Question

Let $f(x)=(2 a+b) \cos ^{-1} x+(a+2 b) \sin ^{-1} x$, where $a, b \in R$ and $a>b$. If domain of $f$ and range of $f$ are the same set, then find the value of $\pi(a-b)$.

Answer 1

Given, $f ( x )=(2 a + b ) \cos ^{-1} x +( a +2 b ) \sin ^{-1} x$
$=(2 a+b) \cos ^{-1} x+(a+2 b)\left(\frac{\pi}{2}-\cos ^{-1} x\right)=(a-b) \cos ^{-1} x+(a+2 b) \frac{\pi}{2} .$
Clearly, domain of $f(x)=[-1,1]$ and $f(x)$ is a continuous decreasing function on $[-1,1]$.
So, range of $f(x)=\left[f_{\min }(x=1), f_{\max }(x=-1)\right]$
Now, $f _{\min }( x =1)=( a +2 b ) \frac{\pi}{2}$ and $f _{\max }( x =-1)=( a - b ) \pi+( a +2 b ) \frac{\pi}{2}=3 a \frac{\pi}{2}$.
$\Rightarrow $ Range of $f ( x )=\left[( a +2 b ) \frac{\pi}{2}, 3 a \frac{\pi}{2}\right]$
As, Domain of $f$ and range of $f$ are the same set,
So, $(a+2 b) \frac{\pi}{2}=-1$
... (1) and $3 a \frac{\pi}{2}=1$ ....(2)
$\therefore $ On solving (1) and (2) we get
$a =\frac{2}{3 \pi} \text { and } b =\frac{-4}{3 \pi}$
Hence, $\pi(a-b)=\pi\left(\frac{2}{3 \pi}+\frac{4}{3 \pi}\right)=\frac{2}{3}+\frac{4}{3}=\frac{6}{3}=2$.