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CIE 9231 2025 Nov Paper 24 Q2

A Level / CIE / FP2

CIE 9231 2025 Nov Paper 24 Paper · Question 2

题目

Problem

(a) Find the exact values of xx for which cosh⁡2x=6sinh⁡2x\cosh 2x = 6\sinh^2 x, giving your answers in logarithmic form.

[4]

(b) Sketch the curves C1:y=cosh⁡2xC_1: y = \cosh 2x and C2:y=6sinh⁡2xC_2: y = 6\sinh^2 x on the same diagram.

[2]
题目中文翻译

(a) 求满足 cosh⁡2x=6sinh⁡2x\cosh 2x = 6\sinh^2 x 的 xx 的精确值,并用对数形式给出答案。

(4)

(b) 在同一图中画出曲线 C1:y=cosh⁡2xC_1: y = \cosh 2x 和 C2:y=6sinh⁡2xC_2: y = 6\sinh^2 x。

(2)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

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用双曲余弦的倍角公式,把等式化为只含 sinh⁡x\sinh x 的方程。保留平方根的两个符号,并把反双曲正弦写成对数。

答题过程

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Using cosh⁡2x=1+2sinh⁡2x\cosh 2x=1+2\sinh^2 x gives 4sinh⁡2x=14\sinh^2 x=1. Thus sinh⁡x=±12\sinh x=\pm\frac12 and

x= arsinh⁡(±12)= ln⁡(±12+1+14)= ln⁡(5±12).\begin{align*} x=&\,\operatorname{arsinh}\bigg(\pm\frac12\bigg)\\ =&\,\ln\bigg(\pm\frac12+\sqrt{1+\frac14}\bigg)\\ =&\,\boxed{\ln\bigg(\frac{\sqrt5\pm1}{2}\bigg)}. \end{align*}

解法二

思路

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也可把双曲函数全写成指数式。令 u=e2xu=e^{2x},所得方程是一个二次方程。

答题过程

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The equation becomes

e2x+e−2x2=32(e2x−2+e−2x).\frac{e^{2x}+e^{-2x}}2 =\frac32(e^{2x}-2+e^{-2x}).

Hence e2x+e−2x=3e^{2x}+e^{-2x}=3. With u=e2x>0u=e^{2x}>0, we obtain u2−3u+1=0u^2-3u+1=0, so

e2x=3±52=(5±12)2.e^{2x}=\frac{3\pm\sqrt5}{2} =\bigg(\frac{\sqrt5\pm1}{2}\bigg)^2.

Taking logarithms gives x=ln⁡(5±12)x=\ln\big(\frac{\sqrt5\pm1}{2}\big).

(b)

解法一

思路

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两条曲线都是偶函数,先标出各自在 yy 轴上的最低点。再用 (a) 标出两个交点,并检查两条曲线在交点内外的高低关系。

答题过程

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Both curves are symmetric about the yy-axis. The minimum of C1C_1 is (0,1)(0,1) and the minimum of C2C_2 is (0,0)(0,0). Let h=ln⁡(5+12)h=\ln\big(\frac{\sqrt5+1}{2}\big). The curves meet at (−h,32)\big(-h,\frac32\big) and (h,32)\big(h,\frac32\big). Since C2−C1C_2-C_1 in height is 4sinh⁡2x−14\sinh^2x-1, C2C_2 is lower for ∣x∣<h|x|<h and higher for ∣x∣>h|x|>h.