Let f:[4, ∞) → [4, ∞) be defined by f(x)=5^x(x-4). Then, f^-1(x) is

The correct answer is Option (2) → $2+\sqrt{4+\log_5x}$

Clearly, $f: [4, ∞) → [4, ∞)$ is a bijection. So, it is invertible.

Let $f(x) = y$. Then,

$5^{x(x-4)}=y$

$⇒x^2 - 4x = \log_5 y$

$⇒x^2-4x-\log_5 y = 0$

$⇒x=\frac{4±\sqrt{16+ 4 \log_5 y}}{2}$

$⇒f^{-1}(y)=2+\sqrt{4 +\log_5 y}$

Hence, $f^{-1}(x)=2+\sqrt{4 +\log_5 x}$

We know that if g(x) is inverse of a bijection f(x), then

$fog (x)=x⇒f(g(x))=x$

This relation suggests the following method for finding the inverse of a bijection.