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

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

statement (A) is wrong because at $x=\pi ,f(x)$ is not differentiable.
$I=\underset{0}{\overset{\pi }{\int }}f(x)dx={\int }_0^{\pi }{\sin }^{-1}(\cos x){\cos }^{-1}(\sin x)dx$
$I=\underset{0}{\overset{\pi }{\int }}{\sin }^{-1}(-\cos x){\cos }^{-1}(\sin x)dx,2I=0$
$\text{ At }x=\pi ,\text{ is a point of local minima. }$