IAL 2025 Jan Q5 | 國際數學站

Figure 1 In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

Figure 1 shows a sketch of part of the curve with polar equation

\begin{align*} r = \sqrt{3} + \tan \frac{\theta}{2} \qquad 0 \leqslant \theta < \pi \end{align*}

The tangent to the curve at the point $P$ is parallel to the initial line.

(a) Using the identity $\tan \frac{\theta}{2} = \frac{1 - \cos \theta}{\sin \theta}$ or otherwise, determine the exact value of $\theta$ at $P$ .

(4)

The region $R$ , shown shaded in Figure 1, is bounded by the initial line, the curve and the line $OP$ , where $O$ is the pole.

(b) Use algebraic integration to determine the exact area of $R$ , giving your answer in the form $p\ln 2 + q\pi + r$ where $p$ , $q$ and $r$ are constants.

(6)

解答

(a)

Since the tangent is parallel to the initial line, $\frac{\mathrm{d}y}{\mathrm{d}\theta} = 0$ , where $y = r\sin\theta$ .

\begin{align*} y = &\,r\sin\theta\\[4mm] = &\,\left(\sqrt{3} + \tan\frac{\theta}{2}\right)\sin\theta\\[4mm] = &\,\left(\sqrt{3} + \frac{1 - \cos\theta}{\sin\theta}\right)\sin\theta\\[4mm] = &\,\sqrt{3}\sin\theta + 1 - \cos\theta \end{align*}

Differentiating with respect to $\theta$ :

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}\theta} = &\,\sqrt{3}\cos\theta + \sin\theta \end{align*}

Setting $\frac{\mathrm{d}y}{\mathrm{d}\theta} = 0$ :

\begin{align*} \sqrt{3}\cos\theta + \sin\theta = &\,0\\[4mm] \sin\theta = &\,-\sqrt{3}\cos\theta\\[4mm] \tan\theta = &\,-\sqrt{3} \end{align*}

Given the domain $0 \leqslant \theta < \pi$ , the only solution is:

\begin{align*} \theta = &\,\pi - \frac{\pi}{3}\\[4mm] = &\,\frac{2\pi}{3} \end{align*}

(b)

The area of the shaded region $R$ is given by $\frac{1}{2}\int r^2 \,\mathrm{d}\theta$ .

\begin{align*} \int r^2 \,\mathrm{d}\theta = &\,\int \left(\sqrt{3} + \tan\frac{\theta}{2}\right)^2 \,\mathrm{d}\theta\\[4mm] = &\,\int \left(3 + 2\sqrt{3}\tan\frac{\theta}{2} + \tan^2\frac{\theta}{2}\right) \,\mathrm{d}\theta\\[4mm] = &\,\int \left(3 + 2\sqrt{3}\tan\frac{\theta}{2} + \sec^2\frac{\theta}{2} - 1\right) \,\mathrm{d}\theta\\[4mm] = &\,\int \left(2 + 2\sqrt{3}\tan\frac{\theta}{2} + \sec^2\frac{\theta}{2}\right) \,\mathrm{d}\theta\\[4mm] = &\,2\theta - 4\sqrt{3}\ln\left(\cos\frac{\theta}{2}\right) + 2\tan\frac{\theta}{2} \end{align*}

Evaluating from $\theta = 0$ to $\theta = \frac{2\pi}{3}$ :

\begin{align*} \text{Area} = &\,\frac{1}{2}\left[2\theta - 4\sqrt{3}\ln\left(\cos\frac{\theta}{2}\right) + 2\tan\frac{\theta}{2}\right]_0^{\frac{2\pi}{3}}\\[4mm] = &\,\frac{1}{2}\left( \left( 2\left(\frac{2\pi}{3}\right) - 4\sqrt{3}\ln\left(\cos\frac{\pi}{3}\right) + 2\tan\frac{\pi}{3} \right) - (0 - 4\sqrt{3}\ln 1 + 0) \right)\\[4mm] = &\,\frac{1}{2}\left( \frac{4\pi}{3} - 4\sqrt{3}\ln\left(\frac{1}{2}\right) + 2\sqrt{3} \right)\\[4mm] = &\,\frac{1}{2}\left( \frac{4\pi}{3} + 4\sqrt{3}\ln 2 + 2\sqrt{3} \right)\\[4mm] = &\,2\sqrt{3}\ln 2 + \frac{2\pi}{3} + \sqrt{3} \end{align*}