Question

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

• A. $\frac{1}{2} [1 + \sqrt{1 + 4 \log_2 x}]$
• B. $\frac{1}{2} [1 - \sqrt{1 + 4 \log_2 x}]$
• C. $\frac{1}{2} [1 + \sqrt{1 - 4 \log_2 x}]$
• D. $\frac{1}{2} [1 - \sqrt{1 - 4 \log_2 x}]$

Hint:

Since the domain is $x \ge 1$, the inverse function must also return a value $\ge 1$ for any $x \ge 1$. The negative branch yields values $\le 0.5$, which immediately rules it out.

Answer 1

The Correct Option is A

Step 1: Concept
To find the inverse function $f^{-1}(x)$, we set $y = f(x)$ and solve for $x$ in terms of $y$. Since the domain of $f$ is $[1, \infty)$, we choose the branch of the inverse that satisfies $x \ge 1$.

Step 2: Meaning
The given function is $y = 2^{x(x-1)}$. Since both the domain and codomain are $[1, \infty)$, we have $x \ge 1$ and $y \ge 1$.

Step 3: Analysis
Taking the logarithm to the base 2 on both sides gives: \[ \log_2 y = x(x-1) \implies x^2 - x - \log_2 y = 0 \] This is a quadratic equation in $x$. Using the quadratic formula, we obtain: \[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-\log_2 y)}}{2} = \frac{1 \pm \sqrt{1 + 4\log_2 y}}{2} \] Since $x \ge 1$, we must choose the positive square root branch because the negative branch would yield a value less than or equal to $0.5$ for $y \ge 1$.

Step 4: Conclusion
Replacing $y$ with $x$ gives the inverse function: \[ f^{-1}(x) = \frac{1}{2} [1 + \sqrt{1 + 4\log_2 x}] \] Final Answer: (A)