Question

Let $\mathrm{f}\left(\mathrm{x}\right)=2+\sqrt{1-{\mathrm{x}}^2},\left|\mathrm{x}\right|\le 1=2\mathrm{e}(1-\mathrm{x}{)}^2,|\mathrm{x}|>1.$ The points where $\mathrm{f}\left(\mathrm{x}\right)$ is not differentiable are:

• A. $0$only
• B. $–1,1$only
• C. $1$ only
• D. $–1$ only

Answer 1

The correct option is B

$–1,1$only

Explanation for the correct option:

Finding the points where $\mathrm{f}\left(\mathrm{x}\right)$ is not differentiable:

Given that $\mathrm{f}\left(\mathrm{x}\right)=2+\sqrt{1-\mathrm{x}2},\left|\mathrm{x}\right|\le 1$

$$\begin{gathered} f\left(x\right)=2·e^{(1-x{)}^2},\left[x\right)>1. \\ f(1+{)}^{\text{'}}=\underset{h\to 0}{\lim }\left(f\frac{(1+h))-f(1)}{h}\right) \\ =\underset{\ln \to 0}{\lim }\frac{\left[2·e^{h^2}-2\right]}{h.} \\ =\underset{h\to 0}{\lim }\frac{2\left(e^{h^2}-1\right)}{h^2}\times h. \\ =0 \\ \left[\because \frac{e^a-1}{e^x}\Rightarrow 0\right] \end{gathered}$$

$$\begin{gathered} f(-1)=\underset{h\to 0}{\lim }\frac{\left(f\right(1+h)-f(1\left)\right)}{h_0} \\ \left.=\underset{h\to 0}{\lim }\frac{\left[2+\sqrt{1-(1-h{)}^2}-f\left(1\right)\right.}{h_1}\right] \\ =\underset{h\to 0}{\lim }\frac{\left[2-\sqrt{1-(1-h{)}^2-2}\right.}{h} \\ =\underset{h\to 0}{\lim }\frac{\sqrt{2h-h^2}}{h} \end{gathered}$$

$\mathrm{f}(-1)=$Not defined

As $\mathrm{f}(+1)\ne \mathrm{f}(-1),$

$\mathrm{f}\left(1\right)=$does not exist

Similarly, ${\mathrm{f}}^1\left(1\right)$ also does not exist

Hence $\mathrm{f}\left(\mathrm{x}\right)$is not differentiable at

$\mathrm{x}=-1$and $1.$

Hence, option (B) is the correct answer.