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IAL 2024 Jan Q9

A Level / Edexcel / P2

IAL 2024 Jan Paper · Question 9

题目

Problem

In this question you must show detailed reasoning.

Solutions relying entirely on calculator technology are not acceptable.

(i) Solve, for 0≤x<360∘0\leq x<360^\circ, the equation

sin⁡xtan⁡x=5\sin x\tan x=5

giving your answers to one decimal place.

(6)

(ii)

Figure 1

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

y=Asin⁡(2θ−3π8)+2y=A\sin\left(2\theta-\frac{3\pi}{8}\right)+2

where AA is a constant and θ\theta is measured in radians.

The points PP, QQ and RR lie on the curve and are shown in Figure 1.

Given that the yy coordinate of PP is 77

(a) state the value of AA,

(1)

(b) find the exact coordinates of QQ,

(3)

(c) find the value of θ\theta at RR, giving your answer to 33 significant figures.

(4)

解答

(i)

解法一

思路

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把 tan⁡x\tan x 换成 sin⁡xcos⁡x\frac{\sin x}{\cos x}，再用 sin⁡2x=1−cos⁡2x\sin^2x=1-\cos^2x，就能得到关于 cos⁡x\cos x 的二次方程。

答题过程

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Since

tan⁡x=sin⁡xcos⁡x,\begin{align*} \tan x=\frac{\sin x}{\cos x}, \end{align*}

the equation becomes

sin⁡x⋅sin⁡xcos⁡x= 5sin⁡2xcos⁡x= 5sin⁡2x= 5cos⁡x.\begin{align*} \sin x\cdot\frac{\sin x}{\cos x}=&\,5\\ \frac{\sin^2x}{\cos x}=&\,5\\ \sin^2x=&\,5\cos x. \end{align*}

Using sin⁡2x=1−cos⁡2x\sin^2x=1-\cos^2x,

1−cos⁡2x= 5cos⁡xcos⁡2x+5cos⁡x−1= 0.\begin{align*} 1-\cos^2x=&\,5\cos x\\ \cos^2x+5\cos x-1=&\,0. \end{align*}

Solve this quadratic in cos⁡x\cos x:

cos⁡x= −5±52−4(1)(−1)2= −5±292.\begin{align*} \cos x=&\,\frac{-5\pm\sqrt{5^2-4(1)(-1)}}{2}\\ =&\,\frac{-5\pm\sqrt{29}}{2}. \end{align*}

The value −5−292\frac{-5-\sqrt{29}}{2} is less than −1-1, so it is not possible for cos⁡x\cos x.

Thus

cos⁡x=−5+292.\begin{align*} \cos x=\frac{-5+\sqrt{29}}{2}. \end{align*}

For 0≤x<360∘0\leq x<360^\circ,

x=78.9∘, 281.1∘.\begin{align*} x=78.9^\circ,\ 281.1^\circ. \end{align*}

(ii)(a)

解法一

思路

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曲线的中线是 y=2y=2。最高点 PP 的 yy 坐标是 77，所以振幅是 7−2=57-2=5。

答题过程

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The midline is

y=2.\begin{align*} y=2. \end{align*}

Since the maximum value is 77,

A=7−2=5.\begin{align*} A=7-2=5. \end{align*}

(ii)(b)

解法一

思路

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QQ 是低点，所以 sin⁡\sin 部分等于 −1-1。低点的 yy 坐标是 2−5=−32-5=-3。

答题过程

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At QQ, the sine term is at its minimum:

sin⁡(2θ−3π8)=−1.\begin{align*} \sin\left(2\theta-\frac{3\pi}{8}\right)=-1. \end{align*}

So

2θ−3π8=3π2.\begin{align*} 2\theta-\frac{3\pi}{8}=\frac{3\pi}{2}. \end{align*}

Hence

2θ= 3π2+3π8= 12π8+3π8= 15π8θ= 15π16.\begin{align*} 2\theta=&\,\frac{3\pi}{2}+\frac{3\pi}{8}\\ =&\,\frac{12\pi}{8}+\frac{3\pi}{8}\\ =&\,\frac{15\pi}{8}\\ \theta=&\,\frac{15\pi}{16}. \end{align*}

The yy coordinate is

2−5=−3.\begin{align*} 2-5=-3. \end{align*}

Therefore,

Q(15π16,−3).\begin{align*} Q\left(\frac{15\pi}{16},-3\right). \end{align*}

(ii)(c)

解法一

思路

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RR 在 xx-axis 上，所以令 y=0y=0。从图像看，RR 是右边那个截距，所以最后要选对应的较大 θ\theta。

答题过程

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At RR, y=0y=0. Since A=5A=5,

5sin⁡(2θ−3π8)+2= 0sin⁡(2θ−3π8)= −25.\begin{align*} 5\sin\left(2\theta-\frac{3\pi}{8}\right)+2=&\,0\\ \sin\left(2\theta-\frac{3\pi}{8}\right)=&\,-\frac25. \end{align*}

From the position of RR on the diagram, take the solution giving the shown intercept:

2θ−3π8=9.8361…\begin{align*} 2\theta-\frac{3\pi}{8}=9.8361\ldots \end{align*}

So

2θ= 9.8361…+3π8= 11.0142…θ= 5.5071…\begin{align*} 2\theta=&\,9.8361\ldots+\frac{3\pi}{8}\\ =&\,11.0142\ldots\\ \theta=&\,5.5071\ldots \end{align*}

Therefore, to 33 significant figures,

θ=5.51.\begin{align*} \theta=5.51. \end{align*}