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CIE 9709 2024 November Paper 32 Q11

A Level / CIE / P3

CIE 9709 2024 Nov Paper 32 Paper · Question 11

题目

Problem

Let f(x)=2e2xe2x−3ex+2\mathrm{f}(x)=\dfrac{2\mathrm{e}^{2x}}{\mathrm{e}^{2x}-3\mathrm{e}^x+2}.

(a) Find f′(x)\mathrm{f}'(x) and hence find the exact coordinates of the stationary point of the curve with equation y=f(x)y=\mathrm{f}(x).

(5)

(b) Use the substitution u=exu=\mathrm{e}^x and partial fractions to find the exact value of ∫ln⁡3ln⁡5f(x) dx\displaystyle\int_{\ln3}^{\ln5}\mathrm{f}(x)\,\mathrm{d}x.

Give your answer in the form ln⁡a\ln a, where aa is a rational number in its simplest form.

(9)
题目中文翻译

设 f(x)=2e2xe2x−3ex+2\mathrm{f}(x)=\dfrac{2\mathrm{e}^{2x}}{\mathrm{e}^{2x}-3\mathrm{e}^x+2}。

(a) 求 f′(x)\mathrm{f}'(x),并由此求方程为 y=f(x)y=\mathrm{f}(x) 的曲线的驻点的精确坐标。

(5)

(b) 使用代换 u=exu=\mathrm{e}^x 和部分分式,求 ∫ln⁡3ln⁡5f(x) dx\displaystyle\int_{\ln3}^{\ln5}\mathrm{f}(x)\,\mathrm{d}x 的精确值。

答案以 ln⁡a\ln a 的形式表示,其中 aa 为最简形式的有理数。

(9)
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

用商法则求导,分子约去相消项后只需求一个指数值。原分母为零的两点要排除。

答题过程

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The domain excludes ex=1,2e^x=1,2, so x≠0,ln⁡2x\ne0,\ln2.

Write D=e2x−3ex+2D=e^{2x}-3e^x+2, so D′=2e2x−3exD'=2e^{2x}-3e^x. The quotient rule gives

f′(x)=4e2xD−2e2xD′D2.f'(x)=\frac{4e^{2x}D-2e^{2x}D'}{D^2}.

The numerator simplifies as follows:

4e2xD−2e2xD′= 2e2x(2D−D′)= 2e2x(4−3ex).\begin{align*} 4e^{2x}D-2e^{2x}D' =&\,2e^{2x}(2D-D')\\[4mm] =&\,2e^{2x}(4-3e^x). \end{align*}

Hence

f′(x)=2e2x(4−3ex)(e2x−3ex+2)2.\boxed{f'(x)= \frac{2e^{2x}(4-3e^x)} {(e^{2x}-3e^x+2)^2}}.

Since e2x>0e^{2x}>0, the stationary point requires 4−3ex=04-3e^x=0, giving x=ln⁡(4/3)x=\ln(4/3).

At this value,

f(x)=2(16/9)16/9−4+2=−16.f(x)=\frac{2(16/9)}{16/9-4+2} =-16.

Thus the exact stationary point is

(ln⁡43,−16).\boxed{\bigg(\ln\frac43,-16\bigg)}.

解法二

思路

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也可先把函数按 exe^x 做部分分式,再分别求导,最后通分求驻点。

答题过程

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Let u=exu=e^x. Polynomial division and partial fractions give

2u2(u−1)(u−2)=2+8u−2−2u−1.\frac{2u^2}{(u-1)(u-2)} =2+\frac8{u-2}-\frac2{u-1}.

Indeed, the remaining numerator 6u−46u-4 equals 8(u−1)−2(u−2)8(u-1)-2(u-2).

Thus

f(x)=2+8ex−2−2ex−1.f(x)=2+\frac8{e^x-2}-\frac2{e^x-1}.

Differentiating,

f′(x)=−8ex(ex−2)2+2ex(ex−1)2.\boxed{f'(x)=-\frac{8e^x}{(e^x-2)^2} +\frac{2e^x}{(e^x-1)^2}}.

At a stationary point, divide by 2ex>02e^x>0:

1(u−1)2=4(u−2)2.\frac1{(u-1)^2}=\frac4{(u-2)^2}.

Cross-multiplication gives

(u−2)2= 4(u−1)2,u2−4u+4= 4u2−8u+4,3u2−4u= 0.\begin{align*} (u-2)^2=&\,4(u-1)^2,\\[4mm] u^2-4u+4=&\,4u^2-8u+4,\\[4mm] 3u^2-4u=&\,0. \end{align*}

Since u>0u>0, u=4/3u=4/3. Substituting in ff gives y=−16y=-16, so the stationary point is

(ln⁡43,−16).\boxed{\bigg(\ln\frac43,-16\bigg)}.

(b)

解法一

思路

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按题意换元时,微分带来一个 uu 的分母,积分中的分子由 2u22u^2 变成 2u2u。然后拆分两项简单对数。

答题过程

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Let u=exu=e^x, so du=u dx\mathrm{d}u=u\,\mathrm{d}x. The limits become u=3u=3 and u=5u=5.

Hence

I=∫352uu2−3u+2 du.I=\int_3^5\frac{2u}{u^2-3u+2}\,\mathrm{d}u.

Write

2u(u−1)(u−2)=Au−1+Bu−2.\frac{2u}{(u-1)(u-2)} =\frac A{u-1}+\frac B{u-2}.

Then 2u=A(u−2)+B(u−1)2u=A(u-2)+B(u-1). Putting u=1u=1 gives A=−2A=-2, and putting u=2u=2 gives B=4B=4.

On 3⩽u⩽53\leqslant u\leqslant5, both u−1u-1 and u−2u-2 are positive. Therefore

I= ∫35(−2u−1+4u−2) du= [−2ln⁡(u−1)+4ln⁡(u−2)]35= −2ln⁡4+4ln⁡3+2ln⁡2= 4ln⁡3−2ln⁡2= ln⁡814.\begin{align*} I=&\,\int_3^5\bigg(-\frac2{u-1}+\frac4{u-2}\bigg)\,\mathrm{d}u\\ =&\,\big[-2\ln(u-1)+4\ln(u-2)\big]_3^5\\ =&\,-2\ln4+4\ln3+2\ln2\\ =&\,4\ln3-2\ln2\\ =&\,\boxed{\ln\frac{81}{4}}. \end{align*}