Question

The number of points in [–π, π] where f(x) = sin^–1 (sin x) is not differentiable is. ____________.

Answer 1

$f\left(x\right)={\sin }^{-1}\left(\sin x\right)=\left\{\begin{array}{ll}-x-\mathrm{\pi },& -\mathrm{\pi }\le x\le -\frac{\mathrm{\pi }}{2}\\ x,& -\frac{\mathrm{\pi }}{2}\le x\le \frac{\mathrm{\pi }}{2}\\ \mathrm{\pi }-x,& \frac{\mathrm{\pi }}{2}\le x\le \mathrm{\pi }\end{array}\right.$

Let us check the differentiability of the function at $x=-\frac{\mathrm{\pi }}{2}$ and $x=\frac{\mathrm{\pi }}{2}$.

At $x=-\frac{\mathrm{\pi }}{2}$,

$$\begin{gathered} Lf\text{'}\left(-\frac{\mathrm{\pi }}{2}\right)=\underset{x\to -{\frac{\mathrm{\pi }}{2}}^{-}}{\lim }\frac{f\left(x\right)-f\left(-{\displaystyle \frac{\mathrm{\pi }}{2}}\right)}{x-\left(-{\displaystyle \frac{\mathrm{\pi }}{2}}\right)} \\ \Rightarrow Lf\text{'}\left(-\frac{\mathrm{\pi }}{2}\right)=\underset{x\to -\frac{\mathrm{\pi }}{2}}{\lim }\frac{-x-\mathrm{\pi }-\left(-{\displaystyle \frac{\mathrm{\pi }}{2}}\right)}{x+{\displaystyle \frac{\mathrm{\pi }}{2}}} \\ \Rightarrow Lf\text{'}\left(-\frac{\mathrm{\pi }}{2}\right)=\underset{x\to -\frac{\mathrm{\pi }}{2}}{\lim }\frac{-\left(x+{\displaystyle \frac{\mathrm{\pi }}{2}}\right)}{x+{\displaystyle \frac{\mathrm{\pi }}{2}}} \\ \Rightarrow Lf\text{'}\left(-\frac{\mathrm{\pi }}{2}\right)=-1 \end{gathered}$$

$$\begin{gathered} Rf\text{'}\left(-\frac{\mathrm{\pi }}{2}\right)=\underset{x\to -{\frac{\mathrm{\pi }}{2}}^{+}}{\lim }\frac{f\left(x\right)-f\left(-{\displaystyle \frac{\mathrm{\pi }}{2}}\right)}{x-\left(-{\displaystyle \frac{\mathrm{\pi }}{2}}\right)} \\ \Rightarrow Rf\text{'}\left(-\frac{\mathrm{\pi }}{2}\right)=\underset{x\to -\frac{\mathrm{\pi }}{2}}{\lim }\frac{x-\left(-{\displaystyle \frac{\mathrm{\pi }}{2}}\right)}{x+{\displaystyle \frac{\mathrm{\pi }}{2}}} \\ \Rightarrow Rf\text{'}\left(-\frac{\mathrm{\pi }}{2}\right)=\underset{x\to -\frac{\mathrm{\pi }}{2}}{\lim }\frac{x+{\displaystyle \frac{\mathrm{\pi }}{2}}}{x+{\displaystyle \frac{\mathrm{\pi }}{2}}} \\ \Rightarrow Rf\text{'}\left(-\frac{\mathrm{\pi }}{2}\right)=1 \end{gathered}$$

$\therefore Lf\text{'}\left(-\frac{\mathrm{\pi }}{2}\right)\ne Rf\text{'}\left(-\frac{\mathrm{\pi }}{2}\right)$

So, the function f(x) is not differentiable at $x=-\frac{\mathrm{\pi }}{2}$.

At $x=\frac{\mathrm{\pi }}{2}$,

$$\begin{gathered} Lf\text{'}\left(\frac{\mathrm{\pi }}{2}\right)=\underset{x\to {\frac{\mathrm{\pi }}{2}}^{-}}{\lim }\frac{f\left(x\right)-f\left({\displaystyle \frac{\mathrm{\pi }}{2}}\right)}{x-{\displaystyle \frac{\mathrm{\pi }}{2}}} \\ \Rightarrow Lf\text{'}\left(\frac{\mathrm{\pi }}{2}\right)=\underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }\frac{x-{\displaystyle \frac{\mathrm{\pi }}{2}}}{x-{\displaystyle \frac{\mathrm{\pi }}{2}}} \\ \Rightarrow Lf\text{'}\left(\frac{\mathrm{\pi }}{2}\right)=1 \end{gathered}$$

$$\begin{gathered} Rf\text{'}\left(\frac{\mathrm{\pi }}{2}\right)=\underset{x\to {\frac{\mathrm{\pi }}{2}}^{+}}{\lim }\frac{f\left(x\right)-f\left({\displaystyle \frac{\mathrm{\pi }}{2}}\right)}{x-{\displaystyle \frac{\mathrm{\pi }}{2}}} \\ \Rightarrow Rf\text{'}\left(\frac{\mathrm{\pi }}{2}\right)=\underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }\frac{\mathrm{\pi }-x-{\displaystyle \frac{\mathrm{\pi }}{2}}}{x-{\displaystyle \frac{\mathrm{\pi }}{2}}} \\ \Rightarrow Rf\text{'}\left(\frac{\mathrm{\pi }}{2}\right)=\underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }\frac{-\left(\mathrm{x}-{\displaystyle \frac{\mathrm{\pi }}{2}}\right)}{x-{\displaystyle \frac{\mathrm{\pi }}{2}}} \\ \Rightarrow Rf\text{'}\left(\frac{\mathrm{\pi }}{2}\right)=-1 \end{gathered}$$

$\therefore Lf\text{'}\left(\frac{\mathrm{\pi }}{2}\right)\ne Rf\text{'}\left(\frac{\mathrm{\pi }}{2}\right)$

So, the function f(x) is not differentiable at $x=\frac{\mathrm{\pi }}{2}$.

Thus, the function f(x) = sin^–1(sinx), x ∈ [–$\mathrm{\pi }$, $\mathrm{\pi }$] is not differentiable at $x=-\frac{\mathrm{\pi }}{2}$ and $x=\frac{\mathrm{\pi }}{2}$.


The number of points in [–π, π] where f(x) = sin^–1 (sin x) is not differentiable is $\underline{2}$.