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IAL 2024 May R Q9

A Level / Edexcel / P1

IAL 2024 May (R) Paper · Question 9

题目

Problem

Figure 3 shows a sketch of

Figure 3
  • the curve with equation y=tan⁡(x−π6)y=\tan\left(x-\dfrac{\pi}{6}\right) for 0≤x≤2π0\le x\le2\pi
  • part of the straight line ll with equation y=π−xy=\pi-x

(a) State the number of solutions of the equation

(i) tan⁡(x−π6)=π−x\tan\left(x-\dfrac{\pi}{6}\right)=\pi-x in the interval 0≤x≤2π0\le x\le2\pi

(ii) tan⁡(x−π6)=π−x\tan\left(x-\dfrac{\pi}{6}\right)=\pi-x in the interval 0≤x≤100π0\le x\le100\pi

(iii) tan⁡(x−π6)=π+x\tan\left(x-\dfrac{\pi}{6}\right)=\pi+x in the interval 0≤x≤2π0\le x\le2\pi

(3)

The line with equation x=ax=a, shown in Figure 3, is the asymptote to the curve with the smallest positive xx coordinate.

(b) State the value of aa.

(1)

The line with equation x=bx=b, also shown in Figure 3, is the asymptote to the curve with the second smallest positive xx coordinate.

The line ll meets x=ax=a at point PP and meets x=bx=b at point QQ as shown in Figure 3.

(c) Find the midpoint of the line segment PQPQ.

(4)

解答

(a)

解法一

思路

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解的个数就是直线和正切图像的交点个数。正切图像每隔 π\pi 重复一次；在图中 0≤x≤2π0\le x\le2\pi 内可以直接数交点。

答题过程

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Using the graph:

tan⁡(x−π6)=π−x\begin{align*} \tan\left(x-\frac{\pi}{6}\right)=\pi-x \end{align*}

has 33 solutions in 0≤x≤2π0\le x\le2\pi.

Over 0≤x≤100π0\le x\le100\pi, the tangent curve repeats every π\pi, giving one intersection in each main branch, with the endpoint branch included. Hence there are

101\begin{align*} 101 \end{align*}

solutions.

For

tan⁡(x−π6)=π+x,\begin{align*} \tan\left(x-\frac{\pi}{6}\right)=\pi+x, \end{align*}

the corresponding line has positive gradient. From the graph in 0≤x≤2π0\le x\le2\pi, it intersects the curve

2\begin{align*} 2 \end{align*}

times.

Therefore the answers are

(i) 3,(ii) 101,(iii) 2.\begin{align*} \text{(i) }3,\qquad \text{(ii) }101,\qquad \text{(iii) }2. \end{align*}

(b)

解法一

思路

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tan⁡u\tan u 的渐近线在 u=π2+nπu=\frac\pi2+n\pi。这里 u=x−π6u=x-\frac\pi6，所以令 x−π6=π2x-\frac\pi6=\frac\pi2 得到最小正渐近线。

答题过程

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The vertical asymptotes occur when

x−π6=π2+nπ.\begin{align*} x-\frac{\pi}{6}=\frac{\pi}{2}+n\pi. \end{align*}

For the smallest positive value,

x−π6= π2x= π2+π6x= 2π3.\begin{align*} x-\frac{\pi}{6}=&\,\frac{\pi}{2}\\ x=&\,\frac{\pi}{2}+\frac{\pi}{6}\\ x=&\,\frac{2\pi}{3}. \end{align*}

Therefore

a=2π3.\begin{align*} a=\frac{2\pi}{3}. \end{align*}

(c)

解法一

思路

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相邻正切渐近线相差 π\pi，所以 b=a+πb=a+\pi。点 P,QP,Q 都在直线 y=π−xy=\pi-x 上，分别代入 x=a,bx=a,b 求坐标，再用中点公式。

答题过程

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The second smallest positive asymptote is

b= a+π= 2π3+π= 5π3.\begin{align*} b =&\,a+\pi\\ =&\,\frac{2\pi}{3}+\pi\\ =&\,\frac{5\pi}{3}. \end{align*}

Since PP lies on x=ax=a and y=π−xy=\pi-x,

P= (2π3,π−2π3)= (2π3,π3).\begin{align*} P =&\,\left(\frac{2\pi}{3},\pi-\frac{2\pi}{3}\right)\\ =&\,\left(\frac{2\pi}{3},\frac{\pi}{3}\right). \end{align*}

Since QQ lies on x=bx=b and y=π−xy=\pi-x,

Q= (5π3,π−5π3)= (5π3,−2π3).\begin{align*} Q =&\,\left(\frac{5\pi}{3},\pi-\frac{5\pi}{3}\right)\\ =&\,\left(\frac{5\pi}{3},-\frac{2\pi}{3}\right). \end{align*}

The midpoint of PQPQ is

(2π3+5π32,π3−2π32)= (7π6,−π6).\begin{align*} \left( \frac{\frac{2\pi}{3}+\frac{5\pi}{3}}{2}, \frac{\frac{\pi}{3}-\frac{2\pi}{3}}{2} \right) =&\,\left(\frac{7\pi}{6},-\frac{\pi}{6}\right). \end{align*}