## Task A-3.1.

Determine the smallest and largest value that the expression $\sin ^{2} x \cos 2 x$ achieves for $x \in\left[0, \frac{\pi}{2}\right]$.
## Solution.

Since $\cos 2 x=1-2 \sin ^{2} x$, we have

$$
\sin ^{2} x \cos 2 x=\sin ^{2} x\left(1-2 \sin ^{2} x\right)
$$

With the substitution $t=\sin ^{2} x$, we seek the minimum and maximum of the quadratic function

$$
f(t)=t(1-2 t)
$$

where $t \in[0,1]$ since $x \in\left[0, \frac{\pi}{2}\right]$.

![](https://cdn.mathpix.com/cropped/2024_05_30_a374c05c2f1967f14be6g-13.jpg?height=626&width=603&top_left_y=1389&top_left_x=658)

$\begin{array}{ll}\text { The maximum is } \frac{1}{8} \text { (the roots are } 0 \text { and } \frac{1}{2} \text {, so the vertex is at } t=\frac{1}{4} \text { ), } & 1 \text { point } \\ \text { and the minimum is }-1 \text { and is achieved at the boundary for } t=1 . & 1 \text { point }\end{array}$

Note: Points should be awarded similarly for a solution that uses $\sin ^{2} x+\cos ^{2} x=1 \text { and } \cos 2 x=2 \cos ^{2} x-1$, and the expression $\left(1-\cos ^{2} x\right)\left(2 \cos ^{2} x-1\right)$ is considered.

Note: If a student concludes from $0 \leqslant \sin x \leqslant 1 \text { and } -1 \leqslant \cos 2 x$ that $\sin ^{2} x \cos 2 x \geqslant-1 \text { and that the value } -1$ is achieved (for $x=\frac{\pi}{2}$), 2 points should be awarded. For determining the maximum, points are awarded as in the solution, without the need to explicitly state that $t \in[0,1]$ and the conclusion for the minimum (which together are worth 2 points).