曲线的方程由 $y = x + \sin^2 x$ 给出，其中 $0 \le x \le \pi$。 写出 $\frac{\mathrm{d}y}{\mathrm{d}x}$ 的表达式。 求该曲线的驻点。 确定该驻点的性质。

1. Differentiate the expression term by term

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}}{\mathrm{d}x}(x) + \frac{\mathrm{d}}{\mathrm{d}x}(\sin^2 x)$
$\frac{\mathrm{d}y}{\mathrm{d}x} = 1 + 2\sin x \cos x$

1. Set the derivative to zero to find the x-coordinate of the stationary point.

From part (i), $\frac{\mathrm{d}y}{\mathrm{d}x} = 1 + 2\sin x \cos x = 1 + \sin 2x$.
For stationary points, $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$:
$1 + \sin 2x = 0$
$\sin 2x = -1$

2. Solve the trigonometric equation for $x$ in the given domain.

Given $0 \le x \le \pi$, the domain for $2x$ is $0 \le 2x \le 2\pi$.
$2x = \frac{3\pi}{2}$
$x = \frac{3\pi}{4}$

3. Find the corresponding y-coordinate.

Substitute $x = \frac{3\pi}{4}$ into the equation of the curve:
$y = \frac{3\pi}{4} + \sin^2\left(\frac{3\pi}{4}\right)$
$y = \frac{3\pi}{4} + \left(\frac{1}{\sqrt{2}}\right)^2$
$y = \frac{3\pi}{4} + \frac{1}{2}$
The stationary point is $\left(\frac{3\pi}{4}, \frac{3\pi}{4} + \frac{1}{2}\right)$.

1. Recall the first derivative and the $x$-coordinate of the stationary point from earlier parts

From (i), $\frac{\mathrm{d}y}{\mathrm{d}x} = 1 + 2\sin x \cos x = 1 + \sin 2x$.
From (ii), the stationary point is at $x = \frac{3\pi}{4}$.

2. Apply the first derivative test to check the sign of the gradient around the stationary point

Consider values of $x$ slightly less than and slightly greater than $\frac{3\pi}{4}$ (approx $2.356$):
When $x = 2.3$, $\frac{\mathrm{d}y}{\mathrm{d}x} = 1 + \sin(4.6) > 0$
When $x = 2.4$, $\frac{\mathrm{d}y}{\mathrm{d}x} = 1 + \sin(4.8) > 0$

Alternatively, observe that since $-1 \le \sin 2x \le 1$, we have $\frac{\mathrm{d}y}{\mathrm{d}x} = 1 + \sin 2x \ge 0$ for all $x$.

3. Conclude the nature of the stationary point

Since the gradient is positive on both sides of $x = \frac{3\pi}{4}$ (it does not change sign), the stationary point is a point of inflexion.