Question

If the area bounded by the curve $y=\left|{\cos }^{-1}(\sin x)\right|-\left|{\sin }^{-1}(\cos x)\right|$ and $x$-axis from $\frac{3\pi }{2}\le x\le 2\pi$, is equal to $\frac{{\pi }^2}{k}$, where $k\in N$, then find $k$.

Answer 1

Answer: 4

We have $y=\left|{\cos }^{-1}(\sin x)\right|-\left|{\sin }^{-1}(\cos x)\right|=\left|\frac{\pi }{2}-{\sin }^{-1}(\sin x)\right|-\left|\frac{\pi }{2}-{\cos }^{-1}(\cos x)\right|$
$=\left|\frac{\pi }{2}-(x-2\pi )\right|-\left|\frac{\pi }{2}-(2\pi -x)\right|=\left|\frac{5\pi }{2}-x\right|-\left|x-\frac{3\pi }{2}\right|=\left(\frac{5\pi }{2}-x\right)-\left(x-\frac{3\pi }{2}\right)=4\pi -2x$
$($ For $x\in \left[\frac{3\pi }{2},2\pi \right],{\sin }^{-1}(\sin x)=x-2\pi$ and ${\cos }^{-1}(\cos x)=2\pi -x)$
Clearly required area $=$ Area of shaded triangle in figure
$=\frac{1}{2}\times \left(\frac{\pi }{2}\right)\times \pi =\frac{{\pi }^2}{4}=\frac{{\pi }^2}{k}\text{ (Given) }$
Hence $k=4$