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IAL 2023 Oct Q10

A Level / Edexcel / P1

IAL 2023 Oct Paper · Question 10

题目

Problem

Figure 4 shows a sketch of part of the curve C1C_1 with equation

y=3cos⁡(xn),x≥0,\begin{align*} y=3\cos\left(\frac{x}{n}\right),\qquad x\ge0, \end{align*}

Figure 4

where nn is a constant.

The curve C1C_1 cuts the positive xx-axis for the first time at point P(270,0)P(270,0), as shown in Figure 4.

(a) (i) State the value of nn

(ii) State the period of C1C_1

(2)

The point QQ, shown in Figure 4, is a minimum point of C1C_1.

(b) State the coordinates of QQ.

(2)

The curve C2C_2 has equation y=2sin⁡x+ky=2\sin x+k, where kk is a constant.

The point R(a,125)R\left(a,\dfrac{12}{5}\right) and the point S(−a,−35)S\left(-a,-\dfrac35\right) both lie on C2C_2.

Given that aa is a constant less than 9090,

(c) find the value of kk.

(2)

解答

(a)

解法一

思路

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cos⁡u\cos u 第一次等于 00 时是 u=90∘u=90^\circ。题目说第一次正 xx 截距是 x=270x=270，所以令 xn=90∘\frac{x}{n}=90^\circ。

答题过程

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For

y=3cos⁡(xn),\begin{align*} y=3\cos\left(\frac{x}{n}\right), \end{align*}

the first positive xx-intercept occurs when

xn=90∘.\begin{align*} \frac{x}{n}=90^\circ. \end{align*}

Since this happens at x=270x=270,

270n= 90n= 3.\begin{align*} \frac{270}{n}=&\,90\\ n=&\,3. \end{align*}

The period is

360n=360(3)=1080.\begin{align*} 360n=360(3)=1080. \end{align*}

Therefore

n=3,period=1080.\begin{align*} n=3,\qquad \text{period}=1080. \end{align*}

(b)

解法一

思路

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当 n=3n=3 时，曲线是 y=3cos⁡(x3)y=3\cos\left(\frac{x}{3}\right)。最小值为 −3-3。图中标出的 QQ 是后面的那个最低点，对应 x=1620x=1620。

答题过程

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The minimum value of 3cos⁡(x3)3\cos\left(\frac{x}{3}\right) is −3-3.

The minimum point shown on the graph is at

x=1620.\begin{align*} x=1620. \end{align*}

Therefore

Q=(1620,−3).\begin{align*} Q=(1620,-3). \end{align*}

(c)

解法一

思路

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两个点都在 y=2sin⁡x+ky=2\sin x+k 上。利用 sin⁡(−a)=−sin⁡a\sin(-a)=-\sin a，把两个点分别代入后相加，可以直接消去 sin⁡a\sin a，求出 kk。

答题过程

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Since R(a,125)R\left(a,\frac{12}{5}\right) lies on C2C_2,

2sin⁡a+k=125.\begin{align*} 2\sin a+k=\frac{12}{5}. \end{align*}

Since S(−a,−35)S\left(-a,-\frac35\right) lies on C2C_2,

2sin⁡(−a)+k=−35.\begin{align*} 2\sin(-a)+k=-\frac35. \end{align*}

Using sin⁡(−a)=−sin⁡a\sin(-a)=-\sin a,

−2sin⁡a+k=−35.\begin{align*} -2\sin a+k=-\frac35. \end{align*}

Add the two equations:

(2sin⁡a+k)+(−2sin⁡a+k)= 125−352k= 95k= 910.\begin{align*} (2\sin a+k)+(-2\sin a+k) =&\,\frac{12}{5}-\frac35\\ 2k=&\,\frac95\\ k=&\,\frac9{10}. \end{align*}