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CIE 9709 2026 June Paper 32 Q6

A Level / CIE / P3

CIE 9709 2026 June Paper 32 Paper · Question 6

题目

Problem

The equation of a curve is y=xtan⁡−13xy=x\tan^{-1}3x for −π≤x≤π-\pi\le x\le\pi.

(a) Find the exact value of the gradient of the curve when x=13x=\frac13.

(4)

(b) Find the exact value of ∫013y dx\displaystyle\int_0^{\frac13} y\,\mathrm{d}x.

(6)
题目中文翻译

一条曲线的方程为 y=xtan⁡−13xy=x\tan^{-1}3x,其中 −π≤x≤π-\pi\le x\le\pi。

(a) 求当 x=13x=\frac13 时曲线斜率的精确值。

(4)

(b) 求 ∫013y dx\displaystyle\int_0^{\frac13} y\,\mathrm{d}x 的精确值。

(6)
原卷通用作答要求

Answer all questions. Write your answers on the answer booklet. Follow the instructions on the front cover of the answer booklet. If you need additional answer paper, ask the invigilator for a continuation booklet.

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。在答题册上作答,并遵守答题册封面的指示。如需额外答题纸,请向监考人员索取续答册。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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使用乘积法则与反正切的链式求导。最后代入 x=1/3x=1/3。

答题过程

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Differentiating gives

dydx=arctan⁡(3x)+3x1+9x2.\frac{\mathrm{d}y}{\mathrm{d}x} =\arctan(3x)+\frac{3x}{1+9x^2}.

At x=1/3x=1/3, the gradient is π/4+1/2\pi/4+1/2.

(b)

解法一

思路

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以反正切为需要求导的部分,使用分部积分。剩余有理式先作除法,再积分。

答题过程

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Set u=arctan⁡(3x)u=\arctan(3x) and dv=x dx\mathrm{d}v=x\,\mathrm{d}x, so du=3/(1+9x2) dx\mathrm{d}u=3/(1+9x^2)\,\mathrm{d}x and v=x2/2v=x^2/2. Integration by parts gives

I= ∫01/3xarctan⁡(3x) dx= [x22arctan⁡(3x)]01/3 −32∫01/3x21+9x2 dx.\begin{align*} I=&\,\int_0^{1/3}x\arctan(3x)\,\mathrm{d}x\\[4mm] =&\,\bigg[\frac{x^2}{2}\arctan(3x)\bigg]_0^{1/3}\\[4mm] &\,\hspace{2pt}-\frac32\int_0^{1/3}\frac{x^2}{1+9x^2}\,\mathrm{d}x. \end{align*}

The remaining fraction is

x21+9x2=19−19(1+9x2).\frac{x^2}{1+9x^2}=\frac19-\frac1{9(1+9x^2)}.

Using ∫(1+9x2)−1 dx=13arctan⁡(3x)+C\int (1+9x^2)^{-1}\,\mathrm{d}x=\frac13\arctan(3x)+C, we obtain

I=[x22arctan⁡(3x)−x6+118arctan⁡(3x)]01/3.I=\bigg[\frac{x^2}{2}\arctan(3x)-\frac x6+\frac1{18}\arctan(3x)\bigg]_0^{1/3}.

Evaluating at 1/31/3 and 00 gives

π72−118+π72=π36−118.\frac{\pi}{72}-\frac1{18}+\frac{\pi}{72} =\frac{\pi}{36}-\frac1{18}.