Question

Let $f(x)=(2a+b){\cos }^{-1}x+(a+2b){\sin }^{-1}x$, 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

Answer: 2

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