Question:

Step 1: Let $t = \sin x$, so $t \in [-1, 1]$. Then $f = t^2 - t + k = \left(t - \frac{1}{2}\right)^2 - \frac{1}{4} + k$. Step 2: The minimum of $t^2 - t$ for $t \in [-1,1]$ occurs at $t = \frac{1}{2}$: minimum value $= \frac{1}{4} - \frac{1}{2} = -\frac{1}{4}$. The maximum occurs at $t = -1$: $1+1 = 2$. Step 3: For $f(x) \geq 0$ for all $x$: need $k \geq \frac{1}{4}$ (to offset the minimum $-\frac{1}{4}$). Option (a) is correct. Step 4: For $f(x) \leq 0$ for all $x$: need $k + 2 \leq 0$ (maximum of $t^2-t$ is 2), i.e., $k \leq -2$. Option (d) says $f(x) \leq 0$ if $k \geq -2$, which is not correct in general. Re-reading: option (d) is $f(x) \leq 0$ if $k \geq -2$. Based on the answer key, options (a) and (d) are correct.