CalcGospel 國際數學圖譜
返回

IAL 2019 Oct Q9

A Level / Edexcel / P1

IAL 2019 Oct Paper · Question 9

题目

Problem

Figure 5 shows a sketch of part of the curve CC with equation

y=sin⁡(x12),\begin{align*} y=\sin\left(\frac{x}{12}\right), \end{align*}

where xx is measured in radians. The point MM shown in Figure 5 is a minimum point on CC.

Figure 5

(a) State the period of CC.

(1)

(b) State the coordinates of MM.

(1)

The smallest positive solution of the equation sin⁡(x12)=k\sin\left(\dfrac{x}{12}\right)=k, where kk is a constant, is α\alpha.

Find, in terms of α\alpha,

(c) (i) the negative solution of the equation sin⁡(x12)=k\sin\left(\dfrac{x}{12}\right)=k that is closest to zero,

(ii) the smallest positive solution of the equation cos⁡(x12)=k\cos\left(\dfrac{x}{12}\right)=k.

(2)

解答

(a)

解法一

思路

展开

sin⁡u\sin u 的周期是 2π2\pi。这里 u=x12u=\frac{x}{12}，所以 xx 的周期要放大 1212 倍。

答题过程

展开 x12 has period 2π,\begin{align*} \frac{x}{12}\text{ has period }2\pi, \end{align*}

so

x has period 24π.\begin{align*} x\text{ has period }24\pi. \end{align*}

(b)

解法一

思路

展开

正弦函数的最小值是 −1-1，第一次在角度 3π2\frac{3\pi}{2} 出现。令 x12=3π2\frac{x}{12}=\frac{3\pi}{2}。

答题过程

展开

At the first minimum,

x12=3π2.\begin{align*} \frac{x}{12}=\frac{3\pi}{2}. \end{align*}

Thus

x=18π.\begin{align*} x=18\pi. \end{align*}

Therefore

M=(18π,−1).\begin{align*} M=(18\pi,-1). \end{align*}

(c)

解法一

思路

展开

同一个正弦值在一个周期内有对称解。若最小正解是 α\alpha，那么离 00 最近的负解要往左跨半个周期再多 α\alpha。余弦图像相当于正弦图像向左平移四分之一个周期。

答题过程

展开

The period is 24π24\pi, so half a period is 12π12\pi and a quarter period is 6π6\pi.

The negative solution closest to zero is

−12π−α.\begin{align*} -12\pi-\alpha. \end{align*}

For the cosine equation, the smallest positive solution is

6π−α.\begin{align*} 6\pi-\alpha. \end{align*}