Exercise 9.6-Additional Questions - Chapter 9 - Limits and Continuity - 11th Maths Guide Samacheer Kalvi Solutions

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Additional Questions
Question 1.
Is the function $f(x)=|x|$ differentiable at the origin. Justify your answer.
Solution:
$
\begin{aligned}
f(x) & =|x| ; f(0)=0 \\
\lim _{h \rightarrow 0^{+}} f(x) & =\lim _{h \rightarrow 0}|0+h|=0 \\
\lim _{h \rightarrow 0^{-}} f(x) & =\lim _{h \rightarrow 0}|0-h|=0 \\
\lim _{h \rightarrow 0^{+}} f(x) & =\lim _{h \rightarrow 0^{-}} f(x) \\
f(x) & =|x| \text { is continuous at } x=0 \\
\text { R } f^{\prime}(0) & =\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{h}{h}=1 \\
\text { L } f^{\prime}(0) & =\lim _{h \rightarrow 0^{-}} \frac{f(0-h)-f(0)}{-h}=\frac{1-h}{-h}=-1
\end{aligned}
$
R $f^{\prime}(0) \neq \mathrm{L} f^{\prime}(0) \therefore f(x)$ is not differentiable at $x=0$
Question 2.
Discuss the differentiability of the functions:
(i) $f(x)\left\{\begin{array}{c}1,0 \leq x \leq 1 \\ x, x>1 \quad \text { atx }=1\end{array}\right.$
(ii) $f(x)\left\{\begin{array}{lc}2 x-3, & 0 \leq x \leq 2 \text { at } x=2, x=4 \\ x^2-3, & 2 Solution:

$(i)$
$
\begin{aligned}
f(x) & =\left\{\begin{array}{c}
1,0 \leq x \leq 1 \\
x, x>1
\end{array} \text { at } x=1\right. \\
f^{\prime}(1) & =\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}=\lim _{h \rightarrow \infty} \frac{1+h-1}{h}=1 \\
f^{\prime}(1) & =1 \\
f^{\prime}(1) & =\mathrm{R} f^{\prime}(1)
\end{aligned}
$
$\therefore f(x)$ is differentiable at $x=1$
(ii)
$
\begin{aligned}
f(x) & =\left\{\begin{array}{ll}
2 x-3, & 0 \leq x \leq 2 \\
x^2-3, & 2 \end{array} \quad \text { at } x=2\right. \\
\mathrm{L} f^{\prime}(2) & =\lim _{h \rightarrow 0} \frac{f(2-h)-f(2)}{-h} \\
\lim _{h \rightarrow 0} \frac{[2(2-h)-3]-(4-3)}{h} & =\lim _{h \rightarrow 0} \frac{4-2 h-3-1}{-h}=\lim _{h \rightarrow 0} \frac{-2 h}{-h}=2 \\
f^{\prime}(2) & =\lim _{h \rightarrow 0} \frac{4+4 h+h^2-3-1}{h}=\lim _{h \rightarrow 0} \frac{\left[(2+h)^2-3\right]-[4-3]}{h} \\
& =\lim _{h \rightarrow 0} \frac{4+4 h+h^2-3-1}{h}=\lim _{h \rightarrow 0} \frac{h(4+h)}{h}=4 \\
& f^{\prime}(2) \neq \mathrm{R} f^{\prime}(2)
\end{aligned}
$
$\therefore \mathrm{f}(2)$ is not differentiable at $\mathrm{x}=2$. Similarly, it can be proved for $\mathrm{x}=4$.