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

Differentiability Excercise: 9.1

Differentiability Exercise 9.1 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 = 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,
$\mathrm{f}(\mathrm{x})=\{\begin{array}{c}-(x-3),x<3\\ x-3,x\ge 3\end{array}$
But $f\left(3\right)=3-3=0$
$LHL=\underset{x\to 3^{-}}{lim}f\left(x\right)$
$=\underset{h\to 0}{lim}f(3-h)$
$=\underset{h\to 0}{lim}3-(3-h)$
$=3-(3-0)$
$=0$
Now Consider,
$RHL=\underset{x\to 3^{+}}{lim}f\left(x\right)$
$=\underset{h\to 0}{lim}f(3+h)$
$=\underset{h\to 0}{lim}3+h-3$
$=3+0-3$
$=0$
$LHL=RHL=f\left(3\right)$
Since f(x) is continuous at x = 3, we have to find its differentiability using the formula,
$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$
Left Hand Derivate (LHD) =$=\underset{h\to 0}{lim}\frac{f(c+h)-f\left(c\right)}{h}$, denoted by L f `(c)
Right Hand Derivate (RHD =$=\underset{h\to 0}{lim}\frac{f(c+h)-f\left(c\right)}{h}$, denoted by R f `(c)
$\begin{array}{rl}& \text{ LHD at }\mathrm{x}=3:\underset{x\to 3^3}{lim}\frac{f(x)-f(3)}{x-3}\\ & =\underset{h\to 0}{lim}\frac{f(3-h)-f(3)}{3-h-3}\\ & =\underset{h\to 0}{lim}\frac{3-(3-h)-0}{-h}\\ & =\underset{h\to 0}{lim}\frac{h}{-h}\\ & =-1\end{array}$
$\begin{array}{rl}& \mathrm{RHD}\text{ at }\mathrm{x}=3:\underset{x\to 3^{+}}{lim}\frac{f(x)-f(3)}{x-3}\\ & =\underset{h\to 0}{lim}\frac{f(3+h)-f(3)}{3+h-3}\\ & =\underset{h\to 0}{lim}\frac{3+h-3-0}{h}\\ & =\underset{h\to 0}{lim}\frac{h}{h}\\ & =1\end{array}$
LHD at x = 3 $\ne$ RHD at x = 3
Hence, f(x) is continuous but not differentiable at x = 3

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:
$f\left(x\right)=$ $\{\begin{array}{cc}x+x+1,& 1<x<0\\ 1,& 0\le \mathrm{x}\le 1\\ -x-x+1,& 1<x<2\end{array}$
$\Rightarrow \mathrm{f}(\mathrm{x})=\{\begin{array}{c}2x+1,-1<x<0\\ 1,0\le \mathrm{x}\le 1\\ -2x+1,1<x<2\end{array}$
We know that a polynomial and a constant function is continuous and differentiable everywhere. So f(x) is continuous and differentiable for x $ϵ$ (-1, 0) and x $ϵ$ (0, 1) and (1, 2).
Continuity at $x=0:$
$\begin{array}{rl}& \underset{x\to 0^{-}}{lim}f(x)=\underset{x\to 0^{-}}{lim}(2x+1)=1\\ & \underset{x\to 0^{+}}{lim}f(x)=\underset{x\to 0^{-}}{lim}1=1\\ & \mathrm{f}(0)=2(0)+1=1\end{array}$
Since, $\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}f\left(x\right)=f\left(0\right),f\left(x\right)$is continuous at x = 0.
Continuity at x = 1:
$\begin{array}{rl}& \underset{x\to 1^{-}}{lim}f(x)=\underset{x\to 1^{-}}{lim}1=1\\ & \underset{x\to 1^{+}}{lim}f(x)=\underset{x\to 1^{-}}{lim}1=1\\ & \mathrm{f}(1)=1\end{array}$
Since, $\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}f\left(x\right)=f\left(0\right),f\left(x\right)$ is continuous at x = 1.
Differentiability at x = 0:
$\begin{array}{rl}& \text{ LHD at }\mathrm{x}=0:\underset{x\to 0^{-}}{lim}\frac{f(x)-f(0)}{x-0}\\ & =\underset{x\to 0^{-}}{lim}\frac{2x+1-1}{x-0}\\ & =\underset{x\to 0^{-}}{lim}\frac{2x}{x}\\ & =2\end{array}$
$\begin{array}{rl}& \text{ RHD at }\mathrm{x}=0:\underset{x\to 0^{+}}{lim}\frac{f(x)-f(0)}{x-0}\\ & =\underset{x\to 0^{+}}{lim}\frac{1-1}{x}\\ & =\underset{x\to 0^{+}}{lim}\frac{0}{h}\\ & =0\end{array}$
LHD at x = 0 $\ne$ RHD at x = 0
Hence, f(x) is continuous but not differentiable at x = 0.

Differentiability at x = 1:
$LHDatx=1:\underset{x\to 1^{-}}{lim}\frac{f\left(x\right)-f\left(1\right)}{x-1}$
$\underset{x\to 1^{-}}{lim}\frac{1-1}{x-1}$
$=0$
$RHDatx=1:\underset{x\to 1^{+}}{lim}\frac{f\left(x\right)-f\left(1\right)}{x-1}$
$=\underset{x\to 1^{+}}{lim}\frac{2x+1-1}{x-1}$
$=\underset{x\to 1^{+}}{lim}\frac{2x}{x-1}$
$=\frac{1}{0}$
$=\mathrm{\infty }$

LHD at x = 1 $\ne$ 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,

$f^{\prime }\left(x\right)=\underset{x\to 0^{-}}{lim}\frac{f\left(x\right)-f\left(0\right)}{x-0}$

$LHDatx=0:\underset{x\to 0^{-}}{lim}\frac{f\left(x\right)-f\left(0\right)}{x-0}$

$\underset{x\to 0}{lim}\frac{f(0-h)-f\left(0\right)}{(0-h)-0}$

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

$=\underset{h\to 0}{lim}-{(-h)}^{m-1}\sin \frac{1}{h}$ $\sin \left(\frac{1}{0}\right)=\sin (\mathrm{\infty }),$

$\Rightarrow NotDefined$ $[0<m<1]$

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

$\Rightarrow$Not defined

(LHD at x = 0) $\ne$ (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)=$$\{\begin{array}{c}{x}^m\sin \left(\frac{1}{x}\right),x\ne 0\\ 0,x=0\end{array}$
Solution:
Differentiability at x = 0:
$LHDatx=0:\underset{x\to 0^{-}}{lim}\frac{f\left(x\right)-f\left(0\right)}{x-0}$
$=\underset{h\to 0}{lim}\frac{f(0-h)-f\left(0\right)}{(0-h)-0}$
$=\underset{h\to 0}{lim}\frac{{(0-h)}^m\sin \left(\frac{1}{-h}\right)}{-h}$
$=\underset{h\to 0}{lim}{(-h)}^{m-1}\sin \left(\frac{1}{-h}\right)$ $\{\frac{a^m}{a^n}\Rightarrow a^{m-n}\}$
$=\underset{h\to 0}{lim}{(-h)}^{m-1}\sin \left(\frac{1}{h}\right)$ $\{\sin (-\theta )\Rightarrow -\sin \theta \}$
$\begin{array}{rl}& \text{ RHD at }x=0:\underset{x\to 0^{+}}{lim}\frac{f(x)-f(0)}{x-0}\\ & =\underset{h\to 0}{lim}\frac{f(0+h)-f(0)}{(0+h)-0}\\ & =\underset{h\to 0}{lim}\frac{h^m\sin \left(\frac{1}{h}\right)}{h}\\ & \Rightarrow \{\text{ when }-1\le k\le 1\}\end{array}$ $\{\begin{array}{c}(0{)}^{m-1}=0\\ \sin \left(\frac{1}{0}\right)=\sin (\mathrm{\infty })\end{array}$
$\Rightarrow 0xk$
$\Rightarrow 0$
$(LHDatx=0)=(RHDatx=0)$

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)=$$\{\begin{array}{c}{x}^m\sin \left(\frac{1}{x}\right),x\ne 0\\ 0,x=0\end{array}$
Solution:
$\begin{array}{rl}& \mathrm{LHL}=\underset{x\to 0^{-}}{lim}\mathrm{f}(\mathrm{x})\\ & =\underset{h\to 0}{lim}\mathrm{f}(0-\mathrm{h})\\ & =\underset{h\to 0}{lim}(-h{)}^m\sin \left(\frac{1}{-h}\right)\end{array}$
As we know $(0{)}^1=0\text{ and }\sin \left(\frac{1}{0}\right)=\sin (\mathrm{\infty })\text{, }$
$\Rightarrow notdefinedasm\le 0$
$\begin{array}{rl}& \mathrm{RHL}=\underset{x\to 0^{+}}{lim}\mathrm{f}(\mathrm{x})\\ & =\underset{h\to 0}{lim}\mathrm{f}(0+\mathrm{h})\\ & =\underset{h\to 0}{lim}(h{)}^m\sin \left(\frac{1}{h}\right)\end{array}$
$\Rightarrow notdefinedasm\le 0$
Since RHL and LHL are not defined, f (x) is not continuous.
$\begin{array}{rl}& \mathrm{LHD}\text{ at }\mathrm{x}=0:\underset{x\to 0^{-}}{lim}\frac{f(x)-f(0)}{x-0}\\ & =\underset{h\to 0}{lim}\frac{f(0-h)-f(0)}{(0-h)-0}\\ & =\underset{h\to 0}{lim}\frac{(-h{)}^m\sin \left(\frac{-1}{h}\right)}{-h}\\ & =\underset{h\to 0}{lim}-(-h{)}^{m-1}\sin \left(\frac{1}{h}\right)\\ & \text{ Since }\{\frac{a^m}{a^n}\Rightarrow a^{m-n}\}\text{ and }\{\sin (-\theta )\Rightarrow -\sin \theta \}\text{, }\end{array}$
$\Rightarrow notdefinedasm\le 0$
$\begin{array}{rl}& \mathrm{RHD}\text{ at }\mathrm{x}=0:\underset{x\to 0^{-}}{lim}\frac{f(x)-f(0)}{x-0}\\ & =\underset{h\to 0}{lim}\frac{f(0+h)-f(0)}{(0+h)-0}\\ & =\underset{h\to 0}{lim}\frac{(h{)}^m\sin \left(\frac{1}{h}\right)}{h}\\ & =\underset{h\to 0}{lim}(h{)}^{m-1}\sin \left(\frac{1}{h}\right)\\ & \text{ Since }\{\frac{a^m}{a^n}\Rightarrow a^{m-n}\}\text{ and }\{\sin (-\theta )\Rightarrow -\sin \theta \}\text{, }\end{array}$
$\Rightarrow notdefinedasm\le 0$
LHD and RHD does not exist .
Hence, f(x) is neither continuous nor differentiable at x = 0.

Differentiability Exercise 9.1 Question 8

Differentiability Exercise 9.1 Question 9

Now consider,

$\begin{array}{rl}& \mathrm{RHL}=\underset{x\to 1^{+}}{lim}\mathrm{f}(\mathrm{x})\\ & =\underset{h\to 0}{lim}\mathrm{f}(1+\mathrm{h})\\ & =\underset{h\to 0}{lim}-(2(1+\mathrm{h})-3)\\ & =\underset{h\to 0}{lim}-(2+2\mathrm{h}-3)\\ & =\underset{h\to 0}{lim}-(-1+2\mathrm{h})\\ & =1\end{array}$

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

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

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

$=-1$

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

Differentiability Exercise 9.1 Question 10