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FP2 2025 Jan Q5

A Level / Edexcel / FP2

IAL 2025 Jan Paper · Question 5

题目

Problem

5.

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

r=3+tan⁡θ20⩽θ<π\begin{align*} r=\sqrt3+\tan\frac{\theta}{2} \qquad 0\leqslant\theta<\pi \end{align*}

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

(a) Using the identity tan⁡θ2=1−cos⁡θsin⁡θ\tan\frac{\theta}{2}=\frac{1-\cos\theta}{\sin\theta} or otherwise, determine the exact value of θ\theta at PP.

(4)

The region RR, shown shaded in Figure 1, is bounded by the initial line, the curve and the line OPOP, where OO is the pole.

(b) Use algebraic integration to determine the exact area of RR, giving your answer in the form pln⁡2+qπ+rp\ln2+q\pi+r where pp, qq and rr are constants.

(6)
题目中文翻译

本题必须展示所有计算步骤。

完全依赖计算器技术得出的解答不予接受。

图 1 是具有以下极坐标方程的曲线的一部分：

r= 3+tan⁡θ2,0⩽θ<π.\begin{align*} r=&\,\sqrt3+\tan\frac\theta2, \qquad0\leqslant\theta<\pi. \end{align*}

曲线在点 PP 处的切线平行于极轴。

(a) 使用恒等式 tan⁡θ2=1−cos⁡θsin⁡θ\tan\frac\theta2=\frac{1-\cos\theta}{\sin\theta} 或其他方法，求点 PP 对应的 θ\theta 的精确值。

(4)

图 1 中的阴影区域 RR 由极轴、曲线及直线 OPOP 围成，其中 OO 是极点。

(b) 使用代数积分，求 RR 的精确面积，答案写成 pln⁡2+qπ+rp\ln2+q\pi+r 的形式，其中 pp、qq、rr 为常数。

(6)

解答

(a)

解法一

思路

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切线平行于 initial line，表示该点的切线是水平的，所以可令 dydθ=0\frac{\mathrm{d}y}{\mathrm{d}\theta}=0。极坐标中 y=rsin⁡θy=r\sin\theta，题目给的半角公式正好能把 yy 化得很简单。

答题过程

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Since

y=rsin⁡θ,\begin{align*} y=r\sin\theta, \end{align*}

we have

y= (3+tan⁡θ2)sin⁡θ= (3+1−cos⁡θsin⁡θ)sin⁡θ= 3sin⁡θ+1−cos⁡θ\begin{align*} y =&\, \bigg(\sqrt3+\tan\frac{\theta}{2}\bigg)\sin\theta\\[3mm] =&\, \bigg(\sqrt3+\frac{1-\cos\theta}{\sin\theta}\bigg)\sin\theta\\[3mm] =&\, \sqrt3\sin\theta+1-\cos\theta \end{align*}

Differentiate:

dydθ= 3cos⁡θ+sin⁡θ\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}\theta} =&\, \sqrt3\cos\theta+\sin\theta \end{align*}

The tangent at PP is parallel to the initial line, so

3cos⁡θ+sin⁡θ= 0sin⁡θ= −3cos⁡θtan⁡θ= −3\begin{align*} \sqrt3\cos\theta+\sin\theta =&\, 0\\[2mm] \sin\theta =&\, -\sqrt3\cos\theta\\[2mm] \tan\theta =&\, -\sqrt3 \end{align*}

Since 0⩽θ<π0\leqslant\theta<\pi, the solution is in the second quadrant. Hence

θ=2π3\begin{align*} \boxed{\theta=\frac{2\pi}{3}} \end{align*}

解法二

思路

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也可以先对 y=(3+tan⁡θ2)sin⁡θy=(\sqrt3+\tan\frac\theta2)\sin\theta 直接求导，再令 t=tan⁡θ2t=\tan\frac\theta2。把 sin⁡θ\sin\theta 和 cos⁡θ\cos\theta 都换成 tt，水平切线条件便成为二次方程。由角度范围可知 t⩾0t\geqslant0，据此选择根。

答题过程

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Differentiate y=rsin⁡θy=r\sin\theta directly:

dydθ= (3+tan⁡θ2)cos⁡θ +12sec⁡2θ2sin⁡θ.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}\theta} =&\,\bigg(\sqrt3+\tan\frac\theta2\bigg)\cos\theta\\[3mm] &\,\hspace{2pt}+\frac12\sec^2\frac\theta2\sin\theta. \end{align*}

At PP, this derivative is zero. Let t=tan⁡θ2t=\tan\frac\theta2. Since 0⩽θ<π0\leqslant\theta<\pi, we have t⩾0t\geqslant0 and

sin⁡θ= 2t1+t2,cos⁡θ= 1−t21+t2,sec⁡2θ2= 1+t2.\begin{align*} \sin\theta=&\,\frac{2t}{1+t^2},\\[3mm] \cos\theta=&\,\frac{1-t^2}{1+t^2},\\[3mm] \sec^2\frac\theta2=&\,1+t^2. \end{align*}

The horizontal tangent condition becomes

(3+t)1−t21+t2+t= 0.\begin{align*} (\sqrt3+t)\frac{1-t^2}{1+t^2}+t=&\,0. \end{align*}

As 1+t2>01+t^2>0, multiplying through gives

(3+t)(1−t2)+t(1+t2)= 0,3+2t−3t2= 0,3t2−2t−3= 0.\begin{align*} (\sqrt3+t)(1-t^2)+t(1+t^2)=&\,0,\\[3mm] \sqrt3+2t-\sqrt3t^2=&\,0,\\[3mm] \sqrt3t^2-2t-\sqrt3=&\,0. \end{align*}

Thus

t= 2±4+1223= 3or−13.\begin{align*} t=&\,\frac{2\pm\sqrt{4+12}}{2\sqrt3}\\[3mm] =&\,\sqrt3\quad\text{or}\quad-\frac1{\sqrt3}. \end{align*}

Only t=3t=\sqrt3 is non-negative. Therefore

θ= 2arctan⁡3=2π3.\begin{align*} \theta=&\,2\arctan\sqrt3 =\boxed{\frac{2\pi}{3}}. \end{align*}

(b)

解法一

思路

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区域由 initial line、曲线和 OPOP 围成，所以面积是

12∫02π/3r2 dθ\begin{align*} \frac12\int_0^{2\pi/3}r^2\,\mathrm{d}\theta \end{align*}

关键是把 tan⁡2θ2\tan^2\frac{\theta}{2} 改成 sec⁡2θ2−1\sec^2\frac{\theta}{2}-1，这样就能直接积分。

答题过程

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The required area is

Area= 12∫02π/3(3+tan⁡θ2)2 dθ\begin{align*} \text{Area} =&\, \frac12\int_0^{2\pi/3} \bigg(\sqrt3+\tan\frac{\theta}{2}\bigg)^2 \,\mathrm{d}\theta \end{align*}

Expand the integrand:

(3+tan⁡θ2)2= 3+23tan⁡θ2+tan⁡2θ2= 3+23tan⁡θ2+sec⁡2θ2−1= 2+23tan⁡θ2+sec⁡2θ2\begin{align*} \bigg(\sqrt3+\tan\frac{\theta}{2}\bigg)^2 =&\, 3+2\sqrt3\tan\frac{\theta}{2} +\tan^2\frac{\theta}{2}\\[3mm] =&\, 3+2\sqrt3\tan\frac{\theta}{2} +\sec^2\frac{\theta}{2}-1\\[3mm] =&\, 2+2\sqrt3\tan\frac{\theta}{2} +\sec^2\frac{\theta}{2} \end{align*}

Therefore

Area= 12∫02π/3(2+23tan⁡θ2+sec⁡2θ2) dθ\begin{align*} \text{Area} =&\, \frac12\int_0^{2\pi/3} \bigg( 2+2\sqrt3\tan\frac{\theta}{2} +\sec^2\frac{\theta}{2} \bigg)\,\mathrm{d}\theta \end{align*}

Integrate:

∫(2+23tan⁡θ2+sec⁡2θ2) dθ= 2θ−43ln⁡(cos⁡θ2)+2tan⁡θ2\begin{align*} \int \bigg( 2+2\sqrt3\tan\frac{\theta}{2} +\sec^2\frac{\theta}{2} \bigg)\,\mathrm{d}\theta =&\, 2\theta -4\sqrt3\ln\bigg(\cos\frac{\theta}{2}\bigg) +2\tan\frac{\theta}{2} \end{align*}

So

Area= 12[2θ−43ln⁡(cos⁡θ2)+2tan⁡θ2]02π/3= 12[4π3−43ln⁡(cos⁡π3)+2tan⁡π3]= 12[4π3−43ln⁡12+23]= 23ln⁡2+2π3+3\begin{align*} \text{Area} =&\, \frac12\bigg[ 2\theta -4\sqrt3\ln\bigg(\cos\frac{\theta}{2}\bigg) +2\tan\frac{\theta}{2} \bigg]_0^{2\pi/3}\\[3mm] =&\, \frac12\bigg[ \frac{4\pi}{3} -4\sqrt3\ln\bigg(\cos\frac{\pi}{3}\bigg) +2\tan\frac{\pi}{3} \bigg]\\[3mm] =&\, \frac12\bigg[ \frac{4\pi}{3} -4\sqrt3\ln\frac12 +2\sqrt3 \bigg]\\[3mm] =&\, \boxed{ 2\sqrt3\ln2+\frac{2\pi}{3}+\sqrt3 } \end{align*}