Correct Option: C Step-by-step Solution For the equation $\sqrt{x^2 - 4x + 3} = x-k$ to have a valid real solution, two conditions must be met: Domain of the square root:$x^2 - 4x + 3 \ge 0 \implies (x-1)(x-3) \ge 0$. This implies $x \le 1$ or $x \ge 3$. Non-negativity of the result: Since the square root is non-negative, $x-k \ge 0 \implies$ $x \ge k$. 1. Solving the Equation Square both sides to eliminate the radical: $$(\sqrt{x^2 - 4x + 3})^2 = (x-k)^2$$$$x^2 - 4x + 3 = x^2 - 2kx + k^2$$$$-4x + 3 = -2kx + k^2$$ Rearrange to solve for $x$: $$(2k-4)x = k^2 - 3$$2. Analyzing the Coefficient of $x$ Case 1: $2k-4 = 0$ ($k=2$) Substituting $k=2$ results in $0 = 1$, which is a contradiction. There is no solution for $k=2$. Case 2: $2k-4 \neq 0$ ($k \neq 2$) We find a unique candidate for $x$:$$x = \frac{k^2 - 3}{2k-4}$$3. Testing the Validity of $x$ The solution must satisfy $x \ge k$: $$\frac{k^2 - 3}{2k-4} \ge k \implies \frac{k^2 - 3 - k(2k-4)}{2k-4} \ge 0$$$$\frac{-k^2 + 4k - 3}{2k-4} \ge 0 \implies \frac{k^2 - 4k + 3}{k-2} \le 0$$$$\frac{(k-1)(k-3)}{k-2} \le 0$$ Using sign analysis (the "wavy curve" method) for the critical points $1, 2, 3$: $(-\infty, 1]$: Satisfied ($\le 0$)$(1, 2)$: Not satisfied ($> 0$)$(2, 3]$: Satisfied ($\le 0$)$(3, \infty)$: Not satisfied ($> 0$) Thus, $x \ge k$ is satisfied when $k \in (-\infty, 1] \cup (2, 3]$. 4. Verifying the Domain Condition Finally, we check if $x = \frac{k^2-3}{2k-4}$ falls within $x \le 1$ or $x \ge 3$ for these ranges of $k$:If $k = 1$: $x = \frac{1-3}{2-4} = 1$. (Valid, as $x \le 1$) If $k = 3$: $x = \frac{9-3}{6-4} = 3$. (Valid, as $x \ge 3$) For any $k$ in the intervals $(-\infty, 1]$ and $(2, 3]$, the resulting $x$ remains within the valid domain. Conclusion The equation has exactly one real solution when $k \in (-\infty, 1] \cup (2, 3]$. The maximum possible value of $k$ is $3$. Final Answer: $\boxed{C}$