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P3 2023 May Q8

A Level / Edexcel / P3

IAL 2023 May Paper · Question 8

题目

Problem

Figure 3 shows a sketch of the curve CC with equation y=f(x)y=f(x), where

f(x)=(2x+1)3e−4xf(x)=(2x+1)^3e^{-4x}

(a) Show that

f′(x)=A(2x+1)2(1−4x)e−4xf'(x)=A(2x+1)^2(1-4x)e^{-4x}

where AA is a constant to be found.

(4)

(b) Hence find the exact coordinates of the two stationary points on CC.

(3)

The function gg is defined by

g(x)=8f(x−2)g(x)=8f(x-2)

(c) Find the coordinates of the maximum stationary point on the curve with equation y=g(x)y=g(x).

(2)
题目中文翻译

图 3 给出了曲线 CC 的示意图,其方程为 y=f(x)y=f(x),其中

f(x)=(2x+1)3e−4xf(x)=(2x+1)^3e^{-4x}

(a) 证明

f′(x)=A(2x+1)2(1−4x)e−4xf'(x)=A(2x+1)^2(1-4x)e^{-4x}

其中 AA 是需要求出的常数。

(b) 由此求曲线 CC 上两个驻点的精确坐标。

函数 gg 定义为

g(x)=8f(x−2)g(x)=8f(x-2)

(c) 求曲线 y=g(x)y=g(x) 上极大驻点的坐标。

解答

(a)

解法一

思路

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f(x)f(x) 是两个函数的乘积:(2x+1)3(2x+1)^3 和 e−4xe^{-4x},所以用乘积法则求导。求导后把共同因子 (2x+1)2e−4x(2x+1)^2e^{-4x} 提出来,再整理成题目给出的形式。

答题过程

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We have

f(x)=(2x+1)3e−4x.f(x)=(2x+1)^3e^{-4x}.

Using the product rule,

f′(x)= 6(2x+1)2e−4x−4(2x+1)3e−4x= 2(2x+1)2e−4x(3−2(2x+1))= 2(2x+1)2e−4x(1−4x).\begin{align*} f'(x)=&\,6(2x+1)^2e^{-4x} -4(2x+1)^3e^{-4x} \\[2mm] =&\,2(2x+1)^2e^{-4x} \big(3-2(2x+1)\big) \\[2mm] =&\,2(2x+1)^2e^{-4x}(1-4x). \end{align*}

Therefore

f′(x)=A(2x+1)2(1−4x)e−4x,f'(x)=A(2x+1)^2(1-4x)e^{-4x},

where

A=2.\boxed{A=2}.

(b)

解法一

思路

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驻点满足 f′(x)=0f'(x)=0。由于 e−4xe^{-4x} 永远不为 0,所以只需要令代数因子为 0。求出两个 xx 值后,代回原函数得到对应的 yy 坐标。

答题过程

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Stationary points occur when

f′(x)=0.f'(x)=0.

From part (a),

2(2x+1)2(1−4x)e−4x=0.2(2x+1)^2(1-4x)e^{-4x}=0.

Since 2≠02\ne0 and e−4x≠0e^{-4x}\ne0,

(2x+1)2(1−4x)=0.(2x+1)^2(1-4x)=0.

So

2x+1=0or1−4x=0.2x+1=0 \qquad\text{or}\qquad 1-4x=0.

Hence

x=−12orx=14.x=-\frac12 \qquad\text{or}\qquad x=\frac14.

When x=−12x=-\frac12,

f(−12)=03e2=0.f\bigl(-\frac12\bigr)=0^3e^2=0.

When x=14x=\frac14,

f(14)= (2⋅14+1)3e−1= (32)31e= 278e.\begin{align*} f\bigl(\frac14\bigr) =&\,\bigl(2\cdot\frac14+1\bigr)^3e^{-1} \\[2mm] =&\,\bigl(\frac32\bigr)^3\frac1e \\[2mm] =&\,\frac{27}{8e}. \end{align*}

Therefore the two stationary points are

(−12,0)and(14,278e).\boxed{\bigl(-\frac12,0\bigr)} \qquad\text{and}\qquad \boxed{\bigl(\frac14,\frac{27}{8e}\bigr)}.

(c)

解法一

思路

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g(x)=8f(x−2)g(x)=8f(x-2) 表示把 y=f(x)y=f(x) 向右平移 2 个单位,再把 yy 坐标乘以 8。原曲线的极大驻点是 (14,278e)\bigl(\frac14,\frac{27}{8e}\bigr),所以直接变换这个点。

答题过程

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The maximum stationary point on y=f(x)y=f(x) is

(14,278e).\bigl(\frac14,\frac{27}{8e}\bigr).

The function

g(x)=8f(x−2)g(x)=8f(x-2)

shifts the graph of y=f(x)y=f(x) 2 units to the right and multiplies all yy-coordinates by 8.

So the xx-coordinate becomes

14+2=94,\frac14+2=\frac94,

and the yy-coordinate becomes

8⋅278e=27e.8\cdot\frac{27}{8e}=\frac{27}{e}.

Therefore the maximum stationary point on y=g(x)y=g(x) is

(94,27e).\boxed{\bigl(\frac94,\frac{27}{e}\bigr)}.