题目
Problem
Let f(x)=e2x−3ex+22e2x.
(a) Find f′(x) and hence find the exact coordinates of the stationary point of the curve with equation y=f(x).
(5)
(b) Use the substitution u=ex and partial fractions to find the exact value of ∫ln3ln5f(x)dx.
Give your answer in the form lna, where a is a rational number in its simplest form.
(9)
题目中文翻译
设 f(x)=e2x−3ex+22e2x。
(a) 求 f′(x),并由此求方程为 y=f(x) 的曲线的驻点的精确坐标。
(5)
(b) 使用代换 u=ex 和部分分式,求 ∫ln3ln5f(x)dx 的精确值。
答案以 lna 的形式表示,其中 a 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
展开
The domain excludes ex=1,2, so x=0,ln2.
Write D=e2x−3ex+2, so D′=2e2x−3ex. The quotient rule gives
f′(x)=D24e2xD−2e2xD′.
The numerator simplifies as follows:
4e2xD−2e2xD′==2e2x(2D−D′)2e2x(4−3ex).
Hence
f′(x)=(e2x−3ex+2)22e2x(4−3ex).
Since e2x>0, the stationary point requires 4−3ex=0, giving x=ln(4/3).
At this value,
f(x)=16/9−4+22(16/9)=−16.
Thus the exact stationary point is
(ln34,−16).
解法二
思路
展开
也可先把函数按 ex 做部分分式,再分别求导,最后通分求驻点。
答题过程
展开
Let u=ex. Polynomial division and partial fractions give
(u−1)(u−2)2u2=2+u−28−u−12.
Indeed, the remaining numerator 6u−4 equals 8(u−1)−2(u−2).
Thus
f(x)=2+ex−28−ex−12.
Differentiating,
f′(x)=−(ex−2)28ex+(ex−1)22ex.
At a stationary point, divide by 2ex>0:
(u−1)21=(u−2)24.
Cross-multiplication gives
(u−2)2=u2−4u+4=3u2−4u=4(u−1)2,4u2−8u+4,0.
Since u>0, u=4/3. Substituting in f gives y=−16, so the stationary point is
(ln34,−16).
(b)
解法一
思路
展开
按题意换元时,微分带来一个 u 的分母,积分中的分子由 2u2 变成 2u。然后拆分两项简单对数。
答题过程
展开
Let u=ex, so du=udx. The limits become u=3 and u=5.
Hence
I=∫35u2−3u+22udu.
Write
(u−1)(u−2)2u=u−1A+u−2B.
Then 2u=A(u−2)+B(u−1). Putting u=1 gives A=−2, and putting u=2 gives B=4.
On 3⩽u⩽5, both u−1 and u−2 are positive. Therefore
I=====∫35(−u−12+u−24)du[−2ln(u−1)+4ln(u−2)]35−2ln4+4ln3+2ln24ln3−2ln2ln481.