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IAL 2025 June FP3 Q1

A Level / Edexcel / FP3

IAL 2025 June Paper · Question 1

题目

Problem

(a) Use the definition of coshx\cosh x in terms of exponentials to show that

2cosh5xcoshx=cosh6x+cosh4x2\cosh 5x \cosh x = \cosh 6x + \cosh 4x

(b) Hence determine the exact values of xx for which

cosh6x+cosh4x=8coshx\cosh 6x + \cosh 4x = 8\cosh x

giving your answers in terms of natural logarithms in simplest form.

(2)
(4)
题目中文翻译

(a) 使用 coshx\cosh x 关于指数的定义证明

2cosh5xcoshx=cosh6x+cosh4x2\cosh 5x \cosh x = \cosh 6x + \cosh 4x

(b) 由此求出使

cosh6x+cosh4x=8coshx\cosh 6x + \cosh 4x = 8\cosh x

成立的 xx 的精确值。

答案用最简自然对数形式表示。

解答

(a)

解法一

思路

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按题意把两个双曲余弦都写成指数形式。乘开后,四个指数项自然配成 e6x+e6xe^{6x}+e^{-6x}e4x+e4xe^{4x}+e^{-4x},再分别还原成双曲余弦。

答题过程

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Using

coshu=eu+eu2,\cosh u=\frac{e^u+e^{-u}}{2},

we have

2cosh5xcoshx=2(e5x+e5x2)×(ex+ex2)=12(e6x+e4x+e4x+e6x)=e6x+e6x2+e4x+e4x2=cosh6x+cosh4x,\begin{align*} 2\cosh5x\cosh x =&\,2\bigg(\frac{e^{5x}+e^{-5x}}{2}\bigg)\\ &\,\hspace{2pt}\times \bigg(\frac{e^x+e^{-x}}{2}\bigg)\\ =&\,\frac12\big( e^{6x}+e^{4x}+e^{-4x}+e^{-6x}\big)\\ =&\,\frac{e^{6x}+e^{-6x}}{2} +\frac{e^{4x}+e^{-4x}}{2}\\ =&\,\cosh6x+\cosh4x, \end{align*}

as required.

(b)

解法一

思路

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“Hence” 要先用 (a) 把左边改写成 2cosh5xcoshx2\cosh5x\cosh x。由于实数范围内 coshx>0\cosh x>0,可安全约去 coshx\cosh x,再把 cosh5x=4\cosh5x=4 写成关于 e5xe^{5x} 的二次方程求解。

答题过程

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Using the result from part (a),

2cosh5xcoshx=8coshx.2\cosh5x\cosh x=8\cosh x.

Since coshx>0\cosh x>0 for all real xx, division by 2coshx2\cosh x gives

cosh5x=4.\cosh5x=4.

Using the exponential definition,

e5x+e5x2=4.\frac{e^{5x}+e^{-5x}}{2}=4.

Multiplying by 2e5x2e^{5x},

e10x8e5x+1=0.e^{10x}-8e^{5x}+1=0.

Let u=e5xu=e^{5x}, where u>0u>0. Then

u28u+1=0,u=4±15.\begin{align*} u^2-8u+1=&\,0,\\ u=&\,4\pm\sqrt{15}. \end{align*}

Both roots are positive, so

5x=ln(4±15).5x=\ln\big(4\pm\sqrt{15}\big).

Therefore,

x=15ln(4±15).\boxed{x=\frac15\ln\big(4\pm\sqrt{15}\big)}.

Equivalently, since

(4+15)(415)=1,\big(4+\sqrt{15}\big)\big(4-\sqrt{15}\big)=1,

the answers may be written as

x=±15ln(4+15).\boxed{x=\pm\frac15\ln\big(4+\sqrt{15}\big)}.

解法二

思路

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官方评分资料也接受使用反双曲余弦。由 cosh5x=4\cosh5x=4,先利用双曲余弦的偶函数性质写出两个实数分支,再用 arcoshu=ln(u+u21)\operatorname{arcosh}u=\ln(u+\sqrt{u^2-1}) 化成题目要求的自然对数形式。

答题过程

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From part (a),

cosh5x=4.\cosh5x=4.

Hence

5x=±arcosh4.5x=\pm\operatorname{arcosh}4.

Using

arcoshu=ln(u+u21),\operatorname{arcosh}u =\ln\big(u+\sqrt{u^2-1}\big),

we obtain

5x=±ln(4+421)=±ln(4+15).\begin{align*} 5x =&\,\pm\ln\big(4+\sqrt{4^2-1}\big)\\ =&\,\pm\ln\big(4+\sqrt{15}\big). \end{align*}

Therefore,

x=±15ln(4+15).\boxed{x=\pm\frac15\ln\big(4+\sqrt{15}\big)}.