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

To find the stationary points, we first calculate the derivative of the function:

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

Setting this equal to zero:

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

Since $e^x$ is never zero, we need to solve:

$\cos x + \sin x = 0$

This can be rearranged to:

$\sin x = -\cos x$

Using the identity $\tan x = -1$, we find that:

$x = \frac{3\pi}{4} + n\pi$

For $0 \leq x \leq 2\pi$, the relevant solutions are:

1. $x = \frac{3\pi}{4}$
2. $x = \frac{7\pi}{4}$

Next, we can evaluate the function at these points to find the coordinates of the stationary points:

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

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

So, the coordinates of the stationary points are roughly:

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