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Q.
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
Inverse Trigonometric Functions
Solution:
$\Theta$ Domain of $f ( x )=[-1,1]$
$\Theta-1 \leq x \leq 1 \Rightarrow-\frac{\pi}{2} \leq \sin ^{-1} x \leq \frac{\pi}{2} \Rightarrow-\frac{a \pi}{2} \leq a \sin ^{-1} x \leq \frac{a \pi}{2} $
$ \& \frac{\pi}{4} \leq \cot ^{-1} x \leq \frac{3 \pi}{4} \Rightarrow-\frac{3 \pi}{4} b \leq-b \cot ^{-1} x \leq-\frac{\pi}{4} b $
$\therefore-\frac{a \pi}{2}-\frac{3 \pi}{4} b \leq a \sin ^{-1} x-b \cot ^{-1} x \leq \frac{a}{2}-\frac{\pi}{4} b$
$\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 $