Let $f(x) = e^x \sin x$ - HSC - SSCE Mathematics Advanced - Question 30 - 2023 - Paper 1

To find the stationary points, we first need to compute the derivative of the function:

$f'(x) = e^x \cos x + e^x \sin x = e^x(\cos x + \sin x)$

Setting the derivative to zero:

$f'(x) = 0 \implies e^x(\cos x + \sin x) = 0$

Since $e^x$ is never zero, we have:

$\cos x + \sin x = 0$

This simplifies to:

$\tan x = -1$

Thus, the solutions within the interval $[0, 2\pi]$ are:

$x = \frac{3\pi}{4} + n\pi, \text{ for } n \in \mathbb{Z}$

Calculating these solutions:

• When $n = 0$: $x = \frac{3\pi}{4}$
• When $n = 1$: $x = \frac{7\pi}{4}$

Now substituting the values back into $f(x)$:

• For $x = \frac{3\pi}{4}$:

$f\left(\frac{3\pi}{4}\right) = e^{\frac{3\pi}{4}} \sin\left(\frac{3\pi}{4}\right) = e^{\frac{3\pi}{4}} \frac{\sqrt{2}}{2}$

• For $x = \frac{7\pi}{4}$:

$f\left(\frac{7\pi}{4}\right) = e^{\frac{7\pi}{4}} \sin\left(\frac{7\pi}{4}\right) = e^{\frac{7\pi}{4}} \left(-\frac{\sqrt{2}}{2}\right)$

Thus, the coordinates of the stationary points are:

• $\left(\frac{3\pi}{4}, e^{\frac{3\pi}{4}} \frac{\sqrt{2}}{2}\right)$
• $\left(\frac{7\pi}{4}, -e^{\frac{7\pi}{4}} \frac{\sqrt{2}}{2}\right)$