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CIE 9709 2023 June Paper 12 Q8

A Level / CIE / P1

CIE 9709 2023 June Paper 12 Paper · Question 8

题目

Problem

The diagram shows the graph of y=f(x)y = f(x) where the function ff is defined by

f(x)=3+2sin⁡14xfor 0≤x≤2π.f(x) = 3 + 2\sin\frac14 x \quad \text{for } 0 \leq x \leq 2\pi.

(a) On the diagram above, sketch the graph of y=f−1(x)y = f^{-1}(x).

[2]

(b) Find an expression for f−1(x)f^{-1}(x).

[2]

(c)

The diagram above shows part of the graph of the function g(x)=3+2sin⁡14xg(x) = 3 + 2\sin\frac14 x for −2π≤x≤2π-2\pi \leq x \leq 2\pi.

Complete the sketch of the graph of g(x)g(x) on the diagram above and hence explain whether the function gg has an inverse.

[2]

(d) Describe fully a sequence of three transformations which can be combined to transform the graph of y=sin⁡xy = \sin x for 0≤x≤12π0 \leq x \leq \frac12\pi to the graph of y=f(x)y = f(x), making clear the order in which the transformations are applied.

[6]
题目中文翻译

图中显示 y=f(x)y = f(x) 的图像,其中函数 ff 定义为

f(x)=3+2sin⁡14x其中 0≤x≤2π.f(x) = 3 + 2\sin\frac14 x \quad \text{其中 } 0 \leq x \leq 2\pi.

(a) 在上图中,画出 y=f−1(x)y = f^{-1}(x) 的图像。

[2]

(b) 求 f−1(x)f^{-1}(x) 的表达式。

[2]

(c)

上图显示函数 g(x)=3+2sin⁡14xg(x) = 3 + 2\sin\frac14 x 在 −2π≤x≤2π-2\pi \leq x \leq 2\pi 上的部分图像。

在上图中补全 g(x)g(x) 的图像,并由此说明函数 gg 是否有反函数。

[2]

(d) 完整描述一组三个变换,它们可以组合起来把 0≤x≤12π0 \leq x \leq \frac12\pi 上的 y=sin⁡xy = \sin x 图像变换为 y=f(x)y = f(x) 的图像,并清楚说明这些变换的应用顺序。

[6]

解答

(a)

解法一

思路

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将原曲线关于 y=xy=x 反射。原图的两个端点交换横纵坐标后,给出反函数曲线的起点和终点。

答题过程

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Sketch the reflection of y=f(x)y=f(x) in the line y=xy=x. The endpoint (0,3)(0,3) maps to (3,0)(3,0), and (2π,5)(2\pi,5) maps to (5,2π)(5,2\pi). The reflected branch increases and becomes steeper as it approaches the upper endpoint.

(b)

解法一

思路

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反解时先分离正弦。原来的角 x/4x/4 位于反正弦的主值区间内。

答题过程

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From y=3+2sin⁡(x/4)y=3+2\sin(x/4), we obtain sin⁡(x/4)=(y−3)/2\sin(x/4)=(y-3)/2. Since 0≤x/4≤π/20\le x/4\le\pi/2,

f−1(x)=4sin⁡−1(x−32),3≤x≤5.f^{-1}(x)=4\sin^{-1}\Big(\frac{x-3}{2}\Big),\qquad3\le x\le5.

(c)

解法一

思路

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把图像向左延伸到 −2π-2\pi。减去中线高度 33 后,剩下的是奇函数,因此左右两段关于 (0,3)(0,3) 中心对称。

答题过程

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The missing portion runs from (−2π,1)(-2\pi,1) to (0,3)(0,3), joining the given portion smoothly. Since x/4x/4 ranges from −π/2-\pi/2 to π/2\pi/2, the sine function is strictly increasing throughout this interval. Thus gg is one-to-one and has an inverse.

(d)

解法一

思路

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横向伸缩把定义域的长度变成四倍,纵向伸缩把振幅变成两倍。最后把中线抬高 33。

答题过程

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Start with y=sin⁡xy=\sin x on [0,π/2][0,\pi/2]. Apply these transformations in order:

  1. Stretch by scale factor 44 in the xx-direction, giving y=sin⁡(x/4)y=\sin(x/4) on [0,2π][0,2\pi].
  2. Stretch by scale factor 22 in the yy-direction, giving y=2sin⁡(x/4)y=2\sin(x/4).
  3. Translate by (03)\begin{pmatrix}0\\3\end{pmatrix}, giving y=3+2sin⁡(x/4)y=3+2\sin(x/4).

解法二

思路

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若先平移再纵向伸缩,平移量也会被放大。因此先上移 3/23/2,再把纵坐标变为两倍。

答题过程

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Starting with y=sin⁡xy=\sin x on [0,π/2][0,\pi/2], apply these transformations in order:

  1. Stretch by scale factor 44 in the xx-direction, giving y=sin⁡(x/4)y=\sin(x/4) on [0,2π][0,2\pi].
  2. Translate by (03/2)\begin{pmatrix}0\\3/2\end{pmatrix}, giving y=sin⁡(x/4)+3/2y=\sin(x/4)+3/2.
  3. Stretch by scale factor 22 in the yy-direction, giving y=2(sin⁡(x/4)+3/2)=3+2sin⁡(x/4)y=2\bigl(\sin(x/4)+3/2\bigr)=3+2\sin(x/4).