题目

Problem

(a) Find the quotient and remainder when $x^2$ is divided by $1+4x^2$ .

(2)

(b) Find the exact value of $\displaystyle\int_0^{0.5}x\tan^{-1}(2x)\,\mathrm{d}x$ .

(6)

题目中文翻译

(a) 求 $x^2$ 除以 $1+4x^2$ 所得的商和余数。

(2)

(b) 求 $\displaystyle\int_0^{0.5}x\tan^{-1}(2x)\,\mathrm{d}x$ 的精确值。

(6)

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

展开

直接比较最高次项，再确定常数余式。

答题过程

展开

Since

x^2=\frac14(1+4x^2)-\frac14,

the quotient is $\boxed{\frac14}$ and the remainder is $\boxed{-\frac14}$ .

(b)

解法一

思路

展开

分部积分使反正切被求导，余下分式正好使用(a)的除法结果。

答题过程

展开

For integration by parts, take $u=\tan^{-1}(2x)$ and $\mathrm{d}v=x\,\mathrm{d}x$ . Then

\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{2}{1+4x^2}, \qquad v=\frac{x^2}{2}.

Hence

\begin{align*} I=&\,\bigg[\frac{x^2}{2}\tan^{-1}(2x)\bigg]_0^{1/2}\\[4mm] &\,\hspace{2pt}-\int_0^{1/2}\frac{x^2}{1+4x^2}\,\mathrm{d}x. \end{align*}

By (a),

\frac{x^2}{1+4x^2}=\frac14-\frac1{4(1+4x^2)}.

Therefore

I=\bigg[\frac{x^2}{2}\tan^{-1}(2x) -\frac x4+\frac18\tan^{-1}(2x)\bigg]_0^{1/2}.

The lower-limit contribution is zero. At the upper limit, $\tan^{-1}(1)=\pi/4$ , so

\begin{align*} I=&\,\frac18\frac\pi4-\frac18+\frac18\frac\pi4\\[4mm] =&\,\boxed{\frac\pi{16}-\frac18}. \end{align*}

CIE 9709 2025 June Paper 32 Q10

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程