity,
$\left ( LHL\: at\: x=2 \right )=\left ( RHL\: at\: x=2 \right )$
$f\left ( 2 \right )=2\left ( 2 \right )^{2}-2$
$=8-2$
$=6$
$LHL=\lim_{x\rightarrow 2^{-}}f\left ( x \right )$
$=\lim_{h\rightarrow 0}f\left ( 2-h \right )$
$=\lim_{h\rightarrow 0}\left [ 2\left ( 2-h^{2} \right )-\left ( 2-h \right ) \right ]$
$=8-2$
$=6$
Now consider,
$RHL=\lim_{x\rightarrow 2^{+}}f\left ( x \right )$
$=\lim_{h\rightarrow 0}f\left ( 2+h \right )$
$=\lim_{h\rightarrow 0}5\: \left ( 2+h \right )-4$
$=10-4$
$=6$
$LHL=RHL=f\left ( 2 \right )$
Since f(x) is continuous at x = 2, 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}=2: \lim _{x \rightarrow 2^{-}} \frac{f(x)-f(2)}{x-2} \\ &=\lim _{h \rightarrow 0} \frac{f(2-h)-f(2)}{2-h-2} \\ &=\lim _{h \rightarrow 0} \frac{\left[2(2-h)^{2}-(2-h)\right]-[8-2]}{-h} \end{aligned}
\begin{aligned} &=\lim _{h \rightarrow 0} \frac{8-8 h+2 h^{2}-2+h-6}{-h} \\ &=\lim _{h \rightarrow 0} \frac{2 h^{2}-7 h}{-h} \\ &=\lim _{h \rightarrow 0} \frac{h(2 h-7)}{-h} \\ &=\lim _{h \rightarrow 0}(7-2 h) \\ &=7 \end{aligned}
\begin{aligned} &\mathrm{RHD} \text { at } \mathrm{x}=2: \lim _{x \rightarrow 2^{+}} \frac{f(x)-f(2)}{x-2} \\ &=\lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{2+h-2} \\ &=\lim _{h \rightarrow 0} \frac{[5(2+h)-4]-[10-4]}{h} \\ &=\lim _{h \rightarrow 0} \frac{10+5 h-4-6}{h} \\ &=5 \end{aligned}
LHD at x = 2 $\neq$ RHD at x = 2
Hence, f(x) is continuous but not differentiable at x = 2.

Differentiability Exercise 9.2 Question 5

Answer: f(x) is continuous on (-1,2) but not differentiable at x = 0,1.

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: $f\left ( x \right )=|x|+|x-1|$
Solution:
The given function f(x) can be defined as:
$f\left ( x \right )=$ $\left\{\begin{array}{cc} x+x+1, & 1
$\Rightarrow \mathrm{f}(\mathrm{x})=\left\{\begin{array}{c} 2 x+1,-1
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

Answer: f(x) is continuous but not differentiable at x = 0; if 0 < m < 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 )=$$\left\{\begin{array}{c} x^{m} \sin \left(\frac{1}{x}\right), x \neq 0 \\ 0, x=0 \end{array}\right.$
Solution:
Now we have to check for continuity at x = 2.For continuity,
$(LHL at \: x = 2) = (RHL at \: x = 2)$
$f\left ( 0 \right )=0$
$LHL=\lim_{x\rightarrow 0^{-}}f\left ( x \right )$
$=\lim_{h\rightarrow 0^{-}}f\left ( 0-h \right )$
$=\lim_{h\rightarrow 0^{-}}\left ( -h \right )^{m}\sin \left ( \frac{1}{-h} \right )$
$=\lim_{h\rightarrow 0^{-}}\left ( -h \right )^{m}\sin \left ( \frac{1}{h} \right )$
As we know $\left ( 0 \right )^{1}=0$ and $\sin \frac{1}{0}$ $=\sin (\infty)=\mathrm{K}, \quad\{\sin (-\theta) \Rightarrow-\sin \theta\},\{\text { when }-1 \leq k \leq 1\}$
$\Rightarrow 0\: x\: k$
$\Rightarrow 0$
Now consider,
$RHL=\lim_{x\rightarrow 0^{+}}f\left ( x \right )$
$=\lim_{h\rightarrow 0}f\left ( 0+h \right )$
$=\lim_{h\rightarrow 0}\left ( 0+h \right )^{m}\sin \left ( \frac{1}{0+h} \right )$
$=\lim_{h\rightarrow 0}\left ( -h \right )^{m}\sin \left ( \frac{1}{h} \right )$
As we know $\left ( a \right )^{0}=1$ and $\sin \frac{1}{0}=\sin (\infty),\{\text { when }-1 \leq k \leq 1\}$
$\Rightarrow 0\; X \: k$
$\Rightarrow 0$$LHL = RHL = f (0)$

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

Answer: f(x) is differentiable at x = 0, if m > 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

Answer: f(x) is neither continuous nor differentiable at x = 0 for m $\leq$ 0

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 8

Answer: $a=3,b=5$
Hint: For differentiability, LHD = RHD where LHD is left hand derivative and RHD is right hand derivative. Also, LHD and RHD should exist in given limit.
Given: $f\left ( x \right )=$ $\left\{\begin{array}{c} x^{2}+3 x+a, \text { if } x \leq 1 \\ b x+2, \text { if } x>1 \end{array}\right.$
Solution:
\begin{aligned} &\text { LHD 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{\left[(1-h)^{2}+3(1-h)+a\right]-[1+3+a]}{-h} \\ &\therefore\left\{(a-b)^{2}=a^{2}+b^{2}-2 a b\right\} \end{aligned}
\begin{aligned} &=\lim _{h \rightarrow 0} \frac{\left[1+h^{2}-2 h+3-3 h+a\right]-[4+a]}{-h} \\ &=\lim _{h \rightarrow 0} \frac{h^{2}-5 h}{-h} \\ &=\lim _{h \rightarrow 0} \frac{h(h-5)}{-h} \\ &=\lim _{h \rightarrow 0}-(h-5) \\ &=5 \end{aligned}
\begin{aligned} &\mathrm{RHD} \text { at } \mathrm{x}=0: \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{[b(1+h)+2]-(b+2)}{h} \\ &=\lim _{h \rightarrow 0} \frac{[b(1+h)+2]-(b+2)}{h} \\ &=\lim _{h \rightarrow 0} \frac{b h}{h} \\ &=\mathrm{b} \end{aligned}
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) \\ &\quad \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\} \end{aligned}
\begin{aligned} &=\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 \\ &=\mathrm{b}+2 \end{aligned}
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,$a=3 and \; b=5$

Differentiability Exercise 9.1 Question 9

Answer: f(x) is continuous but not differentiable at x = 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 = 3 exist and are equal to each other, then f is said to be continuous at x = 3.
Given: $f\left ( x \right )=$$\left\{\begin{array}{c} |2 x-3|[x], \quad x \geq 1 \\ \sin \left(\frac{n x}{2}\right), \mathrm{x}<1 \end{array}\right.$
Solution:
$f(x)=\left\{\begin{array}{r} |2 x-3|[x], \quad x \geq \frac{3}{2} \\ -(2 x-3), 1 \leq x \leq \frac{3}{2} \\ \sin \left(\frac{n x}{2}\right), x<1 \end{array}\right.$
For continuity at x = 1,
\begin{aligned} &\mathrm{LHL}=\lim _{x \rightarrow 1^{-}} \mathrm{f}(\mathrm{x}) \\ &=\lim _{h \rightarrow 0} \mathrm{f}(1-\mathrm{h}) \\ &=\lim _{h \rightarrow 0} \sin \left(\frac{n(1-h)}{2}\right) \\ &=\sin \left(\frac{n}{2}\right) \end{aligned} $\left [ \frac{n}{2}=90^{o} ,\sin 90^{o}=1\right ]$$=1$

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

Answer: $a=\frac{-1}{2},b=\frac{-3}{2}$
Hint: For differentiability, LHD = RHD where LHD is left hand derivative and RHD is right hand derivative. Also, LHD and RHD should exist in given limit.
Given: $f\left ( x \right )=$ $\left\{\begin{array}{c} a x^{2}-b, \text { if }|x|<1 \\ \frac{1}{|x|}, \text { if }|x| \geq 1 \end{array}\right.$
Solution:
$\Rightarrow\left\{\begin{array}{c} \frac{-1}{x}, x \leq 1 \\ a x^{2}-b,-1
$LHL\; at\: x=1:\lim_{x\rightarrow 1^{-}}f\left ( x \right )$
\begin{aligned} &=\lim _{h \rightarrow 0} f(1-h) \\ &=\lim _{h \rightarrow 0} a(1-h)^{2}-b \\ &\therefore\left\{(a-b)^{2}=a^{2}+b^{2}-2 a b\right\} \\ &=\lim _{h \rightarrow 0} a\left(1+h^{2}-2 h\right)-b \\ &=a-b \end{aligned}
$RHL\; at\: x=1:\lim_{x\rightarrow 1^{+}}f\left ( x \right )$
$=\lim_{h\rightarrow 0}f\left ( 1+h \right )$
$=\lim_{h\rightarrow 0}\frac{1}{1+h}$
$=1$
Since f(x) is continuous, so LHS = RHS.
$a-b=1$ ....(1)
\begin{aligned} &\text { LHD } \Rightarrow \lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1} \\ &\Rightarrow \lim _{h \rightarrow 0} f \frac{f(1-h)-1}{1-h-1} \end{aligned}
\begin{aligned} &\Rightarrow \lim _{h \rightarrow 0} \frac{a(1-h)^{2}-b-1}{-h} \\ &\therefore\left\{(a-b)^{2}=a^{2}+b^{2}-2 a b\right\} \\ &=\lim _{h \rightarrow 0} \frac{a h^{2}-2 a h}{-h} \\ &=\lim _{h \rightarrow 0}(2 a-a h) \\ &=2 \mathrm{a} \end{aligned}\begin{aligned} &\mathrm{LHD} \Rightarrow \lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1} \\ &\Rightarrow \lim _{h \rightarrow 0} f \frac{f(1+h)-1}{1+h-1} \\ &\Rightarrow \lim _{h \rightarrow 0} \frac{\frac{1}{1+h}-1}{h} \\ &=\lim _{h \rightarrow 0} \frac{-h}{(1+h) h} \\ &=\lim _{h \rightarrow 0} \frac{-1}{1+h} \\ &=-1 \end{aligned}

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

Answer: $a=3,b=5$
Hint: For differentiability, LHD = RHD where LHD is left hand derivative and RHD is right hand derivative. Also, LHD and RHD should exist in given limit.
Given: $f\left ( x \right )=$$\left\{\begin{array}{c} x^{2}+3 x+a, \text { if } x \leq 1 \\ b x+2, \text { if } x>1 \end{array}\right.$
Solution:
\begin{aligned} &\text { LHD 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{\left[(1-h)^{2}+3(1-h)+a\right]-[1+3+a]}{-h} \\ &\therefore\left\{(a-b)^{2}=a^{2}+b^{2}-2 a b\right\} \end{aligned}
\begin{aligned} &=\lim _{h \rightarrow 0} \frac{\left[1+h^{2}-2 h+3-3 h+a\right]-[4+a]}{-h} \\ &=\lim _{h \rightarrow 0} \frac{h^{2}-5 h}{-h} \\ &=\lim _{h \rightarrow 0} \frac{h(h-5)}{-h} \\ &=\lim _{h \rightarrow 0}-(h-5) \\ &=5 \end{aligned}
$RHD\; at\: x=0:\lim_{x\rightarrow 1^{+}}\frac{f\left ( x \right )-f\left ( 1 \right )}{x-1}$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{1+h-1} \\ &=\lim _{h \rightarrow 0} \frac{[b(1+h)+2]-(b+2)}{h} \\ &=\lim _{h \rightarrow 0} \frac{[b(1+h)+2]-(b+2)}{h} \\ &=\lim _{h \rightarrow 0} \frac{b h}{h} \\ &=b \end{aligned}

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 $b=5$

Class 12 chapter 9, Differentiability consists of only two exercises, ex 9.1 and ex 9.2. the first exercise, ex 9.1 consists of 13 questions. The various topics that this exercise includes are differentiability at a point, meaning and definition of differentiability at a point, valuable results on differentiability and differentiability in a set. The basics of it were already taught in the previous academic year.

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