Question

If domain and range of the function $f(x)=a{\sin }^{-1}x-b{\cot }^{-1}x$ are same, where $a,b$ are non-negative real numbers then $\pi (a+b)$ is equal to

• A. 0
• B. 1
• C. 2
• D. 4

Answer 1

Answer: (B)

$\Theta$ Domain of $f(x)=[-1,1]$
$\Theta -1\le x\le 1\Rightarrow -\frac{\pi }{2}\le {\sin }^{-1}x\le \frac{\pi }{2}\Rightarrow -\frac{a\pi }{2}\le a{\sin }^{-1}x\le \frac{a\pi }{2}$
$\&\frac{\pi }{4}\le {\cot }^{-1}x\le \frac{3\pi }{4}\Rightarrow -\frac{3\pi }{4}b\le -b{\cot }^{-1}x\le -\frac{\pi }{4}b$
$\therefore -\frac{a\pi }{2}-\frac{3\pi }{4}b\le a{\sin }^{-1}x-b{\cot }^{-1}x\le \frac{a}{2}-\frac{\pi }{4}b$
$\Theta -\frac{a\pi }{2}-\frac{3\pi }{4}b=-1$
$\&\frac{a}{2}-\frac{\pi }{4}b=1$
$\therefore b=0\text{ and }a=\frac{2}{\pi }$
$\therefore (a+b)\pi =2$