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CIE 9709 2024 June Paper 11 Q2

A Level / CIE / P1

CIE 9709 2024 June Paper 11 Paper · Question 2

题目

Problem

The diagram shows two curves. One curve has equation y=sin⁡xy = \sin x and the other curve has equation y=f(x)y = f(x).

(a) In order to transform the curve y=sin⁡xy = \sin x to the curve y=f(x)y = f(x), the curve y=sin⁡xy = \sin x is first reflected in the xx-axis.

Describe fully a sequence of two further transformations which are required.

[4]

(b) Find f(x)f(x) in terms of sin⁡x\sin x.

[2]
题目中文翻译

图中显示两条曲线。其中一条曲线的方程为 y=sin⁡xy = \sin x,另一条曲线的方程为 y=f(x)y = f(x)。

(a) 为了将曲线 y=sin⁡xy = \sin x 变换为曲线 y=f(x)y = f(x),先将曲线 y=sin⁡xy = \sin x 关于 xx 轴反射。

完整描述接下来还需要的两个变换。

[4]

(b) 用 sin⁡x\sin x 表示 f(x)f(x)。

[2]

解答

(a)

解法一

思路

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反射已经完成。图中 ff 的最高值是 11,最低值是 −5-5,所以振幅是 33,中线是 y=−2y=-2。

答题过程

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After the reflection, stretch by scale factor 33 in the yy-direction. Then translate by (0−2)\begin{pmatrix}0\\-2\end{pmatrix}. The stretch gives amplitude 33, and the translation moves the midline to y=−2y=-2.

解法二

思路

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若先平移再伸缩,原先的竖直位移也会放大三倍。

答题过程

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After the reflection, translate by (0−2/3)\begin{pmatrix}0\\-2/3\end{pmatrix}, then stretch by scale factor 33 in the yy-direction. The resulting height is 3(−sin⁡x−2/3)=−3sin⁡x−23(-\sin x-2/3)=-3\sin x-2.

(b)

解法一

思路

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先把正弦取负,再按 (a) 的伸缩和平移改变高度。

答题过程

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The reflection gives −sin⁡x-\sin x, the stretch gives −3sin⁡x-3\sin x, and the translation subtracts 22. Thus f(x)=−3sin⁡x−2f(x)=-3\sin x-2.