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CIE 9709 2026 June Paper 13 Q8

A Level / CIE / P1

CIE 9709 2026 June Paper 13 Paper · Question 8

题目

Problem

The diagram shows part of the curve with equation y=25x3y=\frac{2}{5x^3}.

The shaded region is bounded by the curve, the xx-axis and the lines x=1x=1 and x=tx=t. It is given that the area of the shaded region is 19\frac{1}{9}.

Find the value of tt.

(5)
题目中文翻译

图中所示的是曲线 y=25x3y=\frac{2}{5x^3} 的一部分。

阴影区域由该曲线、xx 轴及直线 x=1x=1 和 x=tx=t 围成。已知阴影区域的面积为 19\frac{1}{9}。

求 tt 的值。

(5)
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

解法一

思路

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图中 tt 在 11 右侧。用面积条件列定积分,积分后解出 t2t^2。

答题过程

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The area is

 ∫1t25x3 dx= [−15x2]1t= 15−15t2.\begin{align*} &\,\int_1^t\frac{2}{5x^3}\,\mathrm{d}x\\ =&\,\Big[-\frac1{5x^2}\Big]_1^t\\ =&\,\frac15-\frac1{5t^2}. \end{align*}

Equating this to 1/91/9 gives 1/t2=4/91/t^2=4/9, so t2=9/4t^2=9/4. Since t>1t>1, t=3/2t=3/2.