CalcGospel 國際數學圖譜
返回

IAL 2026 Jan Q9

A Level / Edexcel / FP2

IAL 2026 Jan Paper · Question 9

题目

Problem

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

r=1+sin⁡θ,−π2⩽θ⩽3π2\begin{align*} r=1+\sin\theta, \qquad -\frac{\pi}{2}\leqslant\theta\leqslant\frac{3\pi}{2} \end{align*}

and a sketch of the curve DD with polar equation

r=1+cos⁡2θ,−π2⩽θ⩽π2\begin{align*} r=1+\cos2\theta, \qquad -\frac{\pi}{2}\leqslant\theta\leqslant\frac{\pi}{2} \end{align*}

The curves intersect at the pole OO and at the point PP as shown.

(a) Determine the polar coordinates of PP.

(3)

The finite region RR, shown shaded in Figure 1, is bounded by CC and DD.

(b) Use algebraic integration to show that the area of RR is

aπ+b3\begin{align*} a\pi+b\sqrt3 \end{align*}

where aa and bb are rational numbers to be determined.

(6)

解答

(a)

解法一

思路

展开

交点满足两个 rr 相等。题目说除了极点外还有点 PP，所以解出非极点对应的角度和半径。

答题过程

展开

At an intersection,

1+sin⁡θ=1+cos⁡2θ\begin{align*} 1+\sin\theta=1+\cos2\theta \end{align*}

so

sin⁡θ=cos⁡2θsin⁡θ=1−2sin⁡2θ2sin⁡2θ+sin⁡θ−1=0\begin{align*} \sin\theta ={}& \cos2\theta\\[3mm] \sin\theta ={}& 1-2\sin^2\theta\\[3mm] 2\sin^2\theta+\sin\theta-1 ={}& 0 \end{align*}

Factorising,

(2sin⁡θ−1)(sin⁡θ+1)=0\begin{align*} (2\sin\theta-1)(\sin\theta+1)=0 \end{align*}

The solution sin⁡θ=−1\sin\theta=-1 gives the pole. For point PP,

sin⁡θ=12\begin{align*} \sin\theta=\frac12 \end{align*}

From the diagram, PP is in the first quadrant, so

θ=π6\begin{align*} \theta=\frac{\pi}{6} \end{align*}

Then

r=1+sin⁡π6=32\begin{align*} r = 1+\sin\frac{\pi}{6} = \frac32 \end{align*}

Therefore

P=(32,π6)\begin{align*} P=\left(\frac32,\frac{\pi}{6}\right) \end{align*}

(b)

解法一

思路

展开

极坐标面积公式是 12∫r2 dθ\dfrac12\int r^2\,\mathrm{d}\theta。

这个区域分成两段：从 −π2-\dfrac{\pi}{2} 到 π6\dfrac{\pi}{6} 用曲线 CC，从 π6\dfrac{\pi}{6} 到 π2\dfrac{\pi}{2} 用曲线 DD。

展开平方后用恒等式把 sin⁡2θ\sin^2\theta 和 cos⁡22θ\cos^22\theta 降幂。

答题过程

展开

For curve CC,

(1+sin⁡θ)2=1+2sin⁡θ+sin⁡2θ=1+2sin⁡θ+12−12cos⁡2θ=32+2sin⁡θ−12cos⁡2θ\begin{align*} (1+\sin\theta)^2 ={}& 1+2\sin\theta+\sin^2\theta\\[3mm] ={}& 1+2\sin\theta+\frac12-\frac12\cos2\theta\\[3mm] ={}& \frac32+2\sin\theta-\frac12\cos2\theta \end{align*}

Therefore

∫(1+sin⁡θ)2 dθ=32θ−2cos⁡θ−14sin⁡2θ\begin{align*} \int(1+\sin\theta)^2\,\mathrm{d}\theta ={}& \frac32\theta-2\cos\theta-\frac14\sin2\theta \end{align*}

For curve DD,

(1+cos⁡2θ)2=1+2cos⁡2θ+cos⁡22θ=1+2cos⁡2θ+12+12cos⁡4θ=32+2cos⁡2θ+12cos⁡4θ\begin{align*} (1+\cos2\theta)^2 ={}& 1+2\cos2\theta+\cos^22\theta\\[3mm] ={}& 1+2\cos2\theta+\frac12+\frac12\cos4\theta\\[3mm] ={}& \frac32+2\cos2\theta+\frac12\cos4\theta \end{align*}

Therefore

∫(1+cos⁡2θ)2 dθ=32θ+sin⁡2θ+18sin⁡4θ\begin{align*} \int(1+\cos2\theta)^2\,\mathrm{d}\theta ={}& \frac32\theta+\sin2\theta+\frac18\sin4\theta \end{align*}

So the area of RR is

Area=12[32θ−2cos⁡θ−14sin⁡2θ]−π/2π/6 +12[32θ+sin⁡2θ+18sin⁡4θ]π/6π/2\begin{align*} \text{Area} ={}& \frac12 \left[ \frac32\theta-2\cos\theta-\frac14\sin2\theta \right]_{-\pi/2}^{\pi/6}\\[3mm] &\,\hspace{2pt}+ \frac12 \left[ \frac32\theta+\sin2\theta+\frac18\sin4\theta \right]_{\pi/6}^{\pi/2} \end{align*}

Evaluate the first part:

[32θ−2cos⁡θ−14sin⁡2θ]−π/2π/6=(π4−3−38)−(−3π4)=π−938\begin{align*} \left[ \frac32\theta-2\cos\theta-\frac14\sin2\theta \right]_{-\pi/2}^{\pi/6} ={}& \left( \frac{\pi}{4}-\sqrt3-\frac{\sqrt3}{8} \right) -\left( -\frac{3\pi}{4} \right)\\[3mm] ={}& \pi-\frac{9\sqrt3}{8} \end{align*}

Evaluate the second part:

[32θ+sin⁡2θ+18sin⁡4θ]π/6π/2=3π4−(π4+32+316)=π2−9316\begin{align*} \left[ \frac32\theta+\sin2\theta+\frac18\sin4\theta \right]_{\pi/6}^{\pi/2} ={}& \frac{3\pi}{4} -\left( \frac{\pi}{4}+\frac{\sqrt3}{2}+\frac{\sqrt3}{16} \right)\\[3mm] ={}& \frac{\pi}{2}-\frac{9\sqrt3}{16} \end{align*}

Hence

Area=12(π−938)+12(π2−9316)=3π4−27332\begin{align*} \text{Area} ={}& \frac12 \left( \pi-\frac{9\sqrt3}{8} \right) + \frac12 \left( \frac{\pi}{2}-\frac{9\sqrt3}{16} \right)\\[3mm] ={}& \frac{3\pi}{4}-\frac{27\sqrt3}{32} \end{align*}

Therefore

a=34,b=−2732\begin{align*} a=\frac34, \qquad b=-\frac{27}{32} \end{align*}