Question

If $f\left(x\right)={\sin }^{-1}\left(\cos x\right){\cos }^{-1}\left(\sin x\right)\forall x\in [0,2\pi ],$ then which of the following statement(s) is/are correct?

• A. $f\left(x\right)$ is differentiable in $x\in [0,2\pi ]$
• B. Range of $f\left(x\right)$ is $[-\frac{{\pi }^2}{4},\frac{{\pi }^2}{4}]$
• C. $x=\pi$ is a point of global minima as well as local minima.
• D. $\underset{0}{\overset{\pi }{\int }}f\left(x\right)dx=0$

Answer 1

Answer: (B), (C), (D)

$\underset{0}{\overset{\pi }{\int }}f\left(x\right)dx=0$

Given : $f\left(x\right)={\sin }^{-1}\left(\cos x\right){\cos }^{-1}\left(\sin x\right)$
$\Rightarrow f\left(x\right)=(\frac{\pi }{2}-{\cos }^{-1}(\cos x\left)\right)(\frac{\pi }{2}-{\sin }^{-1}(\sin x\left)\right)\Rightarrow f\left(x\right)=\{\begin{array}{c}(\frac{\pi }{2}-x)(\frac{\pi }{2}-x);0\le x\le \frac{\pi }{2}\\ (\frac{\pi }{2}-x)(\frac{\pi }{2}-(\pi -x\left)\right);\frac{\pi }{2}\le x\le \pi \\ (\frac{\pi }{2}-(2\pi -x\left)\right)(\frac{\pi }{2}-(\pi -x\left)\right);\pi \le x<\frac{3\pi }{2}\\ (\frac{\pi }{2}-(2\pi -x\left)\right)(\frac{\pi }{2}-(x-2\pi \left)\right);\frac{3\pi }{2}\le x\le 2\pi \end{array}$

$\Rightarrow f\left(x\right)=\{\begin{array}{c}{(\frac{\pi }{2}-x)}^2;0\le x\le \frac{\pi }{2}\\ -{(\frac{\pi }{2}-x)}^2;\frac{\pi }{2}\le x\le \pi \\ (\pi -x{)}^2-\frac{{\pi }^2}{4};\pi \le x<\frac{3\pi }{2}\\ \frac{{\pi }^2}{4}-(2\pi -x{)}^2;\frac{3\pi }{2}\le x\le 2\pi \end{array}$

$f\left(x\right)$ is non differentiable at $x=\pi$
Range of $f\left(x\right)$ is $[-\frac{{\pi }^2}{4},\frac{{\pi }^2}{4}]$
$x=\pi$ is a point of global minima as well as local minima.

Now,
$\underset{0}{\overset{\pi }{\int }}f\left(x\right)dx=\underset{0}{\overset{\pi /2}{\int }}{(\frac{\pi }{2}-x)}^{2}dx-\underset{\pi /2}{\overset{\pi }{\int }}{(\frac{\pi }{2}-x)}^{2}dx=0$