Exercise 10.1 - Chapter 10 - Differentiability and Methods of Differentiation - 11th Maths Guide Samacheer Kalvi Solutions

You can Download the Exercise 10.1 - Chapter 10 - Differentiability and Methods of Differentiation - 11th Maths Guide Samacheer Kalvi Solutions with expert answers for all chapters. Perfect for Tamil & English Medium students to revise the syllabus and score more marks in board exams. Download and share it with your friends

Share this to Friend on WhatsApp

Differentiability and Methods of Differentiation
Ex 10.1
Question 1.
Find the derivatives of the following functions using first principle.
(i) $f(x)=6$
Solution:
Given $f(x)=6$
$f(x+h)=6$
$
\begin{aligned}
f^{\prime}(x) & =\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\
& =\lim _{h \rightarrow 0} \frac{6-6}{h} \\
& =\lim _{h \rightarrow 0} \frac{0}{h}
\end{aligned}
$
[ $\mathrm{h} \rightarrow 0$ means $\mathrm{h}$ is very nears to zero from left to right but not zero]
(ii) $f(x)=-4 x+7$
Solution:
Given $f(x)=-4 x+7$
$
\begin{aligned}
& \mathrm{f}(\mathrm{x}+\mathrm{h})=-4(\mathrm{x}+\mathrm{h})+7 \\
& =-4 \mathrm{x}-4 \mathrm{~h}+7 \\
& f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\
& =\lim _{h \rightarrow 0} \frac{(-4 x-4 h+7)-(-4 x+7)}{h} \\
& =\lim _{h \rightarrow 0} \frac{-4 x-4 h+\not \lambda+A x-\not 7}{h} \\
& =\lim _{h \rightarrow 0} \frac{-4 h}{h} \\
& =\lim _{h \rightarrow 0}(-4) \\
&
\end{aligned}
$
(iii) $f(x)=-x^2+2$
Given $f(x)=-x^2+2$
$
f(x+h)=-(x+h)^2+2
$

$
\begin{aligned}
=-\mathrm{x}^2-\mathrm{h}^2 & -2 \mathrm{xh}+2 \\
f^{\prime}(x) & =\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\
& =\lim _{h \rightarrow 0} \frac{\left(-x^2-h^2-2 x h+2\right)-\left(-x^2+2\right)}{h} \\
& =\lim _{h \rightarrow 0} \frac{-x^2-h^2-2 x h+22+x^2-\not 2}{h} \\
& =\lim _{h \rightarrow 0} \frac{-h(h+2 x)}{h} \\
& =-(0+2 x) \\
& =-2 x
\end{aligned}
$
Question 2.
Find the derivatives from the left and from the right at $\mathrm{x}=1$ (if they exist) of the following functions. Are the functions differentiable at $\mathrm{x}=1$ ?
(i) $f(x)=|x-1|$
Solution:

$\begin{aligned}
& f^{\prime}(a)=\lim _{x \rightarrow 0} \frac{f(x)-f(a)}{x-a} \\
& f^{\prime}\left(1^{-}\right)=\lim _{x \rightarrow 1^{-}} \frac{|x-1|-|(1)-1|}{x-1} \\
&=\lim _{x \rightarrow 1^{-}} \frac{|x-1|-0}{x-1} \\
&=\lim _{x \rightarrow 1^{-}} \frac{-(x-1)}{x-1} \\
&=-1 \\
& f^{\prime}\left(1^{+}\right)=\lim _{x \rightarrow 1^{+}} \frac{|x-1|-|(1)-1|}{x-1} \\
&=\lim _{x \rightarrow 1^{+}} \frac{|x-1|-0}{x-1} \\
&=\lim _{x \rightarrow 1^{+}} \frac{+(x-1)}{(x-1)} \\
&=+1 \\
& f^{\prime}\left(1^{-}\right) \neq f^{\prime}\left(1^{+}\right) \\
& f^{\prime}(1) \text { does not exist } \\
& \therefore \mathrm{S}^{\prime} \text { is not differentiable at } \mathrm{x}=1 .
\end{aligned}$

(ii) $f(x)=\sqrt{1-x^2}$
Solution:
$
\begin{aligned}
f^{\prime}(a) & =\lim _{x \rightarrow 0} \frac{f(x)-f(a)}{x-a} \\
f^{\prime}\left(1^{-}\right) & =\lim _{x \rightarrow 1^{-}} \frac{\sqrt{1-x^2-0}}{\sqrt{1-x^2-1}} \\
& =f^{\prime}(x) \rightarrow-\infty \text { as } x \rightarrow 1^{-}
\end{aligned}
$
$\therefore$ ' $\mathrm{f}$ ' is not differentiable at $\mathrm{x}=1$.
(iii)
$
f(x)= \begin{cases}x, & x \leq 1 \\ x^2, & x>1\end{cases}
$
Solution:
$
\begin{aligned}
f^{\prime}(a) & =\lim _{x \rightarrow 0} \frac{f(x)-f(a)}{x-a} \\
f^{\prime}\left(1^{-}\right) & =\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1} \\
& =\lim _{x \rightarrow 1^{-}} \frac{x-1}{x-1} \\
& =1 \\
f^{\prime}\left(1^{+}\right) & =\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1} \\
& =\lim _{x \rightarrow 1^{+}} \frac{x^2-1}{x-1} \\
& =\lim _{x \rightarrow 1^{+}} \frac{(x+1)(x-1)}{(x-1)} \\
& =\lim _{x \rightarrow 1^{+}} x+1 \\
& =1+1 \\
& =2 \\
f^{\prime}\left(1^{-}\right) & \neq f^{\prime}\left(1^{+}\right), f^{\prime}(x) \text { is not differentiate }
\end{aligned}
$
' $\mathrm{f}$ ' is not differentiable at $\mathrm{x}=1$

Question 3.
Determine whether the following functions is differentiable at the indicated values.
(i) $f(x)=x|x|$ at $x=0$
Solution:
$
\begin{aligned}
f^{\prime}(x) & =\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\
& =\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h} \\
& =\lim _{h \rightarrow 0} \frac{h|h|-0}{h} \\
& =\lim _{h \rightarrow 0}|h| \\
& =0
\end{aligned}
$

Limits exists
Hence ' $f$ ' is differentiable at $x=0$.
(ii) $f(x)=\left|x^2-1\right|$ at $x=1$
Solution:
$
\begin{aligned}
f(x) & = \begin{cases}1-x^2 & \text { if } x \leq 0 \\
x^2-1 & \text { if } x>1\end{cases} \\
f^{\prime}\left(1^{+}\right) & =\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1} \\
& =\lim _{x \rightarrow 1^{+}} \frac{x^2-1-0}{x-1} \\
& =\lim _{x \rightarrow 1^{+}} \frac{(x+1)(x-1)}{(x-1)} \\
f^{\prime}\left(1^{-}\right) & =\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1} \\
& =\lim _{x \rightarrow 1^{-}}-\frac{(x-1)(x+1)}{(x-1)} \\
& =-2 \\
f^{\prime}\left(1^{+}\right) & \neq f^{\prime}\left(1^{-}\right)
\end{aligned}
$
$f(x)$ is not differentiable at $x=1$.
(iii) $f(x)=|x|+|x-1|$ at $x=0,1$
Solution:

$
f(x)=\left\{\begin{array}{ccc}
1-2 x & \text { if } & x<0 \\
1 & \text { if } & 0 \leq x \leq 1 \\
2 x-1 & \text { if } & x>1
\end{array}\right.
$
Clearly $f(x)$ is differentiable at $x$, where $x \neq 0, x \neq 1$.
$
\begin{aligned}
f^{\prime}\left(0^{+}\right) & =\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}=\lim _{x \rightarrow 0^{+}} \frac{1-1}{x}=0 \\
f^{\prime}\left(0^{-}\right) & =\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}=\lim _{x \rightarrow 0^{-}} \frac{1-2 x-1}{x}=-2 \\
f^{\prime}\left(0^{+}\right) & \neq f^{\prime}\left(0^{-}\right)
\end{aligned}
$
$\therefore \mathrm{f}(\mathrm{x})$ is not differentiable at $\mathrm{x}=0$.
$
\begin{aligned}
& f\left(1^{+}\right)=\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}=\lim _{x \rightarrow 1^{+}} \frac{2 x-1-1}{x-1}=-2 \\
& f^{\prime}\left(1^{-}\right)=\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}=\lim _{x \rightarrow 1^{-}} \frac{1-1}{x-1}=0 \\
& f^{\prime}\left(1^{+}\right) \neq f^{\prime}\left(1^{-}\right) \\
& \therefore \mathrm{f}(\mathrm{x}) \text { is not differentiable at } \mathrm{x}=1 \text {. } \\
& \text { (iv) } f(x)=\sin |x| \text { at } x=0 \\
& \text { Solution: } \\
& f(x)=\left\{\begin{array}{clc}
-\sin x & \text { if } & x<0 \\
0 & \text { if } & x=0 \\
\sin x & \text { if } & x>0
\end{array}\right. \\
& f^{\prime}\left(0^{+}\right)=\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0} \\
& =\lim _{x \rightarrow 0^{+}} \frac{\sin x-0}{x .} \\
& =1 \text {. } \\
&
\end{aligned}
$

$
\begin{aligned}
f^{\prime}\left(0^{-}\right) & =\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0} \\
& =\lim _{x \rightarrow 0^{-}} \frac{-\sin x-0}{x} \\
& =-1 \\
f^{\prime}\left(0^{+}\right) & \neq f^{\prime}\left(0^{-}\right)
\end{aligned}
$
$\therefore \mathrm{f}(\mathrm{x})$ is not differentiable at $\mathrm{x}=0$.

Question 4.
Show that the following functions are not differentiable at the indicated value of $x$.
(i)
$
f(x)=\left\{\begin{array}{ll}
-x+2 & ; x \leq 2 \\
2 x-4 & ; x>2
\end{array} ; x=2\right.
$
Solution:
$
\begin{aligned}
f(x) & = \begin{cases}-x+2 \quad ; x \leq 2 \\
2 x-4 \quad ; x>2\end{cases} \\
& =\lim _{x \rightarrow 2^{+}} \frac{2 x-4-0}{x-2} \\
& =\lim _{x \rightarrow 2^{+}} \frac{2(x-2)}{x-2} \\
& =2 \\
f^{\prime}\left(2^{-}\right) & =\lim _{x \rightarrow 2^{-}} \frac{f(x)-f(2)}{x-2} \\
& =\lim _{x \rightarrow 2^{-}} \frac{-x+2-0}{x-2} \\
& =\lim _{x \rightarrow 2^{-}} \frac{-(x-2)}{(x-2)}=\lim _{x \rightarrow 2^{-}}-1 \\
& =-1 \\
f^{\prime}\left(2^{+}\right) & \neq f^{\prime}\left(2^{-}\right)
\end{aligned}
$
$\mathrm{f}(\mathrm{x})$ is not differentiable at $x=2$.
(ii)
$
f(x)=\left\{\begin{array}{cl}
3 x & ; x<0 \quad x=0 \\
-4 x & ; x \geq 0
\end{array}\right.
$
Solution:

$
\begin{aligned}
f^{\prime}\left(0^{+}\right) & =\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0} \\
& =\lim _{x \rightarrow 0^{+}} \frac{-4 x-0}{x-0} \\
& =\lim _{x \rightarrow 0^{+}} \frac{-4 x}{x} \\
& =-4 \\
f^{\prime}\left(0^{-}\right) & =\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0} \\
& =\lim _{x \rightarrow 0^{-}} \frac{3 x-0}{x-0} \\
& =\lim _{x \rightarrow 0^{-}} \frac{3 x}{x} \\
& =3 \\
f^{\prime}\left(0^{+}\right) & \neq f^{\prime}\left(0^{-}\right)
\end{aligned}
$
$f(x)$ is not differentiable at $x=0$.
Question 5.
The graph off is shown below. State with reasons that $\mathrm{x}$ values (the numbers), at which $\mathrm{f}$ is not differentiable.
Solution:

(i) at $\mathrm{x}=-1$ and has vertical tangent at $x=-1$ and $x=$ 8(also At $x=-1$. The graph has shape edge $v$ ] and at $x=8$;The graph has shape peak ${ }^{\wedge}$ ]
(ii) At $x=4$ : The graph $f$ is not differentiable, since at $\mathrm{x}=4$. The graph $f$ ' is not continuous.
(iii) At $x=11$; The graph $f^{\prime}$ is not differentiable, since at $x=11$. The tangent line of the graph is perpendicular.
Question 6.
If $f(x)=|x+100|+x^2$, test whether $f^{\prime}(-100)$ exists.
Solution:

$\begin{aligned}
& f(x)=|\mathrm{x}+100|+\mathrm{x}^2 \\
& f^{\prime}\left(-100^{-}\right)=\lim _{x \rightarrow-100^{-}} \frac{f(x)-f(-100)}{x+100}=\lim _{x \rightarrow-100^{-}} \frac{-(x+100)+x^2-(0+10000)}{x+100} \\
&=\lim _{x \rightarrow-100^{-}} \frac{-x-100+x^2-10000}{x+100}=\lim _{x \rightarrow-100^{-}} \frac{x^2+x-10100}{x+100}=1 \\
&=\lim _{x \rightarrow-100^{-}} \frac{(x+100)(x-101)}{x+100}=-201 \\
& f^{\prime}\left(-100^{+}\right)=\lim _{x \rightarrow-100^{+}} \frac{f(x)-f(-100)}{x+100} \\
&=\lim _{x \rightarrow-100^{+}} \frac{x+100+x^2-10000}{x+100} \\
&=\lim _{x \rightarrow-100^{+}} \frac{x^2+x-9900}{x+100} \\
& \lim _{x \rightarrow-100^{+}} \frac{(x+100)(x-99)}{x+100}=-199 \\
& \Rightarrow f^{\prime}\left(-100^{-}\right) \neq f^{\prime}\left(100^{+}\right) \\
& f^{\prime}(-100) \text { does not exists }
\end{aligned}$

Question 7.
Examine the differentiability of functions in R by drawing the diagrams.
(i) $|\sin x|$
Solution:
$
|\sin x|=f(x)
$

Limit exist and continuous for all $x \in R$ clearly, differentiable at $R-\{n \pi n \in z$ ) Not differentiable at $x$ $=n \pi, n \in z$.
(ii) $|\cos x|$
Solution:
$
|\cos x|=f(x)
$

Limit exist and continuous for all $x \in R$ clearly, differentiable at $R\{(2 n+1) \pi / 2 / n \in z\}$ Not differentiable at $x=(2 n+1) \frac{\pi}{2}, n \in Z$.