RD Sharma Solutions Class 12 Mathematics Chapter 9 MCQ

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The RD Sharma Class 12th MCQ solutions book is the prescribed guide for the Class 12 students to clear their doubts in the Chapter 9 MCQ section.

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The RD Sharma books contain the solutions for every MCQ question given in the textbook. For instance, you can refer to the RD Sharma Class 12th MCQ book for the Differentiability chapter.

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You can find answers to every question asked in the textbook. Moreover, practice questions are also given to work out further.

RD Sharma Solutions Class 12 Mathematics Chapter 9 MCQ

Edited By Kuldeep Maurya | Updated on Jan 20, 2022 03:54 PM IST

The RD Sharma solution books are widely recommended by most of the CBSE schools to their students. The CBSE Mathematics portions are a bit challenging for the students to crack. And not every student is gifted to afford home tuition or extra classes. When it comes to the 9th chapter, Differentiability, it becomes even more difficult to solve. Here is where the RD Sharma Class 12th MCQ solutions will be of great help.

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RD Sharma Class 12 Solutions Chapter 9 MCQ Differentiability - Other Exercise
Differentiability Excercise:MCQ
RD Sharma Chapter wise Solutions

Also Read - RD Sharma Solution for Class 9 to 12 Maths

RD Sharma Class 12 Solutions Chapter 9 MCQ Differentiability - Other Exercise

Differentiability Excercise:MCQ

Differentiability exercise Multiple choice question, question 1

Answer:

(a)
HINTS: Learn the definition of continuity and differentiability
GIVEN: $f(x)=|x|, g(x)=\left|x^{3}\right|$
SOLUTION:
$\begin{aligned} &f(x)=|x|=\left\{\begin{array}{cc} x & x>0 \\ -x & x<0 \end{array}\right. \\ &g(x)=\left|x^{3}\right|=\left\{\begin{array}{cc} x^{3} & x>0 \\ -x^{3} & x<0 \end{array}\right. \end{aligned}$
Now for the continuity of f(x),
Check at x=0
$\lim _{x \rightarrow 0} f(x)=\lim _{h \rightarrow 0} f(0-h)$
$\begin{aligned} \lim _{h \rightarrow 0} f(-h) &=\lim _{h \rightarrow 0} f(-(-h)) \\ &=0 \\ \lim _{x \rightarrow 0} f(x) &=\lim _{h \rightarrow 0} f(0+h) \\ &=\lim _{h \rightarrow 0} h \\ &=0 \\ \end{aligned}$
As $\begin{aligned} \lim _{x \rightarrow 0^{-}} f(x) &=\lim _{x \rightarrow 0^{+}} f(x) \end{aligned}$
therefore f(x) continuous for x=0
For differentiability of f(x) at x=0
LHD at x=0
$\begin{aligned} \lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0} &=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{0-h-0} \\ &=\lim _{h \rightarrow 0} \frac{(-(-h))-0}{-h} \\ &=\lim _{h \rightarrow 0}-1 \\ &=-1 \end{aligned}$
RHD at x=0
$\begin{aligned} \lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0} &=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{0+h-0} \\ &=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h} \\ &=\lim _{h \rightarrow 0} \frac{h-0}{a} \\ &=\lim _{h \rightarrow 0} 1 \\ &=1 \end{aligned}$
As LHD and RHD at x=0
f(x) is not differentiable at x=0
Now, continuity of g(x)
Check at x=0
$\begin{aligned} \lim _{x \rightarrow 0^{-}} g(x) &=\lim _{h \rightarrow 0} g(0-h) \\ &=\lim _{h \rightarrow 0}-(-h)^{3} \\ &=0 \\ \end{aligned}$
$\begin{aligned} \lim _{x \rightarrow 0^{+}} g(x) &=\lim _{h \rightarrow 0} g(0+h) \\ &=\lim _{h \rightarrow 0} h^{3} \\ &=0 \end{aligned}$
Differentiability of g(x) at x=0
LHD of x=0
$\begin{aligned} \lim _{x \rightarrow 0^{-}} \frac{g(x)-g(0)}{x-0} &=\lim _{h \rightarrow 0} \frac{g(0-h)-g(0)}{0-h-0} \\ &=\lim _{h \rightarrow 0} \frac{-(-h)^{3}-0}{-h} \\ &=\lim _{h \rightarrow 0}-h^{2} \\ &=0 \end{aligned}$
RHD at x=0
$\begin{aligned} \lim _{x \rightarrow 0^{+}} \frac{g(x)-g(0)}{x-0} &=\lim _{h \rightarrow 0} \frac{g(0+h)-g(0)}{0+h-0} \\ &=\lim _{h \rightarrow 0} \frac{h^{3}-0}{h} \\ &=\lim _{h \rightarrow 0} h^{2} \\ &=0 \end{aligned}$
As LHD and RHD at x=0
g(x) is differentiable at x=0

Differentiability exercise Multiple choice question, question 2

Answer:

(b)
HINTS: Understand the definition of continuity and differentiability
GIVEN: $f(x)=\sin ^{-1}(\cos x)$
SOLUTION:
Check the continuity at x=0
Let,
$\begin{aligned} \lim _{x \rightarrow 0^{-}} f(x) &=\lim _{h \rightarrow 0} f(0-h) \\ &=\lim _{h \rightarrow 0} \sin ^{-1}(\cos (0-h)\\ &=\lim _{h \rightarrow 0} \sin ^{-1}(\cos h) \\ \end{aligned}$
$\begin{aligned} &=\sin ^{-1}(\cos 0) \\ &=\sin ^{-1}1 \\ &=\frac{\pi}{2} \\ \end{aligned}$
$\begin{aligned} \lim _{x \rightarrow 0^{+}} f(x) &=\lim _{h \rightarrow 0} f(0+h) \\ &=\lim _{h \rightarrow 0} \sin ^{-1}(\cos h) \\ \end{aligned}$
$\begin{aligned} &=\sin ^{-1}(\cos 0) \\ &=\sin ^{-1} 1 \\ &=\frac{\pi}{2} \end{aligned}$
As $\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)$
therefore f(x) is continuous at x=0
check the differentiability at x=0
LHD at x=0

Differentiability exercise Multiple choice question, question 3

Answer:

(a)
HINTS: Understand the definition of differentiability and modulus function/
GIVEN: $f(x)=x|x|$
SOLUTION:
$f(x)=\left\{\begin{matrix} x(-x) & & x<0\\ x(x)& & x\geq 0 \end{matrix}\right.\\$
$=\left\{\begin{matrix} -x^{2} & & x<0\\ x^{2}& & x\geq 0 \end{matrix}\right.$
For $x<0$ and $x>0$ function is differentiable as it is a polynomial function.
Now at x=0
LHD=
$\begin{aligned} \lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0} &=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{0-h-0} \\ &=\lim _{h \rightarrow 0} \frac{f-(-h)^{2}-0}{-h} \\ \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{-\left(-h^{2}\right)-0}{-h} \\ &=\lim _{h \rightarrow 0}-h \\ &=0 \end{aligned}$
RHD at x=0
$\begin{aligned} \lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0} &=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{0+h-0} \\ &=\lim _{h \rightarrow 0} \frac{h^{2}-0}{h} \\ &=\lim _{h \rightarrow 0} \mathrm{~h} \\ &=0 \end{aligned}$
As LHD at x=0 and RHD at x=0
Hence, f(x) is differentiable at $(-\infty ,\infty )$

Differentiability exercise Multiple choice question, question 4

Answer:

: (b)
HINTS: Understand the definition of continuity and differentiability
SOLUTION:
$f(x)= \begin{cases}\frac{1 x+21}{\tan ^{-1}(x+2)} & x \neq-2 \\ 2 & x=-2\end{cases}$
Check the continuity at x=-2
$\begin{aligned} \lim _{x \rightarrow-2} f(x) &=\lim _{h \rightarrow 0} f(-2-h) \\ \end{aligned}$

$\begin{aligned}&=\lim _{h \rightarrow 0} \frac{1}{\frac{\tan^{-1} h}{h}} \\ \end{aligned}$
$\begin{aligned}&=-1 \quad\left[\operatorname{as}\lim_{h \rightarrow 0} \frac{\tan ^{-1} h}{h}=1\right] \\ \end{aligned}$
$\begin{aligned} \lim _{x \rightarrow-2^{+}} f(x) &=\lim _{h \rightarrow 0} f(-2+h) \\ \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{-2+h+2}{\tan ^{-1}(-2+h+2)} \quad[|x+2|=x+2=x>2] \\ \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{h}{\tan ^{-1} h} \\ &=\lim _{h \rightarrow 0} \frac{1}{\frac{\tan ^{-1} h}{h}} \\ &=1 \end{aligned}$
As $\lim _{x \rightarrow-2^{-}} f(x) \text { and } \lim _{x \rightarrow-2^{+}} f(x)$
f(x) is not continuous at x=-2

Tangents and Normals Exercise Multiple Choice Questions Question 5 .

Answer:
$\left ( c \right )\left ( 0,0 \right )$
Hint:
Use differentiation
Given:
The curve $x=at^{2},y=2at$
Solution:
We Have
$\begin{aligned} &x=a t^{2}\\ &\frac{d x}{d t}=2 a t\\ &\text { And }\\ &y=2 a t\\ &\frac{d y}{d t}=2 a \end{aligned}$
$\frac{dy}{dx}=\frac{1}{t}=\infty,$ For tangent to the curve perpendicular to x-axis
$t=0$
So, the point of contact is $\left ( at^{2},2at \right )=\left ( 0,0 \right )$

Differentiability exercise Multiple choice question, question 5

Answer:

(a) ,(c)
HINTS: Understand the definition of continuity and differentiability
GIVEN: $f(x)=(x+|x|)(|x|)$
SOLUTION: $\begin{aligned} f(x) &=(x+|x|)(|x|) \\ f(x) &= \begin{cases}(x+x) x & x>0 \\( x-x)(-x) & x<0\end{cases} \\ &=\left\{\begin{array}{cc} 2 x^{2} & x>0 \\ 0 & x<0 \end{array}\right. \end{aligned}$
Check the continuity of f(x) at $x<0$
$\begin{aligned} &\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{x \rightarrow 0^{+}} 2 h^{2}=0 \\ \end{aligned}$
As
$\begin{aligned} &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x) \end{aligned}$
Hence, f(x) is continuous at x=0
LHD at x=0
$\begin{aligned} \lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0} &=\lim _{h \rightarrow 0} \frac{f(-h)-f(0)}{-h} \\ &=\lim _{h \rightarrow 0} \frac{0-0}{-h} \\ &=0 \end{aligned}$
RHD at x=0
$\begin{aligned} \lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0} &=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h} \\ &=\lim _{h \rightarrow 0} \frac{2 h^{2}-0}{h} \\ &=\lim _{h \rightarrow 0} 2 h \\ &=0 \end{aligned}$
As LHD at x=0 and RHD at x=0
Hence, f(x) is differentiable.
$f^{\prime}(x)= \begin{cases}4 x & x>0 \\ 0 & x<0\end{cases}$
At x=0
$\begin{aligned} \lim _{x \rightarrow 0^{+}} f(x) &=\lim _{h \rightarrow 0} f(0+h) \\ &=\lim _{h \rightarrow 0} 4 h \\ &=0 \\ \end{aligned}$
$\begin{aligned} \lim _{x \rightarrow 0^{-}} f(x) &=\lim _{h \rightarrow 0^{+}} f(x) \end{aligned}$
Hence $f'(x)$ is continuous
Now,
$\begin{aligned} \lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0} &=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{0-h-0} \\ &=\lim _{h \rightarrow 0} \frac{0-0}{-h} \\ &=0 \end{aligned}$
RHD=
$\begin{aligned} \lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0} &=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{0+h-0} \\ &=\lim _{h \rightarrow 0} \frac{4 h-0}{h} \\ &=\lim _{h \rightarrow 0} \frac{4 h}{h} \\ &=4 \end{aligned}$
Hence $f'(x)$ is not differentiable
$f'(x)$ does not exist

Differentiability exercise Multiple choice question, question 6

Answer:

(a)
HINTS: Understand the definition of continuity and differentiability
GIVEN: $f(x)=e^{-\left | x \right |}$
SOLUTION:
$f(x)= \begin{cases}e^{-x} & x>0 \\ e^{x} & x<0\end{cases}$
At x=0
$\begin{aligned} &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0^{+}} f(o-h)=\lim _{h \rightarrow 0^{+}} e^{-x} \\ &\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{x \rightarrow 0^{+}} e^{-h}=1 \end{aligned}$
Hence, f(x) is continuous everywhere
LHD at x=0
$\begin{aligned} \lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0} &=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{0-h-0} \\ &=\lim _{h \rightarrow 0} \frac{e^{-h}-1}{-h} \\ &=\lim _{h \rightarrow 0} \frac{e^{-h}}{-h} \quad\left[\frac{0}{0} \text { form }\right] \\ &=-e^{-0} \\ &=-1 \end{aligned}$
RHD at x=0
$\begin{aligned} \lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0} &=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{0+h-0} \\ &=\lim _{h \rightarrow 0} \frac{e^{-h}-1}{h} \\ &=\lim _{h \rightarrow 0} \frac{-e^{-h}}{1} \quad\left[\frac{0}{0} \text { form }\right] \\ &=-e^{-0} \\ &=-1 \end{aligned}$
LHD $\neq$ RHD
Hence, f(x) is differentiable at x=0

Tangents and Normals Exercise Multiple Choice Questions Question 8 .

Answer:

Answer:
$\left ( d \right )\left ( 6,36 \right )$
Hint:
Use differentiation and slope of tangent is zero
Given:
The curve $y=12x-x^{2}$
Solution:
Let the point be $p\left ( x,y \right )$
$\begin{aligned} &y=12 x-x^{2} \\ &\frac{d y}{d x}=12-2 x \\ &\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=12-2 x_{1} \end{aligned}$
Since slope of tangent is zero

$\begin{aligned} &\text { So, }\left(\frac{d y}{d x}\right)_{x_{1}, y_{1}}=0\\ &12-2 x_{1}=0\\ &2 x_{1}=12\\ &x_{1}=6 \end{aligned}$
Also curve passing through tangent
$\begin{aligned} &y_{1}=12 x_{1}-x_{1}^{2} \\ &y_{1}=12 \times 6-36 \\ &y_{1}=72-36=36 \end{aligned}$
The Points are $\left ( 6,36 \right )$

Differentiability exercise Multiple choice question, question 7

Answer:

: (b)
HINTS: Understand the definition of continuity and differentiability
SOLUTION:
$f(x)=|\cos x|$
let
$\begin{aligned} f(x) &=|x| \\ g(x) &=\cos x \\ h &=\operatorname{fog}(x) \\ &=f(g(x)) \\ &=f(\cos x) \\ &=|\cos x| \end{aligned}$
As cosx and $\begin{aligned} &|x| \end{aligned}$ are continuous function.Hence $\begin{aligned} &|\cos x| \end{aligned}$ is also continuous function For differentiability
$f(x)= \begin{cases}-\cos x & x<(2 n+1) \frac{\pi}{2} \\ 0 & x=(2 n+1) \frac{\pi}{2} \\ \cos x & x>(2 n+1) \frac{\pi}{2}\end{cases}$
At $(2 n+1) \frac{\pi}{2}$
$\begin{aligned} & \lim _{x \rightarrow(2 n+1) \frac{\pi}{2}} \frac{f(x)-f(2 n+1) \frac{\pi}{2}}{x-(2 n+1) \frac{\pi}{2}} \\ \end{aligned}$
$\begin{aligned} =& \lim _{h \rightarrow 0} \frac{f(2 n+1) \frac{\pi}{2}-h-0}{(2 n+1) \frac{\pi}{2}-h-(2 n+1) \frac{\pi}{2}} \\ \end{aligned}$
$\begin{aligned} =& \lim _{h \rightarrow 0} \frac{-\cos (2 n+1) \frac{\pi}{2}-h}{-1} \\ \end{aligned}$
$\begin{aligned}=& \sin (2 n+1) \frac{\pi}{2} \\ & \lim _{x \rightarrow(2 n+1) \frac{\pi^{-1}}{2}^{-1}} \frac{f(x)-f(2 n+1) \frac{\pi}{2}}{x-(2 n+1) \frac{\pi}{2}} \\ \end{aligned}$
$\begin{aligned} =& \lim _{h \rightarrow 0} \frac{f\left((2 n+1) \frac{\pi}{2}-h\right)-0}{(2 n+1) \frac{\pi}{2}+h-(2 n+1) \frac{\pi}{2}} \\ \end{aligned}$
$\begin{aligned} =& \lim _{h \rightarrow 0} \frac{\cos (2 n+1) \frac{\pi}{2}+h}{-1} \\ =&-\sin (2 n+1) \frac{\pi}{2} \end{aligned}$
As LHD $\neq$ RHD
f(x) is not differentiate at $x=(2 n+1) \frac{\pi}{2}$

Tangents and Normals Exercise Multiple Choice Question Question 11 .

Answer:
$\left ( b \right )x+y-1=x-y-2$
Hint:
Use slope of the tangent $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
Given:
$y=x^{2}-3x+2$
Solution:
$y=x^{2}-3x+2$
Let the tangent meet the x-axis at point $\left ( x,o \right )$
$\frac{dy}{dx}=2x-3$
The tangent passes through point $\left ( x,o \right )$
$\begin{aligned} &0=x^{2}-3 x+2 \\ &(x-2)(x-1)=0 \\ &\Rightarrow x=2, x=1 \end{aligned}$
Case 1
When $x=2$
Slope of tangent $\frac{dy}{dx}|\left ( _{2,0} \right )=2\left ( 2 \right )-3=4-3=1$
$\therefore \left ( x_{1},y_{1} \right )=\left ( 2,0 \right )$
Equation of tangent

$\begin{aligned} &y-y_{1}=m\left(x-x_{1}\right) \\ &y-0=1(x-2) \\ &x-y-2=0 \end{aligned}$
Case 2
When $x=1$
Slope of tangent
$\frac{dy}{dx}|\left ( 2,0 \right )=2-3=-1$
$\left ( x_{1}-y_{1} \right )=\left ( 1,0 \right )$
Equation of tangent

$\begin{aligned} &y-y_{1}=m\left(x-x_{1}\right) \\ &y-0=-1(x-1) \\ &\Rightarrow x+y-1=0 \end{aligned}$

Tangents and Normals Exercise Multiple Choice Questions Question 12 .

Answer:
$\left ( b \right )\frac{6}{7}$
Hint:
Use differentiation
Given:
$x=t^{2}+3t-8$
$y=2t^{2}+2t-5$
Solution:
Given curve are $x=t^{2}+3t-8$ (1)
And $y=2t^{2}+2t-5$ (2)
At $\left ( 2,-1 \right )$
From (1) $t^{2}+3t-10=0$
$\Rightarrow t=2 \; or\: t=-5$
From (2) $2t^{2}-2t-4=0$
$\Rightarrow t^{2}-t-2=0$
$\Rightarrow t=2 \: \: or \: \: t=-1$
From both the solution, we get t=2
Differentiating both the equation w.r.t t, we get

$\frac{dx}{dt}=2t+3$ (3)

$\frac{dy}{dt}=4t-2$ (4)

Now, $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

From (3) and (4) we get

$\frac{4t-2}{2t+3}$

$\therefore \frac{dy}{dx}=\frac{4t-2}{2t+3}$ Is the slope of the tangent to the given curve.

$\left|\frac{d y}{d x}\right|_{(2,-1)}=\left|\frac{4 t-2}{2 t+3}\right|_{t-2}=\frac{8-2}{4+3}=\frac{6}{7}$ Is the slope of the tangent to the given curve at (2,-1)

Tangents and Normals Exercise Multiple Choice Questions Question 13 .

Answer:
$\left ( d \right )\left ( 1,2 \right )$ And $\left ( 1,-2 \right )$
Hint:
Use differentiation
Given:
$x^{2}+y^{2}-2x-3=0$
Solution:
$x^{2}+y^{2}-2x-3=0$
Differentiate w.r.t x, we get
$\begin{aligned} &2 x+2 y \frac{d y}{d x}-2=0 \\ &2 y \frac{d y}{d x}=2-2 x \\ &\frac{d y}{d x}=\frac{2(1-x)}{2 y} \\ &\frac{d y}{d x}=\frac{1-x}{y} \end{aligned}$ (1)
If line is parallel to x-axis, angle with x-axis $=\theta =0$
Slope of x-axis $=\tan \theta =\tan 0^{o}=0$
Slope of tangent =Slope of x-axis
$\frac{dy}{dx}=0$
$\frac{1-x}{y}=0$
$x=1$
Find y When x=1
$\begin{aligned} &x^{2}+y^{2}-2 x-3=0 \\ &(1)^{2}+(y)^{2}-2(1)-3=0 \\ &1+y^{2}-2-3=0 \\ &y=\pm 2 \end{aligned}$
Hence, the points are $\left ( 1,2 \right )$ and $\left ( 1,-2 \right )$

Tangents and Normals Exercise Multiple Choice Questions Question 14 .

Answer:
$\left ( c \right )90^{o}$
Hint:
Use differentiation
Given:
$xy=a^{2}$ And $x^{2}-y^{2}=2a^{2}$
Solution:
$\begin{aligned} &x y=a^{2} \text { And } x^{2}-y^{2}=2 a^{2} \\ &x \frac{d y}{d x}+y=0 \text { And } 2 x-2 y \frac{d y}{d x}=0 \\ &\frac{d y}{d x}=-\frac{y}{x} \text { And } \frac{d y}{d x}=\frac{x}{y} \end{aligned}$
$\begin{aligned} &\left(\frac{d y}{d x}\right)_{1}=-\frac{y}{x} \text { And }\left(\frac{d y}{d x}\right)_{2}=\frac{x}{y} \\ &\left(\frac{d y}{d x}\right)_{1} \times\left(\frac{d y}{d x}\right)_{2}=-\frac{y}{x} \times \frac{x}{y}=-1 \end{aligned}$
Hence the intersection angle $=\frac{\pi }{2}$

Differentiability exercise Multiple choice question, question 8

Answer:

(b)
HINTS: Understand the definition of continuity and differentiability
GIVEN: $f(x)=\sqrt{1-\sqrt{1-x^{2}}}$
SOLUTION:The function is defined only when
$\begin{aligned} &1-x^{2}>0 \\ &1>x^{2} \\ &-1<x<1 \end{aligned}$
Now between $x \in[-1,1]$ at x=0
$f(x)=0$
So we Check the continuity at x=0
LHD =
$\begin{aligned} \lim _{x \rightarrow 0^{-}} f(x) &=\lim _{x \rightarrow 0} \sqrt{1-\sqrt{1-x^{2}}} \\ &=0 \end{aligned}$
RHD =
$\begin{aligned} \lim _{x \rightarrow 0^{+}} f(x) &=\lim _{x \rightarrow 0} \sqrt{1-\sqrt{1-x^{2}}} \\ &=0 \end{aligned}$
As LHD = RHD
M(-1,1)

Differentiability exercise Multiple choice question, question 9

Answer:

(b)
HINTS: Understand the definition of differentiability
GIVEN: $f(x)=a|\sin x|+b e^{|x|}+e|x|^{3}$
SOLUTION:The f(x) is differntaible
If x=0
LHD=RHD
$\begin{aligned} &\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}=\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0} \\ \end{aligned}$
$\begin{aligned} &\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h}=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h} \\ \end{aligned}$
$\begin{aligned} &\lim _{h \rightarrow 0} \frac{a\left(\sin (-4)+b e^{-(-h)}+c\left(-(-h)^{3}\right)-b\right.}{-h} \\ \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{a \sinh +b e^{h}+c h^{3}-b}{-h}=f^{\prime}(0) \end{aligned}$
as
$\begin{aligned} &\left | \sin x \right |= \begin{cases}-\sin x & x<0 \\ \sin x & x>0\end{cases} \\ &|x|= \begin{cases}x & x>0 \\ -x & x<0\end{cases} \\ &|x|^{3}= \begin{cases}x^{3} & x>0 \\ -x^{3} & x<0\end{cases} \end{aligned}$
$\begin{aligned} &\lim _{h \rightarrow 0} \frac{a \sinh +b\left(e^{h}-1\right)+c h^{3}-b}{-h}=\lim _{h \rightarrow 0} \frac{a \sinh +b\left(e^{h}-1\right)+c h^{3}}{-h} \\ \end{aligned}$
$\begin{aligned} &=a \lim _{h \rightarrow 0} \frac{\sinh }{-h}-b \lim _{h \rightarrow 0} \frac{\left(e^{h}-1\right)}{h}+c \lim _{h \rightarrow 0} \frac{+h^{3}}{-h} \\ \end{aligned}$
$\begin{aligned} &=a \lim _{h \rightarrow 0} \frac{\sinh }{-h}-b \lim _{h \rightarrow 0} \frac{\left(e^{h}-1\right)}{h}+c \lim _{h \rightarrow 0} \frac{+h^{3}}{-h}-\frac{h^{3}}{h} \\ \end{aligned}$ $\begin{aligned} &-a \times 1-b \times 1+0=a \times 1+b \times 1 \quad\left[\text { as } \lim _{h \rightarrow 0} \frac{\sinh }{-h}=1 \text { and } \lim _{h \rightarrow 0} \frac{\left(e^{h}-1\right)}{h}=1\right] \\\end{aligned}$ $\begin{aligned} &-a=a \quad \&-b=b \\ &\quad a=0 \quad \& \quad b=0 \end{aligned}$

Differentiability exercise Multiple choice question, question 10

Answer:

(b)
HINTS: Understand the definition of continuity and differentiability
$f(x)=x^{2}+\frac{x^{2}}{1+x^{2}}+\frac{x^{2}}{\left(1+x^{2}\right)^{2}}+\ldots$
GIVEN: then at x=0
SOLUTION:
$f(x)=\sum_{n=0}^{\infty} \frac{x^{2}}{\left(1+x^{2}\right)^{n}}$
Now f(x) is G.P with $\mathrm{r}=\frac{1}{1+x^{2}}$
So,
$\begin{aligned} &f(x)=\frac{x^{2}\left(\frac{1}{1+x^{2}}\right)^{n}-1}{\frac{1}{1+x^{2}}-1}\left[a=x^{2} \text { and } s=\frac{a^{n-1}}{n-1}\right] \\ \end{aligned}$
$\begin{aligned} &f(x)=\frac{\frac{x^{2}-\left(1+x^{2}\right)^{n}}{\left(1+x^{2}\right)^{n}}}{\frac{1-1-x^{2}}{\left(1+x^{2}\right)}} \\ \end{aligned}$
$\begin{aligned} &f(x)=-1\left(1+x^{2}\right)^{-n+1}+x^{-2}\left(1+x^{2}\right) \\ &x=0 \\ \end{aligned}$
$\begin{aligned} &\lim _{x \rightarrow 0} f(x)=\lim _{h \rightarrow 0}-\left(1+h^{2}\right)^{-n-1}+h^{-2}\left(1+h^{2}\right)=\infty \end{aligned}$
As LHL $\neq$ RHL
Therefore, f(x) is discontinues at x=0

Differentiability exercise Multiple choice question, question 11

ANSWER: (a),(b)
HINTS: Find LHD and RHD at x=1
GIVEN: $f(x)=|\log x|$
SOLUTION:
$f(x)=|\log x|= \begin{cases}\log _{e} x & 0<x<1 \\ \log _{e} x & x \geq 1\end{cases}$
LHD at x=1
$\begin{aligned} \lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1} &=\lim _{x \rightarrow 0} \frac{f(1-h)-f(1)}{-h} \\ \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0}-\frac{\log (1-h)-f(1)}{-h} \\ \end{aligned}$
$\begin{aligned}\lim _{h \rightarrow 0}-\frac{\log (1-h)}{-h} &=\lim _{h \rightarrow 0} \frac{1}{1-h} \times-1 \\ &=-1 \end{aligned}$
RHD at x=1
$\begin{aligned} \lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1} &=\lim _{x \rightarrow 1} \frac{\log x-\log 1}{x-1} \\ \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0}-\frac{\log (1+h)}{h} \\ &=\lim _{h \rightarrow 0} \frac{1}{1+h} \\ &=1 \end{aligned}$
As LHD $\neq$ RHD at x=1
$f^{\prime}\left(1^{-}\right)=-1 , f^{\prime}\left(1^{+}\right)=1$

Differentiability exercise Multiple choice question, question 12

Answer:

(b)
HINTS: Understand the definition of continuity and differentiability
GIVEN: $f(x)=\left|\log _{e}\right| x \|$
SOLUTION:
$\begin{aligned} &f(x)= \begin{cases}|\log x| & x>0 \\ |\log (-x)| & x \geq 1\end{cases} \\ \end{aligned}$
$\begin{aligned} &f(x)= \begin{cases}\log (-x) & x<-1 \\ -\log (-x) & -1<x<0 \\ -\log x & 0<x<1 \\ \log x & x>1\end{cases} \end{aligned}$
As f(x) is an absolute function .So it is continues for all x.
Now for differentiability
X=1
LHD=
$\begin{aligned} &=\lim _{x \rightarrow-1^{-}} \frac{f(x)-f(1)}{x-1} \\ &=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h} \\ \end{aligned}$
$\begin{aligned}&=\lim _{h \rightarrow 0}-\frac{\log (1-h)-0}{-h} \\ &=\lim _{h \rightarrow 0}-\frac{\log (1-h)}{-h} \\ &=\lim _{h \rightarrow 0} \frac{-1}{1+h} =-1 \end{aligned}$
RHD =
$\begin{aligned} &=\lim _{x \rightarrow-1^{+}} \frac{f(x)-f(-1)}{x-(-1)} \\ &=\lim _{h \rightarrow 0}-\frac{f(-1+h)-f(-1)}{h} \\ \end{aligned}$
$\begin{aligned} &=\lim _{x \rightarrow 1} \frac{-\log (-(-1+h))-0}{h} \\ &=\lim _{h \rightarrow 0}-\frac{\log (1-h)}{h} \\ \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{-1}{1-h}(-1) \\ &=\lim _{h \rightarrow 0} \frac{1}{1-h}\\ &=1 \end{aligned}$
As LHD $\neq$ RHD at x=-1
Now at x=1
LHD=
$\begin{aligned} &=\lim _{x \rightarrow-1^{-}} \frac{f(x)-f(1)}{x-1} \\ &=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h} \\ \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0}-\frac{\log (1-h)-0}{-h} \\ &=\lim _{h \rightarrow 0}-\frac{\log (1-h)}{-h} \\ &=\lim _{h \rightarrow 0} \frac{1}{1-h} \times(-1)=-1 \end{aligned}$
RHD=
$\begin{aligned} &=\lim _{x \rightarrow-1^{+}} \frac{f(x)-f(1)}{x-1} \\ &=\lim _{h \rightarrow 0}-\frac{f(1+h)-f(1)}{h} \\ \end{aligned}$
$\begin{aligned} &=\lim _{x \rightarrow 1} \frac{\log (1+h)-0}{h} \\ &=\lim _{h \rightarrow 0}-\frac{\log (1+h)}{h} \\ &=\lim _{h \rightarrow 0} \frac{1}{1+h}=1 \end{aligned}$
As LHD $\neq$ RHD
Therefore f(x) is not differentiable at $x\pm 1$

Differentiability exercise Multiple choice question, question 14

ANSWER: (c)
HINTS: understand the continuity and differentiability of [x]
GIVEN: $f(x)=x-[x]$
SOLUTION:
$f(x)=x-[x]$
$f(x)=\left\{\begin{array}{cc} x-[x] & x \neq n \text { when } n \in Z \\ 0 & x \in n \end{array}\right.$
Now,
LHL at x=n where $n \in Z \\$
$\begin{aligned} \lim _{x \rightarrow n^{-}} f(x) &=\lim _{x \rightarrow n} n-(n-1) \quad\{[n]=n-1\\ &=\lim _{x \rightarrow n} 1 \\ &=1 \end{aligned}$
RHL at z=n
$\begin{aligned} \lim _{x \rightarrow n^{+}} f(x)=& \lim _{x \rightarrow n} n-n=0 \quad\{[n]=n\\ & \lim _{x \rightarrow n^{-}} f(x) \text { and } \lim _{x \rightarrow n^{+}} 1 f(x) \end{aligned}$
This f is not continues at integer points. Hence f is continuous at non interger points only

Differentiability exercise Multiple choice question, question 15

Answer:

(d)
HINTS: As x=1 put LHD=RHD
GIVEN: $f(x)=x-[x]$
SOLUTION:
$f(x)= \begin{cases}a x^{2}+1 & x>1 \\ x+\frac{1}{2} & x \leq 1\end{cases}$
As f(x) is derivable at x=1
LHD=RHD at x=1
$\begin{aligned} \lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1} &=\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1} \\ \end{aligned}$
$\begin{aligned} \lim _{h \rightarrow 0} \frac{f(1-h)-f(x)}{-h} &=\lim _{x \rightarrow 0} \frac{a(1+h)^{2}+1-(h)}{h} \\ \end{aligned}$
$\begin{aligned}\lim _{x \rightarrow 0} \frac{-h}{h} &=\lim _{h \rightarrow 0} \frac{a\left[1+h^{2}+2 h\right]-1-\frac{3}{2}}{h} \\ \end{aligned}$
$\begin{aligned} &1 =\lim _{h \rightarrow 0} \frac{a\left[1+h^{2}+2 h\right]-\frac{1}{2}}{h} \\ &1-2 =\lim _{h \rightarrow 0} \frac{a-\frac{1}{2}}{h} \\ &-1=\lim _{h \rightarrow 0} \frac{a-\frac{1}{2}}{h} \end{aligned}$
$\begin{aligned} &\lim _{x \rightarrow 1} \frac{x+\frac{1}{2}-\frac{3}{2}}{x-1}=\lim _{x \rightarrow 1} \frac{a x^{2}+1-\frac{3}{2}}{x-1} \\ \end{aligned}$
$\begin{aligned}&\lim _{x \rightarrow 1} \frac{x-1}{x-1}=\lim _{x \rightarrow 1} \frac{a x^{2}-\frac{1}{2}}{x-1} \\ \end{aligned}$
$\begin{aligned} &\pm=\lim _{x \rightarrow 1} \frac{a x^{2}-\frac{1}{2}}{x-1}\end{aligned}$
As , at x=1 denominator becomes 0.So f limits exist so it must be $\frac{0}{0}$ form
$\begin{aligned} &1=\lim _{x \rightarrow 1} \frac{2 a x}{1} \\ &1=2 a \times 1 \\ &2 a=1 \\ &a=\frac{1}{2} \end{aligned}$

Differentiability exercise Multiple choice question, question 16

Answer:

(b)
HINTS: check the continuity and differentiability at $x=n\pi$
GIVEN: $f(x)=\left | \sin x \right |$
SOLUTION:
$\begin{aligned} &f(x)=|x|, g(x)=\sin x \\ &h(x)=\operatorname{fog}(x) \end{aligned}$
as $\begin{aligned} |x|\end{aligned}$ and sin x are continuous. So composition of function , $\begin{aligned} |\sin x|\end{aligned}$ is continuous
$x=n\pi$
Now , we know $f(x)=\left | \sin x \right |$ is not differentiable where x=0 , $x=n\pi$
so $\begin{aligned} |\sin x|\end{aligned}$ is every were continuous but not differentiable at $x=n\pi$

Differentiability exercise Multiple choice question, question 18

Answer:

(b)
HINTS: composition of function is continuous of function
GIVEN: $f(x)=1+|\cos x|$
SOLUTION:
Let $g(x)=1+|x|, h(x)=\cos x$ as $1+|x|$ and cos x are continuous function
Therefore, composition of function ,that is is also continuous.
$f(x)=g o h(x)=g(h(x))$
Now,
cos x is differentiable
$g(x)=1+|x|= \begin{cases}1+x & x>1 \\ 1-x & x<1\end{cases}$
At, x=1
LHD= $\lim _{x \rightarrow 1} \frac{g(x)-g(1)}{x-1}$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{g(1-h)-g(1)}{-h} \\ &=\lim _{h \rightarrow 0} \frac{1-(1-h)-2}{-h} \\ &=\infty \end{aligned}$
$\begin{aligned} &\lim _{h \rightarrow 0} \frac{g(x)-g(1)}{x-1}=\lim _{h \rightarrow 0} \frac{g(1+h)-g(1)}{h} \\ \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{1-(1+h)-2}{h} \\ &=\lim _{h \rightarrow 0} \frac{h}{h} \\ &=1\end{aligned}$
As LHD $\neq$ RHD
g(x) is not differentiable at x=1
therefore f(x) is also with differentiation at x=1
at x=1
$\begin{aligned} &1+\cos x=1 \\ &\cos x=0 \\ &x=n \pi \end{aligned}$
Hence f(x) is continuous everywhere.

Differentiability exercise Multiple choice question, question 19

Answer:

(b)
HINTS: understand the definition of continuity and differentiability
GIVEN: $f(x)=|\cos x|$
SOLUTION:
As $\cos x$ & $\left | x \right |$ are continuous function .Hence $\cos x$ is also continuous function.
For differentiability ,
$\begin{aligned} &f(x)= \begin{cases}-\cos x & x<(2 n+1) \frac{\pi}{2} \\ 0 & x=(2 n+1) \frac{\pi}{2} \\ \cos x & x>(2 n+1) \frac{\pi}{2}\end{cases} \\ \end{aligned}$
at $\begin{aligned}&(2 n+1) \frac{\pi}{2} & \end{aligned}$
$\begin{aligned} &\lim _{x \rightarrow(2 n+1) \frac{\pi}{2}} \frac{f(x)-f(2 n+1) \frac{\pi}{2}}{x-(2 n+1) \frac{\pi}{2}} \\ &=\lim _{h \rightarrow 0} \frac{f(2 n+1) \frac{\pi}{2}-h-0}{(2 n+1) \frac{\pi}{2}-h-(2 n+1) \frac{\pi}{2}} \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{-\cos (2 n+1) \frac{\pi}{2}-h}{-1} \\ &=\sin (2 n+1) \frac{\pi}{2} \end{aligned}$
$\begin{aligned} &\lim _{x \rightarrow(2 n+1) \frac{\pi^{-1}}{2}^{-1}} \frac{f(x)-f(2 n+1) \frac{\pi}{2}}{x-(2 n+1) \frac{\pi}{2}} \\ &==\lim _{h \rightarrow 0} \frac{f\left((2 n+1) \frac{\pi}{2}-h\right)-0}{(2 n+1) \frac{\pi}{2}+h-(2 n+1) \frac{\pi}{2}} \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{\cos (2 n+1) \frac{\pi}{2}+h}{-1} \\ &=\lim _{h \rightarrow 0} \frac{-\sin (2 n+1) \frac{\pi}{2}+h}{1} \\ &=-\sin (2 n+1) \frac{\pi}{2} \end{aligned}$
As LHD $\neq$ RHD
f(x) is not differentiate at $x=(2 n+1) \frac{\pi}{2}$

Differentiability exercise Multiple choice question, question 21

Answer:

(b)
Hint: LHD = RHD at x = 0
Given : $f(x)=a+b|x|+c|x|^{4}$
Explanation: LHD at x = 0
$\begin{aligned} &\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x-0}=\lim _{h \rightarrow 0} \frac{f(-h)-f(0)}{-h} \\ \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{a+b(-(-h))+c(-h)^{4}-a}{-h} \\ &=\lim _{h \rightarrow 0} \frac{b h+c h^{4}}{-h}=-b\end{aligned}$
RHD at x = 0
$\begin{aligned} &\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x-0}=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h} \\ \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{a+b h+c h^{4}-a}{h} \\ &=\lim _{h \rightarrow 0} \frac{b h+c h^{4}}{h}=b \end{aligned}$
Thus f(x) is differentiable at x = 0
LHD = RHD
$\begin{aligned} &-b=b \\ &2 b=0 \\ &b=0 \end{aligned}$

Differentiability exercise Multiple choice question, question 20

Answer:

(a)
Hint: understand the greatest integer function
Given : $f(x)=\frac{\sin (\pi[x-\pi])}{h+[x]^{2}} \cdots(i)$
Explanation : $\pi[x-\pi]=n \pi$
Taking sin on both sides
$\begin{aligned} &\sin (\pi[x-\pi])=\sin (n \pi) \\ &\sin (\pi[x-\pi])=0 \end{aligned}$
Put in 0
$\begin{aligned} f(x) &=\frac{0}{h+[x]^{2}} \quad\left[a s h+[x]^{2} \neq 0\right] \\ &=0 \end{aligned}$
As f(x) is constant function so it is continuous as well as differentiable for all $x \in R$

Differentiability exercise Multiple choice question, question 22

Answer:

(d)
HINTS: check at x=3 LHL & RHL,LHD & RHD
GIVEN: $f(x)=|3-x|+(3+x)$
SOLUTION:
To check at x=3 we take the interval [2,4]
$\begin{aligned} f(x) &= \begin{cases}3-x+(3+3) & 2<x<3 \\ -3+x+3+4 & 3<x<4\end{cases} \\ &= \begin{cases}-x+9 & 2<x<3 \\ x+4 & 3<x<4\end{cases} \end{aligned}$
As (x) divides the least integer function.
At x=3
$\begin{aligned} &\lim _{x \rightarrow 3^{-}} f(x)=\lim _{h \rightarrow 0}-x+9=-3+9=6 \\ &\lim _{x \rightarrow 3^{-}} f(x)=\lim _{h \rightarrow 0}-x+4=3+4=7 \end{aligned}$
As LHL $\neq$ RHL
So f(x) is not continuous at x=3
So f(x) is also not differentiable at x=3

Differentiability exercise Multiple choice question, question 23

Answer:

(d)
HINTS: check at x=0 LHD & RMD
GIVEN: $f(x)=\left\{\begin{array}{cc} \frac{1}{1+e^{\frac{1}{x}}} & x \neq 0 \\ 0 & x=0 \end{array}\right.$
SOLUTION:
At x=0
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(-h)=\lim _{h \rightarrow 0} \frac{1}{1+e^{\frac{1}{-h}}}=1$
RHL=
$\begin{aligned} &=\lim _{x \rightarrow \sigma^{+}} f(x)=\lim _{h \rightarrow 0} f(h) \\ &=\lim _{h \rightarrow 0} \frac{1}{1+e^{\frac{1}{-h}}}=\frac{1}{1+e^{\infty}}=0 \end{aligned}$
As LHL $\neq$ RHL
f(x) is not continuous, so f(x) is not differentiable

Differentiability exercise Multiple choice question, question 24

Answer:

(a)
HINTS: understand the definition of continuity and differentiability.
GIVEN: $f(x)=\left\{\begin{array}{cl} \frac{1-\cos x}{x \sin x} & x \neq 0 \\ \frac{1}{2} & x=0 \end{array}\right.$
SOLUTION:
At x=0
$\begin{aligned} \lim _{x \rightarrow 0^{-}} f(x) &=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} \frac{1-\cos (-h)}{-n \sin (-h)} \\ \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{1-\cos (-h)}{-h \sin (-h)} \\ &=\lim _{h \rightarrow 0} \frac{1-\cos h}{h \sin h} \\ \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{\pm \sinh }{\sin h+h \cosh } \\ &=\lim _{h \rightarrow 0} \frac{\frac{1}{\sinh +h \cosh }}{\sinh } \\ &=\lim _{h \rightarrow 0} \frac{1}{1+h \operatorname{coth}}\\ &=1 \end{aligned}$
RHL=
$\begin{aligned} \lim _{x \rightarrow 0^{+}} f(x) &=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} \frac{1-\cos -h}{-n \sin -h} \\ \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{0-\sin h}{\sinh +h \cosh } \\ &=\lim _{h \rightarrow 0} \frac{1}{\frac{\sinh }{\sin h}+\frac{\cosh }{\sinh }} \\ &=1\end{aligned}$
As LHL=RHL
f(x) is not continuous, so f(x) is not differentiable
Now,
LHD=
$\begin{aligned} & \lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h} \\ =& \lim _{h \rightarrow 0} \frac{\frac{1-(\cos (-h)}{-h \sin (-h)}-\frac{1}{2}}{-h} \\ \end{aligned}$
$\begin{aligned} =& \lim _{h \rightarrow 0} \frac{\frac{1-\cos h}{h \sinh }-\frac{1}{2}}{-h} \\ =& \lim _{h \rightarrow 0} \frac{1-\cos h}{-\sinh }+\lim _{h \rightarrow 0} \frac{1}{2 h}=\infty \end{aligned}$
RHD=
$\begin{aligned} &\lim _{x \rightarrow 0^{-1}} \frac{f(x)-f(0)}{x-0}=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h} \\ \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{\frac{1-\cos h}{h \sinh }-\frac{1}{2}}{-a} \\ &=\lim _{h \rightarrow 0} \frac{1-\cos h}{-\sinh }-\lim _{h \rightarrow 0} \frac{1}{2 h}=\infty\end{aligned}$
F(x) is continuous and differential

Differentiability exercise Multiple choice question, question 25

Answer:

(b)
HINTS: understand the definition of differentiability.
GIVEN: $f(x)=|x-3| \cos x$
SOLUTION:
$f(x)= \begin{cases}-(x-3) \cos x & x<3 \\ (x-3) \cos x & x>3\end{cases}$
At x=3
$\begin{aligned} &=\lim _{x \rightarrow-1^{-}} \frac{f(x)-f(3)}{x-3} \\ &=\lim _{h \rightarrow 0} \frac{f(3-h)-0}{-h} \\ \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{-(3-h-3) \cos (3-h)}{-h} \\ &=\lim _{h \rightarrow 0} \cos (3-h)=-\cos 3 \end{aligned}$
RHD=
$\begin{aligned} &=\lim _{x \rightarrow-3^{-}} \frac{f(x)-f(3)}{x-3} \\ &=\lim _{h \rightarrow 0} \frac{f(3+h)-f(3)}{x-3} \\ \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{(3+h-3) \cos (3+h)-0}{h} \\ &=\lim _{h \rightarrow 0} \cos (3+h)=\cos 3 \end{aligned}$
So f(x) is not differentiable at x=-3.At other point f(x) is a product of two continuous and differential function (x-3 and cos x).So f(x) is differentiable at R-3

Differentiability exercise Multiple choice question, question 26

Answer:

(a)
HINTS: understand the definition of continuity and differentiability.
GIVEN: $f(x)=\left\{\begin{array}{cc} 1 & x \leq-1 \\ |x| & -1<x<1 \\ 0 & x \geq 1 \end{array}\right.$
SOLUTION:
At x=-1
$\begin{aligned} &\lim _{x \rightarrow-1^{-}} f(x)=\lim _{x \rightarrow-1} 1=1 \\ &\lim _{x \rightarrow-1^{+}} f(x)=\lim _{x \rightarrow-1}|x|=\lim _{x \rightarrow-1}-x=1 \end{aligned}$
So f(x) is continuous at x=-1
Now at x=1
$\begin{aligned} &\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow-1} x=1 \\ &\lim _{x \rightarrow-1^{+}} f(x)=\lim _{x \rightarrow 1} 0=0 \\ \end{aligned}$
As,
$\begin{aligned} &\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x) \end{aligned}$
f(x) is not continuous at x=+1.Hence not differentiable at x=1

Differentiability exercise Multiple choice question, question 26
Edit Q

Question:26

Differentiability exercise Multiple choice question, question 26

Answer:

Differentiability exercise Multiple choice question, question 27

Answer:

(a)
Hint: understand the concept of continuity and differentiability
Given: $f(x)=e^{|x|}$
Explanation:
$f(x)= \begin{cases}e^{x} & x>0 \\ e^{-x} & x<0\end{cases}$
$\begin{aligned} &L H L=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0} e^{-0}=1 \\ &R H L=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0} e^{+0}=1 \end{aligned}$
as LHL = RHL
$\therefore f(x)$ continuous at x
$\begin{aligned} &L H D=\lim _{x \rightarrow 0^{-}}=\frac{f(x)-f(0)}{x-0} \\ \end{aligned}$
$\begin{aligned}&=\lim _{h \rightarrow 0} \frac{f(x)-f(0)}{x-0}=\lim _{h \rightarrow 0} \frac{e^{-(-h)}-1}{-h} \\ \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{e^{h}-1}{-h}=-1 \quad\left[\lim _{x \rightarrow 0} \frac{e^{x}-1}{x}=1\right] \end{aligned}$
$\begin{aligned} &R H D=\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0} \\ \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{e^{h}-1}{h}=1 \\ \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{e^{h}}{1}=1 \end{aligned}$
As LHD $\neq$ RHD
$f(x)$ is not differentiable

RD Sharma Class 12 Solutions Chapter 9 MCQ_{Differentiability for Class 12, consists of two exercises, ex 9.1 and ex 9.2. Next comes the MCQ section, with 28 questions to be answered. The concepts like Differentiability functions, continuous function, and discontinuous function are asked. When the MCQs give an idea about the answer from one side, on the other side, they confuse students with similar options. This makes students lose marks in MCQs. Using the RD Sharma Class 12 Chapter 9 MCQ solutions, the students can clear their doubts and score good marks.}

_{The solutions presented in the RD Sharma Class 12th MCQ book are framed by experts with broad knowledge in the particular domain. Students use this material while doing their homework, assignment, and preparing for their exams. It follows the NCERT pattern making it even easier for the CBSE board students to adapt to it.}

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