Question

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

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

Answer 1

The correct option is C

an empty set

Explanation for the correct option:

Differentibaility of a function:

Given set $S=\left\{t\in R:f\left(x\right)=|x-\pi |(e^{\left|x\right|}-1)\sin x\right\}$ is not differentiable at $t$ .

Therefore, $f\left(x\right)=|x-\pi |(e^{\left|x\right|}-1)\sin x$

Checking the differentiability at $x=\mathrm{\pi }$

By definition of differentiability

Right Hand Derivative

$$\begin{gathered} \Rightarrow \underset{h\to 0}{\lim }\frac{f(\mathrm{\pi }+\mathrm{h})-\mathrm{f}\left(\mathrm{\pi }\right)}{h}=\underset{h\to 0}{\lim }\frac{|\left(\mathrm{\pi }+\mathrm{h}\right)-\pi |(e^{|\mathrm{\pi }+\mathrm{h}|}-1)\sin \left(\mathrm{\pi }+\mathrm{h}\right)-\left\{|\left(\mathrm{\pi }\right)-\pi |(e^{|\mathrm{\pi }|}-1)\sin \left(\mathrm{\pi }\right)\right\}}{h} \\ =\underset{h\to 0}{\lim }\frac{|\left(\mathrm{\pi }+\mathrm{h}\right)-\pi |(e^{|\mathrm{\pi }+\mathrm{h}|}-1)\sin \left(\mathrm{\pi }+\mathrm{h}\right)}{h}\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{\pi }\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{]} \\ =\underset{h\to 0}{\lim }\frac{h(e^{\left(\mathrm{\pi }+\mathrm{h}\right)}-1)\sin \left(\mathrm{\pi }+\mathrm{h}\right)}{h} \\ =\underset{h\to 0}{\lim }\left(-\sin h\right)(e^{\left(\mathrm{\pi }+\mathrm{h}\right)}-1)\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{(}\mathbf{\pi }\mathbf{}\mathbf{+}\mathbf{}\mathbf{h}\mathbf{)}\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}\mathbf{}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{h}\mathbf{]} \\ =0\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{[}\mathbf{}\mathbf{h}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{}\mathbf{\&}\mathbf{}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{]} \end{gathered}$$

Left Hand Derivative

$$\begin{gathered} \Rightarrow \underset{h\to 0}{\lim }\frac{f(\mathrm{\pi }-\mathrm{h})-\mathrm{f}\left(\mathrm{\pi }\right)}{h}=\underset{h\to 0}{\lim }\frac{|\left(\mathrm{\pi }-\mathrm{h}\right)-\pi |(e^{|\mathrm{\pi }-\mathrm{h}|}-1)\sin \left(\mathrm{\pi }-\mathrm{h}\right)-\left\{|\left(\mathrm{\pi }\right)-\pi |(e^{|\mathrm{\pi }|}-1)\sin \left(\mathrm{\pi }\right)\right\}}{h} \\ =\underset{h\to 0}{\lim }\frac{|\left(\mathrm{\pi }-\mathrm{h}\right)-\pi |(e^{|\mathrm{\pi }-\mathrm{h}|}-1)\sin \left(\mathrm{\pi }-\mathrm{h}\right)}{h}\left[\because sin\pi =0\right] \\ =\underset{h\to 0}{\lim }\frac{h(e^{\left(\mathrm{\pi }-\mathrm{h}\right)}-1)\sin \left(\mathrm{\pi }-\mathrm{h}\right)}{h} \\ =\underset{h\to 0}{\lim }\left(\sin h\right)(e^{\left(\mathrm{\pi }-\mathrm{h}\right)}-1) \\ =0\mathbf{}\left[h=0\&sin\left(0\right)=0\right] \end{gathered}$$

Thus, Right Hand Derivative is equal to Left Hand Derivative

Hence, the function is differentiable at $\mathrm{\pi }$

Similarly $f\left(x\right)$ is differentiable at $0$

Therefore, it is differentiable in $\left\{0,\pi \right\}$

Alternatively:

$h\left(x\right)=f\left(x\right)g\left(x\right)$, if $f\left(a\right)=0andatx=a,g\left(x\right)$ is not differentiable then $h\left(x\right)$ is differentiable function at $x=a$

Here,$|x-\mathrm{\pi }|,e^{\left|x\right|}-1$is not differentiable at $x=\mathrm{\pi }\mathrm{and}\mathrm{x}=0\mathrm{respectively}$.

For $x=0,\pi andsinx=0$

Therefore given function is differentibale at $\left\{0,\pi \right\}$

Hence, the correct option is (C).