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: What an inverse function means.
The inverse just undoes the original rule. So we write $y = f(x)$ and then solve to get $x$ back in terms of $y$. That solved form is the inverse.

Step 2: Write the equation.
We are given $y = 2^{x(x-1)}$. We want $x$ alone on one side.

Step 3: Pull down the power with a log.
Since the power is on base 2, take $\log_2$ on both sides. This brings the exponent down: \[ \log_2 y = x(x-1) \]

Step 4: Make it a neat quadratic.
Expand the right side and move everything to one side: \[ x^2 - x - \log_2 y = 0 \] This is a simple quadratic in $x$.

Step 5: Use the quadratic formula.
Here the coefficients are $1$, $-1$, and $-\log_2 y$. So \[ x = \frac{1 \pm \sqrt{1 + 4\log_2 x}}{2} \] (we renamed $y$ as $x$ for the final answer).

Step 6: Pick the right sign.
The domain says $x \ge 1$. The minus sign would give a small value below 1, so it is not allowed. We keep the plus sign. \[ \boxed{ f^{-1}(x) = \tfrac{1}{2}\left[1 + \sqrt{1 + 4\log_2 x}\right] } \]