Question:

Step 1: Use the symmetry condition $f(x) = f(6-x)$. This means the parabola is symmetric about $x = 3$, so the vertex is at $x = 3$, giving $p = 6$. Step 2: Use the minimum value condition. The minimum of $f(x) = x^2 - 6x + q$ occurs at $x = 3$: $$f(3) = 9 - 18 + q = q - 9 = \frac{-81}{4}$$ $$q = 9 - \frac{81}{4} = \frac{36 - 81}{4} = \frac{-45}{4}$$ Step 3: So $f(x) = x^2 - 6x - \frac{45}{4}$. The range of $f(x)$ is $\left[\frac{-81}{4}, \infty\right)$. Step 4: Compute $22 + [f(x)]$. The minimum of $f(x)$ is $\frac{-81}{4} = -20.25$, so $[f(x)]_{\min} = -21$. Thus the minimum of $22 + [f(x)] = 22 + (-21) = 1$. Step 5: The least value of $\tan^{-1}(22 + [f(x)])$ is $\tan^{-1}(1) = \frac{\pi}{4}$. So option (1) is correct. Step 6: For $\text{sgn}(f(x) + k) = 0$, we need $f(x) + k = 0$, i.e., $f(x) = -k$. This has a solution when $-k \geq \frac{-81}{4}$, i.e., $k \leq \frac{81}{4} = 20.25$. The largest integral value of $k$ is 20. So option (3) is correct.