Question

If $f:[2,\infty )\to [1,\infty )$ is defined by $f\left(x\right)=2^{x^4-4x^2},$ then $f^{-1}\left(x\right)$ is

• A. $\sqrt{2+\sqrt{4+{\log }_{2}x}}$
• B. $-\sqrt{2+\sqrt{4+{\log }_{2}x}}$
• C. ${\log }_e(x^4-4x^2)$
• D. ${\log }_2(x^4-4x^2)$

Answer 1

The correct option is A $\sqrt{2+\sqrt{4+{\log }_{2}x}}$

$y=f\left(x\right)=2^{x^4-4x^2}$
$\Rightarrow {\log }_{2}y=x^4-4x^2$
$\Rightarrow (x^2{)}^2-4(x^2)-{\log }_{2}y=0$
$\Rightarrow x^2=\frac{4\pm \sqrt{16+4{\log }_{2}y}}{2}$
$\Rightarrow x^2=2+\sqrt{4+{\log }_{2}y}(\because x\ge 2\text{and}y\ge 1)$
$\Rightarrow x=\sqrt{2+\sqrt{4+{\log }_{2}y}}$
$\Rightarrow f^{-1}\left(y\right)=\sqrt{2+\sqrt{4+{\log }_{2}y}}$

Hence, $f^{-1}\left(x\right)=\sqrt{2+\sqrt{4+{\log }_{2}x}}$