Question:

<p><strong>Step 1: Identify the function and its domain/range.</strong></p><p>The given function is $f:[1,\infty) o [3,\infty)$ defined by $f(x)=(\log_2x)^2+2\log_2x+3$.</p><p><strong>Step 2: Set $y=f(x)$ and make a substitution to simplify the expression.</strong></p><p>Let $y = (\log_2x)^2+2\log_2x+3$.</p><p>Let $t = \log_2x$. Substituting $t$ into the equation, we get a quadratic expression in $t$:</p><p>$y = t^2+2t+3$</p><p><strong>Step 3: Complete the square for the quadratic expression in $t$.</strong></p><p>We can rewrite the expression as:</p><p>$y = (t^2+2t+1)+2$</p><p>$y = (t+1)^2+2$</p><p><strong>Step 4: Solve for $t$ in terms of $y$.</strong></p><p>Subtract 2 from both sides:</p><p>$(t+1)^2 = y-2$</p><p>Take the square root of both sides:</p><p>$t+1 = \pm\sqrt{y-2}$</p><p>Isolate $t$:</p><p>$t = -1 \pm\sqrt{y-2}$</p><p><strong>Step 5: Determine the valid range for $t$ based on the domain of $f(x)$.</strong></p><p>The domain of $f(x)$ is given as $[1,\infty)$.</p><p>Since $x \ge 1$, we have $\log_2x \ge \log_21$.</p><p>$\log_2x \ge 0$</p><p>So, $t = \log_2x \ge 0$. This means $t$ must be non-negative.</p><p><strong>Step 6: Choose the correct sign for the square root based on the range of $t$.</strong></p><p>We have $t = -1 \pm\sqrt{y-2}$.</p><p>Since $t \ge 0$, we must have $-1 \pm\sqrt{y-2} \ge 0$.</p><p>If we choose the negative sign, $t = -1 - \sqrt{y-2}$. Since $\sqrt{y-2} \ge 0$ (because the range of $f(x)$ is $[3,\infty)$, so $y \ge 3 \Rightarrow y-2 \ge 1$), this would imply $t \le -1$, which contradicts $t \ge 0$.</p><p>Therefore, we must choose the positive sign:</p><p>$t = -1 + \sqrt{y-2}$</p><p><strong>Step 7: Substitute back $t = \log_2x$ and solve for $x$.</strong></p><p>$\log_2x = -1 + \sqrt{y-2}$</p><p>To solve for $x$, we use the definition of logarithm ($b^c = a \iff \log_b a = c$):</p><p>$x = 2^{(-1 + \sqrt{y-2})}$</p><p><strong>Step 8: Replace $y$ with $x$ to denote the inverse function.</strong></p><p>The inverse function is $f^{-1}(x) = 2^{-1 + \sqrt{x-2}}$. This matches option (A).</p><p><strong>Step 9: Determine the domain of $f^{-1}(x)$.</strong></p><p>The domain of $f^{-1}(x)$ is the range of the original function $f(x)$.</p><p>The problem states that the range of $f(x)$ is $[3,\infty)$.</p><p>Therefore, the domain of $f^{-1}(x)$ is $[3,\infty)$. This matches option (D).</p><p><strong>Step 10: Verify the range of $f(x)$.</strong></p><p>From $f(x) = (t+1)^2+2$ and $t=\log_2x$, with $x \in [1,\infty)$, we have $t \in [0,\infty)$.</p><p>Thus, $t+1 \in [1,\infty)$.</p><p>$(t+1)^2 \in [1^2,\infty) = [1,\infty)$.</p><p>So, $f(x) = (t+1)^2+2 \in [1+2,\infty) = [3,\infty)$.</p><p>The range is indeed $[3,\infty)$, confirming the domain of $f^{-1}(x)$.</p><p>Both (A) and (D) are correct. In the given options, (B) represents 'A,D'.</p><p><strong>Answer:</strong> The correct options are A and D, which corresponds to option B in the multiple-choice selection.</p> <div class="key-concept"><strong>Key Concept:</strong> To find the inverse of a function, express the independent variable in terms of the dependent variable. When dealing with quadratic forms, complete the square. The specific domain of the original function restricts the range of intermediate variables, which is crucial for selecting the correct branch of the inverse. The domain of the inverse function is precisely the range of the original function.</div> <div class="trap-box"><strong>Trap:</strong> A common mistake is incorrectly choosing the sign when taking the square root, i.e., $-1 - \sqrt{x-2}$, by not considering the restricted range of $t=\log_2x$. Another trap is to conclude the domain of $f^{-1}(x)$ as $[2,\infty)$ simply from the algebraic condition $x-2 \ge 0$, instead of realizing it must be the specified range of $f(x)$, which is $[3,\infty)$.</div>