已知曲线方程为 $y = e^{-2x}\tan x$。

1. Differentiate the function using the product rule.

Let $y = e^{-2x}\tan x$.
Using the product rule:
$\frac{dy}{dx} = \frac{d}{dx}(e^{-2x}) \cdot \tan x + e^{-2x} \cdot \frac{d}{dx}(\tan x)$
$\frac{dy}{dx} = -2e^{-2x}\tan x + e^{-2x}\sec^2 x$

2. Factorise and apply trigonometric identity to show the required result.

Factorising out $e^{-2x}$:
$\frac{dy}{dx} = e^{-2x}(\sec^2 x - 2\tan x)$
Using the identity $\sec^2 x = 1 + \tan^2 x$:
$\frac{dy}{dx} = e^{-2x}(1 + \tan^2 x - 2\tan x)$
$\frac{dy}{dx} = e^{-2x}(1 - 2\tan x + \tan^2 x)$
$\frac{dy}{dx} = e^{-2x}(1 - \tan x)^2$

1. Set the derivative to zero to find the stationary point

For a stationary point, $\frac{dy}{dx} = 0$.
$e^{-2x}(1 - \tan x)^2 = 0$

2. Solve the resulting trigonometric equation for x

Since $e^{-2x} > 0$ for all $x$, we must have:
$(1 - \tan x)^2 = 0$
$1 - \tan x = 0$
$\tan x = 1$
For $0 < x < \frac{\pi}{2}$, the principal value is:
$x = \frac{\pi}{4}$

1. State the condition for a function to be never decreasing

A function is never decreasing if its derivative $\frac{dy}{dx} \ge 0$ for all $x$ in its domain.

2. Analyse the sign of the components of the derivative

From part (a), $\frac{dy}{dx} = e^{-2x}(1 - \tan x)^2$.
For all real $x$, $e^{-2x} > 0$.
For all real $x$ where $\tan x$ is defined, $(1 - \tan x)^2 \ge 0$.
Thus, $\frac{dy}{dx} = e^{-2x}(1 - \tan x)^2 \ge 0$ for all valid $x$.
Since $\frac{dy}{dx} \ge 0$, $y$ is never decreasing.

1. Relate the non-decreasing nature of the curve to the gradient around the stationary point

Since $y$ is never decreasing, $\frac{dy}{dx} \geqslant 0$ for all $x$ in the domain. This means the gradient is positive both immediately before and immediately after the stationary point at $x = \frac{\pi}{4}$.

2. Conclude the nature of the stationary point

Because the gradient does not change sign as it passes through zero, the stationary point is a point of inflexion.