Question

If $f(x)={\tan }^{-1}x-\frac{2}{\pi }{\left({\tan }^{-1}x\right)}^2+\frac{4}{{\pi }^2}{\left({\tan }^{-1}x\right)}^3-\dots \dots \dots \dots \dots \dots$ 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, ${f(x)|}_{\min }=\frac{-\pi }{2}$ at $x=-1$.
${f(x)|}_{\max }=\frac{\pi }{6}$ 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 $a=10$.