Question

If $f ( x )=\tan ^{-1} x -\frac{2}{\pi}\left(\tan ^{-1} x \right)^2+\frac{4}{\pi^2}\left(\tan ^{-1} x \right)^3-\ldots \ldots \ldots \ldots \ldots \ldots$ then the sum of integral values of a for which the equation $f ^2( x )+\left(\sin ^{-1} x \right)^2= a$, possess solution, is

• A. 6
• B. 7
• C. 9
• D. 10

Answer 1

Answer: (D)

Clearly, $f(x)=\frac{\tan ^{-1} x}{1+\frac{2}{\pi} \tan ^{-1} x}$
Now domain of equation $f^2(x)+\left(\sin ^{-1} x\right)^2=a$, is $x \in[-1,1]$.
Now, $f(x)=\frac{\pi}{2}\left(\frac{\frac{2}{\pi} \tan ^{-1} x+1-1}{1+\frac{2}{\pi} \tan ^{-1} x}\right)=\frac{\pi}{2}\left(1-\frac{1}{1+\frac{2}{\pi} \tan ^{-1} x}\right)$
So, $\left.f(x)\right|_{\min }=\frac{-\pi}{2}$ at $x=-1$.
$\left.f(x)\right|_{\max }=\frac{\pi}{6}$ at $x=1$.
So, $( f ( x ))^2 \in\left[0, \frac{\pi^2}{4}\right]$ and minimum and maximum is attained at $x =0$ and $x =-1$.
Also $\left(\sin ^{-1} x\right)^2 \in\left[0, \frac{\pi^2}{4}\right]$
$\therefore a_{\max }=\frac{\pi^2}{2}$ at $x=-1$ and $a_{\min }=0$ at $x=0$.
So, $a \in\left[0, \frac{\pi^2}{2}\right]$
$\therefore a _{\text {integer }}=0,1,2,3,4$
Hence, sum of integral values of $a=10$.