It is given that $\mathrm{f}(x)$ is such that $\mathrm{f}'(x) = 3 \sin x \cos x$ and $\left(\frac{\pi}{2}, 1\right)$ is a point on $\mathrm{f}(x)$. Find an expression for $\mathrm{f}(x)$.

1. Express the derivative using the double angle formula for sine

$\mathrm{f}'(x) = 3 \sin x \cos x = \frac{3}{2} (2 \sin x \cos x) = \frac{3}{2} \sin 2x$

2. Integrate the derivative to find the general form of $\mathrm{f}(x)$

$\mathrm{f}(x) = \int \frac{3}{2} \sin 2x \, dx = \frac{3}{2} \left( -\frac{\cos 2x}{2} \right) + c = -\frac{3}{4} \cos 2x + c$

3. Substitute the given point to find the constant of integration $c$

Since $\left(\frac{\pi}{2}, 1\right)$ is on the curve:
$1 = -\frac{3}{4} \cos\left(2 \cdot \frac{\pi}{2}\right) + c$
$1 = -\frac{3}{4} \cos \pi + c$
$1 = -\frac{3}{4} (-1) + c$
$1 = \frac{3}{4} + c$
$c = \frac{1}{4}$

4. State the final expression for $\mathrm{f}(x)$

$\mathrm{f}(x) = -\frac{3}{4} \cos 2x + \frac{1}{4}$