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FP3 2022 Jan Q3

A Level / Edexcel / FP3

IAL 2022 Jan Paper · Question 3

题目

Problem

3.

(a) Given that

y=arsech⁡(x2),0<x≤2,y=\operatorname{arsech}\bigg(\frac{x}{2}\bigg), \qquad 0<x\le2,

show that

dydx=pxq−x2\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{p}{x\sqrt{q-x^2}}

where pp and qq are constants to be determined.

(4)

In part (b) solutions based entirely on calculator technology are not acceptable.

f(x)=artanh⁡(x)+arsech⁡(x2),0<x≤1\mathrm{f}(x)=\operatorname{artanh}(x) +\operatorname{arsech}\bigg(\frac{x}{2}\bigg), \qquad 0<x\le1

(b) Determine, in simplest form, the exact value of xx for which f′(x)=0\mathrm{f}'(x)=0.

(5)
题目中文翻译

3.

(a) 已知

y=arsech⁡(x2),0<x≤2,y=\operatorname{arsech}\bigg(\frac{x}{2}\bigg), \qquad 0<x\le2,

证明

dydx=pxq−x2\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{p}{x\sqrt{q-x^2}}

其中 pp 和 qq 为待定常数。

(4)

在 (b) 中，完全依赖计算器技术的解答不予接受。

f(x)=artanh⁡(x)+arsech⁡(x2),0<x≤1\mathrm{f}(x)=\operatorname{artanh}(x) +\operatorname{arsech}\bigg(\frac{x}{2}\bigg), \qquad 0<x\le1

(b) 求使 f′(x)=0\mathrm{f}'(x)=0 的 xx 的最简精确值。

(5)

解答

(a)

解法一：对 sech⁡y=x2\operatorname{sech}y=\frac{x}{2} 隐式求导

思路

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先对反双曲正割关系取 sech⁡\operatorname{sech}，得到 sech⁡y=x2\operatorname{sech}y=\frac{x}{2}。隐式求导后，用 tanh⁡2y=1−sech⁡2y\tanh^2y=1-\operatorname{sech}^2y 把结果完全改写成 xx；由定义域可确定 tanh⁡y\tanh y 取非负根。

答题过程

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Since

y=arsech⁡(x2),y=\operatorname{arsech}\bigg(\frac{x}{2}\bigg),

we have

sech⁡y=x2.\operatorname{sech}y=\frac{x}{2}.

Differentiating implicitly with respect to xx,

−sech⁡ytanh⁡ydydx=12.-\operatorname{sech}y\tanh y \frac{\mathrm{d}y}{\mathrm{d}x}=\frac12.

For 0<x≤20<x\le2, y≥0y\ge0, so

tanh⁡y= 1−sech⁡2y= 1−x24= 4−x22.\begin{align*} \tanh y =&\,\sqrt{1-\operatorname{sech}^2y} \\ =&\,\sqrt{1-\frac{x^2}{4}} \\ =&\,\frac{\sqrt{4-x^2}}{2}. \end{align*}

Therefore,

dydx= −12sech⁡ytanh⁡y= −2x4−x2.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,-\frac{1} {2\operatorname{sech}y\tanh y} \\ =&\,-\frac{2}{x\sqrt{4-x^2}}. \end{align*}

Hence

p=−2,q=4.\boxed{p=-2,\qquad q=4}.

解法二：改写成 arcosh⁡\operatorname{arcosh}

思路

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利用 arsech⁡u=arcosh⁡(1/u)\operatorname{arsech}u=\operatorname{arcosh}(1/u)，把函数改写成 arcosh⁡(2/x)\operatorname{arcosh}(2/x)，然后套用反双曲余弦的导数公式。条件 x>0x>0 可用于正确化简根式中的绝对值。

答题过程

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For x>0x>0,

y=arcosh⁡(2x).y=\operatorname{arcosh}\bigg(\frac2x\bigg).

Therefore,

dydx= −2/x2(2/x)2−1= −2/x24−x2/x= −2x4−x2.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,\frac{-2/x^2} {\sqrt{(2/x)^2-1}} \\ =&\,\frac{-2/x^2} {\sqrt{4-x^2}/x} \\ =&\,-\frac{2}{x\sqrt{4-x^2}}. \end{align*}

Thus

p=−2,q=4.\boxed{p=-2,\qquad q=4}.

(b)

解法一

思路

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使用 (a) 的结果求 f′(x)\mathrm{f}'(x) 并令其为零。平方前先依据定义域确认等式两边非负，避免产生未经控制的增根；所得方程是关于 x2x^2 的二次方程，最后用 0<x≤10<x\le1 筛选并检查答案。

答题过程

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Using part (a),

f′(x)=11−x2−2x4−x2.\mathrm{f}'(x) =\frac{1}{1-x^2} -\frac{2}{x\sqrt{4-x^2}}.

For an interior solution, 0<x<10<x<1. Setting f′(x)=0\mathrm{f}'(x)=0 gives

x4−x2=2(1−x2).x\sqrt{4-x^2}=2(1-x^2).

Both sides are non-negative on this domain, so squaring is valid:

x2(4−x2)= 4(1−x2)2,4x2−x4= 4−8x2+4x4,⟹5x4−12x2+4= 0.\begin{align*} x^2(4-x^2) =&\,4(1-x^2)^2, \\ 4x^2-x^4 =&\,4-8x^2+4x^4, \\ \Longrightarrow\quad 5x^4-12x^2+4 =&\,0. \end{align*}

Let u=x2u=x^2. Then

5u2−12u+4= 0,(5u−2)(u−2)= 0.\begin{align*} 5u^2-12u+4=&\,0, \\ (5u-2)(u-2)=&\,0. \end{align*}

so

u=2oru=25.u=2\quad\text{or}\quad u=\frac25.

The condition 0<x≤10<x\le1 excludes x2=2x^2=2 and selects the positive square root. Substitution in the unsquared equation confirms the solution. Hence

x=25=105.\boxed{x=\sqrt{\frac25}=\frac{\sqrt{10}}5}.