Question

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

• A. $f(x)$ is continuous and differentiable in $[0,2 \pi]$.
• B. Range of $f(x)$ is $\left[-\pi^2 / 4, \pi^2 / 4\right]$.
• C. $x=\pi$ is a point of global minima as well as local minima.
• D. $\int\limits_0^\pi f(x) d x=0$

Answer 1

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

$ f(x)=\left(\frac{\pi}{2}-\cos ^{-1}(\cos x)\right)\left(\frac{\pi}{2}-\sin ^{-1}(\sin x)\right)$
So $f(x)=\left(\frac{\pi}{2}-x\right)\left(\frac{\pi}{2}-x\right), 0 \leq x \leq \frac{\pi}{2}=\left(\frac{\pi}{2}-x\right)^2$
Thus, $f(x)=\left(\frac{\pi}{2}-x\right)\left(\frac{\pi}{2}-(\pi-x)\right), \frac{\pi}{2} \leq x \leq \pi=\left(\frac{\pi}{2}-x\right)\left(-\frac{\pi}{2}+x\right)=-\left(\frac{\pi}{2}-x\right)^2$
$f(x)=\left(\frac{\pi}{2}-(2 \pi-x)\right)\left(\frac{\pi}{2}-(\pi-x)\right), \pi \leq x<\frac{3 \pi}{2} $
$=\left(-\frac{\pi}{2}-\pi+x\right)\left(\frac{\pi}{2}-(\pi-x)\right)=-\left(\frac{\pi}{2}+(\pi-x)\right)\left(\frac{\pi}{2}-(\pi-x)\right)$
$\text { so } f(x)=-\left(\frac{\pi^2}{4}-(\pi-x)^2\right)=(\pi-x)^2-\frac{\pi^2}{4}, \pi \leq x<\frac{3 \pi}{2} $
$\text { Now, } f(x)=\left(\frac{\pi}{2}-(2 \pi-x)\right)\left(\frac{\pi}{2}-(x-2 \pi)\right) \frac{3 \pi}{2} \leq x \leq 2 \pi$
$=\left(\frac{\pi}{2}-(2 \pi-x)\right)\left(\frac{\pi}{2}+(2 \pi-x)\right)=\frac{\pi^2}{4}-(2 \pi-x)^2 $

statement (A) is wrong because at $x =\pi, f ( x )$ is not differentiable.
$I =\int\limits_0^\pi f(x) d x=\int_0^\pi \sin ^{-1}(\cos x) \cos ^{-1}(\sin x) d x $
$I=\int\limits_0^\pi \sin ^{-1}(-\cos x) \cos ^{-1}(\sin x) d x, 2 I=0$
$\text { At } x=\pi, \text { is a point of local minima. }$