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CIE 9231 2025 June Paper 24 Q8

A Level / CIE / FP2

CIE 9231 2025 June Paper 24 Paper · Question 8

题目

Problem

The curve CC has equation y=tanhxy = \tanh x for x0x \ge 0.

(a) Sketch CC and state the equation of the asymptote.

[2]

(b) By considering a suitable set of NN rectangles of unit width, use your sketch to show that

r=1Ntanhr>ln(coshN).\sum_{r = 1}^{N} \tanh r > \ln(\cosh N).
[3]

(c) The arc of CC joining the point where x=0x = 0 to the point where x=12ln3x = \frac{1}{2}\ln 3 is rotated through one complete revolution about the xx-axis. The area of the surface generated is denoted by SS.

(i) Use the substitution u=1+sech4xu = \sqrt{1 + \operatorname{sech}^4 x} to show that

S=π542u2u21du.S = \pi\int_{\frac{5}{4}}^{\sqrt{2}} \frac{u^2}{u^2 - 1}\,\mathrm{d}u.
[7]

(ii) Find the exact value of

π542u2u21du.\pi\int_{\frac{5}{4}}^{\sqrt{2}} \frac{u^2}{u^2 - 1}\,\mathrm{d}u.

You need not simplify your answer.

[3]
题目中文翻译

曲线 CC 的方程为 y=tanhxy = \tanh x,其中 x0x \ge 0

(a) 绘制 CC 的草图,并写出渐近线的方程。

(b) 考虑一组合适的、宽度为 1 的 NN 个矩形,利用你的草图证明

r=1Ntanhr>ln(coshN)\sum_{r = 1}^{N} \tanh r > \ln(\cosh N)

(c) 将 CC 上从 x=0x = 0 的点到 x=12ln3x = \frac{1}{2}\ln 3 的点之间的弧段绕 xx 轴旋转一周。所得旋转曲面的面积记为 SS

(i) 使用代换 u=1+sech4xu = \sqrt{1 + \operatorname{sech}^4 x} 证明

S=π542u2u21duS = \pi\int_{\frac{5}{4}}^{\sqrt{2}} \frac{u^2}{u^2 - 1}\,\mathrm{d}u

(ii) 求

π542u2u21du\pi\int_{\frac{5}{4}}^{\sqrt{2}} \frac{u^2}{u^2 - 1}\,\mathrm{d}u

的精确值。答案不必化简。

解答