Question

Let $f:[2,\infty )\to [1,\infty )$defined by $f\left(x\right)=2^{x^4-4x^2}$ and $g:[\frac{\pi }{2},\pi ]\to A$ defined by $g\left(x\right)=\frac{\sin x+4}{\sin x-2}$ be two invertible functions, then

The domain of $f^{-1}{g}^{-1}\left(x\right)$ is

• A. $[-5,\sin 1]$
• B. $[-5,\frac{\sin 1}{2-\sin 1}]$
• C. $[-5,-\frac{(4+\sin 1)}{2-\sin 1}]$
• D. $[-\frac{(4+\sin 1)}{2-\sin 1},-2]$

Answer 1

Answer: (C)

$[-5,-\frac{(4+\sin 1)}{2-\sin 1}]$

$x:x$ in domain of $g^{-1}\left(x\right)$,
$g^{-1}\left(x\right)$ in domain of $f^{-1}\left(x\right)$
$g^{-1}\left(x\right)={\sin }^{-1}\frac{2(x+2)}{x-1},\forall x\in [-5,-2]$ ....(i)
$\Rightarrow {\sin }^{-1}\frac{2(x+2)}{x-1}\ge 1$
ie, $\Rightarrow \sin 1\le \frac{2(x+2)}{x-1}\le 1$
Solving this, we get
$x\le -\frac{(4-\sin 1)}{2-\sin 1}$ or $x>1$ ....(ii)
From Eqs. (i) and (ii), we get
$x\in [-5,-\frac{(4+\sin 1)}{2-\sin 1}]$