Question

Consider, $f(x)=\begin{cases}\frac{3}{2} \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)+3 \tan ^{-1} x, & x>1 \\ \frac{3 \pi}{2} x, & -1 \leq x \leq 1 \\ \cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)+2 \tan ^{-1} x-\frac{3 \pi}{2}, & x<-1\end{cases}$
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

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

$\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 $