Question

Consider, $f(x)=\{\begin{array}{ll}\frac{3}{2}{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)+3{\tan }^{-1}x,& x>1\\ \frac{3\pi }{2}x,& -1\le x\le 1\\ {\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)+2{\tan }^{-1}x-\frac{3\pi }{2},& x<-1\end{array}$
If area enclosed by the graph of $f(x)$ and the $x$-axis from $x=-2$ to $x=2$ is $\left(\frac{p}{q}\right)\pi$, where $p,q\in N$, then find the least value of $|p-q|$.

Answer 1

Answer: 7

$f(x)=\{\begin{array}{ll}\frac{3}{2}\left(\pi -2{\tan }^{-1}x\right)+3{\tan }^{-1}x=\frac{3\pi }{2};& x>1\\ \frac{3\pi }{2}x;& x\in [-1,1]\\ -2{\tan }^{-1}x+2{\tan }^{-1}x-\frac{3\pi }{2}=\frac{-3\pi }{2};& x<-1y\end{array}$

$\text{ Area }=\left(1\times \frac{3\pi }{2}\right)2+\left(\frac{1}{2}\times 1\cdot \frac{3\pi }{2}\right)2$
$=\frac{6\pi }{2}+\frac{3\pi }{2}=\frac{9\pi }{2}=\left(\frac{9}{2}\right)\pi$
$p=9;q=2$
$\therefore p-q=7$