Question

Let $S=\{tϵR:f(x)=|x-\pi |\cdot (e^{|x|}-1\left)\sin |x|\text{is not differentiable at t}\right\}$. Then the set $S$ is equal to

• A. $\left\{\pi \right\}$
• B. $\{0,\pi \}$
• C. $\varphi$ (an empty set)
• D. $\left\{0\right\}$

Answer 1

The correct option is C $\varphi$ (an empty set)

We have to check the differentiability of $f\left(x\right)$ at $x=0$ and $\pi$

1. At $x=0$

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

$=\underset{h\to 0}{lim}\frac{f\left(h\right)-f\left(0\right)}{h}=\frac{|h-\pi |(e^h-1)\sin h-0}{h}=\underset{h\to 0}{lim}\frac{\pi (e^h-1)\sinh }{h}$

$\because f\left(0\right)=0$

$=\underset{h\to 0}{lim}\frac{\pi \times 0\times \sin h}{h}=0\times 1=0$

$\because {lim}_{h\to 0}\frac{\sin h}{h}=1$

For Left-hand limit at $x=0$

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

$=\underset{h\to 0}{lim}\frac{0-(|-h-\pi |(e^h-1)\sin |-h|)}{-h})=\underset{h\to 0}{lim}\frac{\pi (e^h-1)\sin h}{h}$

$=\underset{h\to 0}{lim}\frac{\pi \times 0\times \sin h}{h}=0\times 1=0$

$LHL=RHL⟹f\left(x\right)$ is differentiable at $x=0$

2. At $x=\pi$

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

$=\underset{h\to 0}{lim}\frac{f(\pi +h)-f\left(\pi \right)}{h}=\frac{|\pi +h-\pi |(e^{\pi +h}-1)\sin (\pi +h)-0}{h}=\underset{h\to 0}{lim}\frac{h(e^{\pi }-1)\sin (\pi +h)}{h}$

$\because f\left(\pi \right)=0$

$=\underset{h\to 0}{lim}\frac{h\times (e^{\pi }-1)\times \sin \pi }{h}=0\times 1=0$

For Left-hand limit at $x=\pi$

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

$=\underset{h\to 0}{lim}\frac{0-(|\pi -h-\pi |(e^{\pi -h}-1)\sin |\pi -h|)}{-h})=\underset{h\to 0}{lim}\frac{h(e^{\pi }-1)\sin (\pi +h)}{h}$

$=\underset{h\to 0}{lim}\frac{h\times (e^{\pi }-1)\times \sin \pi }{h}=0$

$LHL=RHL⟹f\left(x\right)$ is differentiable at $x=\pi$

Set $S$ is an empty set.