曲线 $C$ 的极坐标方程为 $r^2=\dfrac{a}{1+\tan2\theta}$，其中 $0\leqslant\theta<\frac{1}{4}\pi$，$a$ 为正常数。

(a) 证明：在 $C$ 上距离极轴最远的点处，

并验证此方程在 $0.5$ 与 $0.6$ 之间有一个根。

(b) 画出 $C$ 的草图。

(c) 使用代换 $v=\tan2\theta$，证明 $C$ 与极轴所围成的面积为

并求此面积的精确值。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

离初始线最远是让纵坐标y=r sinθ最大。对纵坐标求导，提出正因子后可化成指定式。

The perpendicular distance from the initial line is

It is positive for $0<\theta<\pi/4$ and tends to zero at the two ends. At an interior maximum, $\mathrm{d}y/\mathrm{d}\theta=0$.

By the product and chain rules,

Since $r>0$ at the maximum, division by $r$ is valid. Also

Thus the maximum satisfies

as required.

Let the left-hand side be $f(\theta)$. In radians,

The denominator is non-zero and $f$ is continuous on $[0.5,0.6]$. Hence there is a root between $0.5$ and $0.6$.

也可对距离平方求导，因为纵坐标为正，内部最大值对应距离平方的驻点。

The squared distance from the initial line is

It vanishes at the two ends in the limiting sense and is positive in between. At an interior maximum of $y$, differentiating gives

Divide by $2a\sin\theta/(1+\tan2\theta)>0$:

Using $\sec^22\theta/(1+\tan2\theta)=1/[\cos2\theta(\sin2\theta+\cos2\theta)]$ gives the required equation. Its left-hand side has values $0.2354162034\ldots$ and $-0.3785037476\ldots$ at $0.5$ and $0.6$, respectively, and is continuous between them. Thus a root lies in $(0.5,0.6)$.

曲线在第一象限，从初始线上的半径√a处出发，再向极点靠近；画出单一拱形及其水平最高点。

At $\theta=0$, $r=\sqrt a$, giving $(\sqrt a,0)$. As $\theta\to\pi/4^-$, $r\to0$, so the curve approaches the pole.

Sketch a single smooth arch in the first quadrant joining $(\sqrt a,0)$ towards $O$, above the initial line. Mark the initial line and pole clearly, with a horizontal tangent at the highest point identified in (a). The tangent at $(\sqrt a,0)$ has gradient $-1$; near the pole the tangent approaches gradient $1$.

先用极坐标面积公式，再完整处理题目指定的代换和无穷上限。分式分解后积分，对数项先合并再取极限。

The area is

Set $v=\tan2\theta$. Then

Therefore

When $\theta=0$, $v=0$; as $\theta\to\pi/4^-$, $v\to\infty$. Hence

as required.

For partial fractions, write

Equating coefficients in

gives $C+D=0$, $D+E=0$, $C+E=1$, so $C=E=1/2$ and $D=-1/2$. Thus

Therefore

Combine the logarithms before taking the limit:

Since $\tan^{-1}R\to\pi/2$ and the lower endpoint contributes zero,