RD Sharma Solutions Class 12 Mathematics Chapter 9 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:
$\begin{array}{rl}& f(x)=|x|=\{\begin{array}{cc}x& x>0\\ -x& x<0\end{array}\\ & g(x)=\left|x^3\right|=\{\begin{array}{cc}{x}^3& x>0\\ -x^3& x<0\end{array}\end{array}$
Now for the continuity of f(x),
Check at x=0
$\underset{x\to 0}{lim}f(x)=\underset{h\to 0}{lim}f(0-h)$
$\begin{array}{rl}\underset{h\to 0}{lim}f(-h)& =\underset{h\to 0}{lim}f(-(-h))\\ & =0\\ \underset{x\to 0}{lim}f(x)& =\underset{h\to 0}{lim}f(0+h)\\ & =\underset{h\to 0}{lim}h\\ & =0\end{array}$
As $\begin{array}{rl}\underset{x\to 0^{-}}{lim}f(x)& =\underset{x\to 0^{+}}{lim}f(x)\end{array}$
therefore f(x) continuous for x=0
For differentiability of f(x) at x=0
LHD at x=0
$\begin{array}{rl}\underset{x\to 0^{-}}{lim}\frac{f(x)-f(0)}{x-0}& =\underset{h\to 0}{lim}\frac{f(0-h)-f(0)}{0-h-0}\\ & =\underset{h\to 0}{lim}\frac{(-(-h))-0}{-h}\\ & =\underset{h\to 0}{lim}-1\\ & =-1\end{array}$
RHD at x=0
$\begin{array}{rl}\underset{x\to 0^{+}}{lim}\frac{f(x)-f(0)}{x-0}& =\underset{h\to 0}{lim}\frac{f(0+h)-f(0)}{0+h-0}\\ & =\underset{h\to 0}{lim}\frac{f(h)-f(0)}{h}\\ & =\underset{h\to 0}{lim}\frac{h-0}{a}\\ & =\underset{h\to 0}{lim}1\\ & =1\end{array}$
As LHD and RHD at x=0
f(x) is not differentiable at x=0
Now, continuity of g(x)
Check at x=0
$\begin{array}{rl}\underset{x\to 0^{-}}{lim}g(x)& =\underset{h\to 0}{lim}g(0-h)\\ & =\underset{h\to 0}{lim}-(-h{)}^3\\ & =0\end{array}$
$\begin{array}{rl}\underset{x\to 0^{+}}{lim}g(x)& =\underset{h\to 0}{lim}g(0+h)\\ & =\underset{h\to 0}{lim}{h}^3\\ & =0\end{array}$
Differentiability of g(x) at x=0
LHD of x=0
$\begin{array}{rl}\underset{x\to 0^{-}}{lim}\frac{g(x)-g(0)}{x-0}& =\underset{h\to 0}{lim}\frac{g(0-h)-g(0)}{0-h-0}\\ & =\underset{h\to 0}{lim}\frac{-(-h{)}^3-0}{-h}\\ & =\underset{h\to 0}{lim}-h^2\\ & =0\end{array}$
RHD at x=0
$\begin{array}{rl}\underset{x\to 0^{+}}{lim}\frac{g(x)-g(0)}{x-0}& =\underset{h\to 0}{lim}\frac{g(0+h)-g(0)}{0+h-0}\\ & =\underset{h\to 0}{lim}\frac{h^3-0}{h}\\ & =\underset{h\to 0}{lim}{h}^2\\ & =0\end{array}$
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,
$\begin{array}{rl}\underset{x\to 0^{-}}{lim}f(x)& =\underset{h\to 0}{lim}f(0-h)\\ & =\underset{h\to 0}{lim}{\sin }^{-1}(\cos (0-h)\\ & =\underset{h\to 0}{lim}{\sin }^{-1}(\cos h)\end{array}$
$\begin{array}{rl}& ={\sin }^{-1}(\cos 0)\\ & ={\sin }^{-1}1\\ & =\frac{\pi }{2}\end{array}$
$\begin{array}{rl}\underset{x\to 0^{+}}{lim}f(x)& =\underset{h\to 0}{lim}f(0+h)\\ & =\underset{h\to 0}{lim}{\sin }^{-1}(\cos h)\end{array}$
$\begin{array}{rl}& ={\sin }^{-1}(\cos 0)\\ & ={\sin }^{-1}1\\ & =\frac{\pi }{2}\end{array}$
As $\underset{x\to 0^{-}}{lim}f(x)=\underset{x\to 0^{+}}{lim}f(x)$
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:
$f(x)=\{\begin{array}{ccc}x(-x)& & x<0\\ x(x)& & x\ge 0\end{array}$
$=\{\begin{array}{ccc}-x^2& & x<0\\ x^2& & x\ge 0\end{array}$
For $x<0$ and $x>0$ function is differentiable as it is a polynomial function.
Now at x=0
LHD=
$\begin{array}{rl}\underset{x\to 0^{-}}{lim}\frac{f(x)-f(0)}{x-0}& =\underset{h\to 0}{lim}\frac{f(0-h)-f(0)}{0-h-0}\\ & =\underset{h\to 0}{lim}\frac{f-(-h{)}^2-0}{-h}\end{array}$
$\begin{array}{rl}& =\underset{h\to 0}{lim}\frac{-(-h^2)-0}{-h}\\ & =\underset{h\to 0}{lim}-h\\ & =0\end{array}$
RHD at x=0
$\begin{array}{rl}\underset{x\to 0^{+}}{lim}\frac{f(x)-f(0)}{x-0}& =\underset{h\to 0}{lim}\frac{f(0+h)-f(0)}{0+h-0}\\ & =\underset{h\to 0}{lim}\frac{h^2-0}{h}\\ & =\underset{h\to 0}{lim}\mathrm{h}\\ & =0\end{array}$
As LHD at x=0 and RHD at x=0
Hence, f(x) is differentiable at $(-\mathrm{\infty },\mathrm{\infty })$

Differentiability exercise Multiple choice question, question 4

Answer:

: (b)
HINTS: Understand the definition of continuity and differentiability
SOLUTION:
$f(x)=\{\begin{array}{ll}\frac{1x+21}{{\tan }^{-1}(x+2)}& x\ne -2\\ 2& x=-2\end{array}$
Check the continuity at x=-2
$\begin{array}{rl}\underset{x\to -2}{lim}f(x)& =\underset{h\to 0}{lim}f(-2-h)\end{array}$

$\begin{array}{rl}& =\underset{h\to 0}{lim}\frac{1}{\frac{{\tan }^{-1}h}{h}}\end{array}$
$\begin{array}{rl}& =-1[\mathrm{as}\underset{h\to 0}{lim}\frac{{\tan }^{-1}h}{h}=1]\end{array}$
$\begin{array}{rl}\underset{x\to -2^{+}}{lim}f(x)& =\underset{h\to 0}{lim}f(-2+h)\end{array}$
$\begin{array}{rl}& =\underset{h\to 0}{lim}\frac{-2+h+2}{{\tan }^{-1}(-2+h+2)}[|x+2|=x+2=x>2]\end{array}$
$\begin{array}{rl}& =\underset{h\to 0}{lim}\frac{h}{{\tan }^{-1}h}\\ & =\underset{h\to 0}{lim}\frac{1}{\frac{{\tan }^{-1}h}{h}}\\ & =1\end{array}$
As $\underset{x\to -2^{-}}{lim}f(x)\text{ and }\underset{x\to -2^{+}}{lim}f(x)$
f(x) is not continuous at x=-2

Tangents and Normals Exercise Multiple Choice Questions Question 5 .

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

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:$\begin{array}{rl}f(x)& =(x+|x|)(|x|)\\ f(x)& =\{\begin{array}{ll}(x+x)x& x>0\\ (x-x)(-x)& x<0\end{array}\\ & =\{\begin{array}{cc}2x^2& x>0\\ 0& x<0\end{array}\end{array}$
Check the continuity of f(x) at $x<0$
$\begin{array}{rl}& \underset{x\to 0^{+}}{lim}f(x)=\underset{h\to 0}{lim}f(0+h)=\underset{x\to 0^{+}}{lim}2h^2=0\end{array}$
As
$\begin{array}{rl}& \underset{x\to 0^{-}}{lim}f(x)=\underset{x\to 0^{+}}{lim}f(x)\end{array}$
Hence, f(x) is continuous at x=0
LHD at x=0
$\begin{array}{rl}\underset{x\to 0^{-}}{lim}\frac{f(x)-f(0)}{x-0}& =\underset{h\to 0}{lim}\frac{f(-h)-f(0)}{-h}\\ & =\underset{h\to 0}{lim}\frac{0-0}{-h}\\ & =0\end{array}$
RHD at x=0
$\begin{array}{rl}\underset{x\to 0^{+}}{lim}\frac{f(x)-f(0)}{x-0}& =\underset{h\to 0}{lim}\frac{f(0+h)-f(0)}{h}\\ & =\underset{h\to 0}{lim}\frac{2h^2-0}{h}\\ & =\underset{h\to 0}{lim}2h\\ & =0\end{array}$
As LHD at x=0 and RHD at x=0
Hence, f(x) is differentiable.
$f^{\prime }(x)=\{\begin{array}{ll}4x& x>0\\ 0& x<0\end{array}$
At x=0
$\begin{array}{rl}\underset{x\to 0^{+}}{lim}f(x)& =\underset{h\to 0}{lim}f(0+h)\\ & =\underset{h\to 0}{lim}4h\\ & =0\end{array}$
$\begin{array}{rl}\underset{x\to 0^{-}}{lim}f(x)& =\underset{h\to 0^{+}}{lim}f(x)\end{array}$
Hence $f^{\prime }(x)$ is continuous
Now,
$\begin{array}{rl}\underset{x\to 0^{-}}{lim}\frac{f(x)-f(0)}{x-0}& =\underset{h\to 0}{lim}\frac{f(0-h)-f(0)}{0-h-0}\\ & =\underset{h\to 0}{lim}\frac{0-0}{-h}\\ & =0\end{array}$
RHD=
$\begin{array}{rl}\underset{x\to 0^{+}}{lim}\frac{f(x)-f(0)}{x-0}& =\underset{h\to 0}{lim}\frac{f(0+h)-f(0)}{0+h-0}\\ & =\underset{h\to 0}{lim}\frac{4h-0}{h}\\ & =\underset{h\to 0}{lim}\frac{4h}{h}\\ & =4\end{array}$
Hence $f^{\prime }(x)$ is not differentiable
$f^{\prime }(x)$ 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:
$f(x)=\{\begin{array}{ll}{e}^{-x}& x>0\\ e^x& x<0\end{array}$
At x=0
$\begin{array}{rl}& \underset{x\to 0^{-}}{lim}f(x)=\underset{h\to 0^{+}}{lim}f(o-h)=\underset{h\to 0^{+}}{lim}{e}^{-x}\\ & \underset{x\to 0^{+}}{lim}f(x)=\underset{h\to 0}{lim}f(0+h)=\underset{x\to 0^{+}}{lim}{e}^{-h}=1\end{array}$
Hence, f(x) is continuous everywhere
LHD at x=0
$\begin{array}{rl}\underset{x\to 0^{-}}{lim}\frac{f(x)-f(0)}{x-0}& =\underset{h\to 0}{lim}\frac{f(0-h)-f(0)}{0-h-0}\\ & =\underset{h\to 0}{lim}\frac{e^{-h}-1}{-h}\\ & =\underset{h\to 0}{lim}\frac{e^{-h}}{-h}\left[\frac{0}{0}\text{ form }\right]\\ & =-e^{-0}\\ & =-1\end{array}$
RHD at x=0
$\begin{array}{rl}\underset{x\to 0^{+}}{lim}\frac{f(x)-f(0)}{x-0}& =\underset{h\to 0}{lim}\frac{f(0+h)-f(0)}{0+h-0}\\ & =\underset{h\to 0}{lim}\frac{e^{-h}-1}{h}\\ & =\underset{h\to 0}{lim}\frac{-e^{-h}}{1}\left[\frac{0}{0}\text{ form }\right]\\ & =-e^{-0}\\ & =-1\end{array}$
LHD $\ne$ RHD
Hence, f(x) is differentiable at x=0

Tangents and Normals Exercise Multiple Choice Questions Question 8 .

Answer:

Answer:
$\left(d\right)(6,36)$
Hint:
Use differentiation and slope of tangent is zero
Given:
The curve $y=12x-x^2$
Solution:
Let the point be $p(x,y)$
$\begin{array}{rl}& y=12x-x^2\\ & \frac{dy}{dx}=12-2x\\ & {\left(\frac{dy}{dx}\right)}_{(x_1,y_1)}=12-2x_1\end{array}$
Since slope of tangent is zero

$\begin{array}{rl}& \text{ So, }{\left(\frac{dy}{dx}\right)}_{x_1,y_1}=0\\ & 12-2x_1=0\\ & 2x_1=12\\ & x_1=6\end{array}$
Also curve passing through tangent
$\begin{array}{rl}& y_1=12x_1-x_1^2\\ & y_1=12\times 6-36\\ & y_1=72-36=36\end{array}$
The Points are $(6,36)$

Differentiability exercise Multiple choice question, question 7

Answer:

: (b)
HINTS: Understand the definition of continuity and differentiability
SOLUTION:
$f(x)=|\cos x|$
let
$\begin{array}{rl}f(x)& =|x|\\ g(x)& =\cos x\\ h& =\mathrm{fog}(x)\\ & =f(g(x))\\ & =f(\cos x)\\ & =|\cos x|\end{array}$
As cosx and $\begin{array}{rl}& |x|\end{array}$ are continuous function.Hence $\begin{array}{rl}& |\cos x|\end{array}$ is also continuous function For differentiability
$f(x)=\{\begin{array}{ll}-\cos x& x<(2n+1)\frac{\pi }{2}\\ 0& x=(2n+1)\frac{\pi }{2}\\ \cos x& x>(2n+1)\frac{\pi }{2}\end{array}$
At $(2n+1)\frac{\pi }{2}$
$\begin{array}{rl}& \underset{x\to (2n+1)\frac{\pi }{2}}{lim}\frac{f(x)-f(2n+1)\frac{\pi }{2}}{x-(2n+1)\frac{\pi }{2}}\end{array}$
$\begin{array}{rl}=& \underset{h\to 0}{lim}\frac{f(2n+1)\frac{\pi }{2}-h-0}{(2n+1)\frac{\pi }{2}-h-(2n+1)\frac{\pi }{2}}\end{array}$
$\begin{array}{rl}=& \underset{h\to 0}{lim}\frac{-\cos (2n+1)\frac{\pi }{2}-h}{-1}\end{array}$
$\begin{array}{rl}=& \sin (2n+1)\frac{\pi }{2}\\ & \underset{x\to (2n+1){\frac{{\pi }^{-1}}{2}}^{-1}}{lim}\frac{f(x)-f(2n+1)\frac{\pi }{2}}{x-(2n+1)\frac{\pi }{2}}\end{array}$
$\begin{array}{rl}=& \underset{h\to 0}{lim}\frac{f((2n+1)\frac{\pi }{2}-h)-0}{(2n+1)\frac{\pi }{2}+h-(2n+1)\frac{\pi }{2}}\end{array}$
$\begin{array}{rl}=& \underset{h\to 0}{lim}\frac{\cos (2n+1)\frac{\pi }{2}+h}{-1}\\ =& -\sin (2n+1)\frac{\pi }{2}\end{array}$
As LHD $\ne$ RHD
f(x) is not differentiate at $x=(2n+1)\frac{\pi }{2}$

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 $(x,o)$
$\frac{dy}{dx}=2x-3$
The tangent passes through point $(x,o)$
$\begin{array}{rl}& 0=x^2-3x+2\\ & (x-2)(x-1)=0\\ & \Rightarrow x=2,x=1\end{array}$
Case 1
When $x=2$
Slope of tangent $\frac{dy}{dx}|\left({}_{2,0}\right)=2\left(2\right)-3=4-3=1$
$\therefore (x_1,y_1)=(2,0)$
Equation of tangent

$\begin{array}{rl}& y-y_1=m(x-x_1)\\ & y-0=1(x-2)\\ & x-y-2=0\end{array}$
Case 2
When $x=1$
Slope of tangent
$\frac{dy}{dx}|(2,0)=2-3=-1$
$(x_1-y_1)=(1,0)$
Equation of tangent

$\begin{array}{rl}& y-y_1=m(x-x_1)\\ & y-0=-1(x-1)\\ & \Rightarrow x+y-1=0\end{array}$

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 $(2,-1)$
From (1) $t^2+3t-10=0$
$\Rightarrow t=2ort=-5$
From (2) $2t^2-2t-4=0$
$\Rightarrow t^2-t-2=0$
$\Rightarrow t=2ort=-1$
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.

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

Tangents and Normals Exercise Multiple Choice Questions Question 13 .

Answer:
$\left(d\right)(1,2)$And$(1,-2)$
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
$\begin{array}{rl}& 2x+2y\frac{dy}{dx}-2=0\\ & 2y\frac{dy}{dx}=2-2x\\ & \frac{dy}{dx}=\frac{2(1-x)}{2y}\\ & \frac{dy}{dx}=\frac{1-x}{y}\end{array}$ (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
$\begin{array}{rl}& x^2+y^2-2x-3=0\\ & (1{)}^2+(y{)}^2-2(1)-3=0\\ & 1+y^2-2-3=0\\ & y=\pm 2\end{array}$
Hence, the points are $(1,2)$ and $(1,-2)$

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:
$\begin{array}{rl}& xy=a^2\text{ And }{x}^2-y^2=2a^2\\ & x\frac{dy}{dx}+y=0\text{ And }2x-2y\frac{dy}{dx}=0\\ & \frac{dy}{dx}=-\frac{y}{x}\text{ And }\frac{dy}{dx}=\frac{x}{y}\end{array}$
$\begin{array}{rl}& {\left(\frac{dy}{dx}\right)}_1=-\frac{y}{x}\text{ And }{\left(\frac{dy}{dx}\right)}_2=\frac{x}{y}\\ & {\left(\frac{dy}{dx}\right)}_1\times {\left(\frac{dy}{dx}\right)}_2=-\frac{y}{x}\times \frac{x}{y}=-1\end{array}$
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
$\begin{array}{rl}& 1-x^2>0\\ & 1>x^2\\ & -1<x<1\end{array}$
Now between $x\in [-1,1]$ at x=0
$f(x)=0$
So we Check the continuity at x=0
LHD =
$\begin{array}{rl}\underset{x\to 0^{-}}{lim}f(x)& =\underset{x\to 0}{lim}\sqrt{1-\sqrt{1-x^2}}\\ & =0\end{array}$
RHD =
$\begin{array}{rl}\underset{x\to 0^{+}}{lim}f(x)& =\underset{x\to 0}{lim}\sqrt{1-\sqrt{1-x^2}}\\ & =0\end{array}$
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
$\begin{array}{rl}& \underset{x\to 0^{-}}{lim}\frac{f(x)-f(0)}{x-0}=\underset{x\to 0^{+}}{lim}\frac{f(x)-f(0)}{x-0}\end{array}$
$\begin{array}{rl}& \underset{h\to 0}{lim}\frac{f(0-h)-f(0)}{-h}=\underset{h\to 0}{lim}\frac{f(0+h)-f(0)}{h}\end{array}$
$\begin{array}{rl}& \underset{h\to 0}{lim}\frac{a(\sin (-4)+b{e}^{-(-h)}+c(-(-h{)}^3)-b}{-h}\end{array}$
$\begin{array}{rl}& =\underset{h\to 0}{lim}\frac{a\sinh +b{e}^h+c{h}^3-b}{-h}=f^{\prime }(0)\end{array}$
as
$\begin{array}{rl}& |\sin x|=\{\begin{array}{ll}-\sin x& x<0\\ \sin x& x>0\end{array}\\ & |x|=\{\begin{array}{ll}x& x>0\\ -x& x<0\end{array}\\ & |x{|}^3=\{\begin{array}{ll}{x}^3& x>0\\ -x^3& x<0\end{array}\end{array}$
$\begin{array}{rl}& \underset{h\to 0}{lim}\frac{a\sinh +b(e^h-1)+c{h}^3-b}{-h}=\underset{h\to 0}{lim}\frac{a\sinh +b(e^h-1)+c{h}^3}{-h}\end{array}$
$\begin{array}{rl}& =a\underset{h\to 0}{lim}\frac{\sinh }{-h}-b\underset{h\to 0}{lim}\frac{(e^h-1)}{h}+c\underset{h\to 0}{lim}\frac{+h^3}{-h}\end{array}$
$\begin{array}{rl}& =a\underset{h\to 0}{lim}\frac{\sinh }{-h}-b\underset{h\to 0}{lim}\frac{(e^h-1)}{h}+c\underset{h\to 0}{lim}\frac{+h^3}{-h}-\frac{h^3}{h}\end{array}$$\begin{array}{rl}& -a\times 1-b\times 1+0=a\times 1+b\times 1[\text{ as }\underset{h\to 0}{lim}\frac{\sinh }{-h}=1\text{ and }\underset{h\to 0}{lim}\frac{(e^h-1)}{h}=1]\end{array}$$\begin{array}{rl}& -a=a\mathrm{\&}-b=b\\ & a=0\mathrm{\&}b=0\end{array}$

Differentiability exercise Multiple choice question, question 10

Answer:

(b)
HINTS: Understand the definition of continuity and differentiability
$f(x)=x^2+\frac{x^2}{1+x^2}+\frac{x^2}{{(1+x^2)}^2}+\dots$
GIVEN: then at x=0
SOLUTION:
$f(x)=\sum _{n=0}^{\mathrm{\infty }}\frac{x^2}{{(1+x^2)}^n}$
Now f(x) is G.P with $\mathrm{r}=\frac{1}{1+x^2}$
So,
$\begin{array}{rl}& f(x)=\frac{x^2{\left(\frac{1}{1+x^2}\right)}^n-1}{\frac{1}{1+x^2}-1}[a=x^2\text{ and }s=\frac{a^{n-1}}{n-1}]\end{array}$
$\begin{array}{rl}& f(x)=\frac{\frac{x^2-{(1+x^2)}^n}{{(1+x^2)}^n}}{\frac{1-1-x^2}{(1+x^2)}}\end{array}$
$\begin{array}{rl}& f(x)=-1{(1+x^2)}^{-n+1}+x^{-2}(1+x^2)\\ & x=0\end{array}$
$\begin{array}{rl}& \underset{x\to 0}{lim}f(x)=\underset{h\to 0}{lim}-{(1+h^2)}^{-n-1}+h^{-2}(1+h^2)=\mathrm{\infty }\end{array}$
As LHL $\ne$ 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:
$f(x)=|\log x|=\{\begin{array}{ll}{\log }_{e}x& 0<x<1\\ {\log }_{e}x& x\ge 1\end{array}$
LHD at x=1
$\begin{array}{rl}\underset{x\to 1^{-}}{lim}\frac{f(x)-f(1)}{x-1}& =\underset{x\to 0}{lim}\frac{f(1-h)-f(1)}{-h}\end{array}$
$\begin{array}{rl}& =\underset{h\to 0}{lim}-\frac{\log (1-h)-f(1)}{-h}\end{array}$
$\begin{array}{rl}\underset{h\to 0}{lim}-\frac{\log (1-h)}{-h}& =\underset{h\to 0}{lim}\frac{1}{1-h}\times -1\\ & =-1\end{array}$
RHD at x=1
$\begin{array}{rl}\underset{x\to 1^{+}}{lim}\frac{f(x)-f(1)}{x-1}& =\underset{x\to 1}{lim}\frac{\log x-\log 1}{x-1}\end{array}$
$\begin{array}{rl}& =\underset{h\to 0}{lim}-\frac{\log (1+h)}{h}\\ & =\underset{h\to 0}{lim}\frac{1}{1+h}\\ & =1\end{array}$
As LHD $\ne$ RHD at x=1
$f^{\prime }\left(1^{-}\right)=-1,f^{\prime }\left(1^{+}\right)=1$