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IAL 2025 June FP3 Q4

A Level / Edexcel / FP3

IAL 2025 June Paper · Question 4

题目

Problem

y=arsinhx+arsinh(1x),x>0y = \operatorname{arsinh}x + \operatorname{arsinh}\left(\frac{1}{x}\right), \qquad x > 0

(a) Show that

dydx=x1x1+x2\frac{dy}{dx} = \frac{x-1}{x\sqrt{1+x^2}}

(b) Hence determine the exact value of yy for which dydx=0\dfrac{dy}{dx} = 0, giving your answer as a simplified natural logarithm.

(2)
(3)
题目中文翻译 y=arsinhx+arsinh(1x),x>0y = \operatorname{arsinh}x + \operatorname{arsinh}\left(\frac{1}{x}\right), \qquad x > 0

(a) 证明

dydx=x1x1+x2\frac{dy}{dx} = \frac{x-1}{x\sqrt{1+x^2}}

(b) 由此求出使 dydx=0\dfrac{dy}{dx}=0 时的 yy 的准确值,并将答案写成最简自然对数形式。

解答

(a)

解法一

思路

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分别对两个反双曲正弦项求导。第二项要使用链式法则;又因 x>0x>0,所以 1+x2=1+x2/x\sqrt{1+x^{-2}}=\sqrt{1+x^2}/x,可直接合并成目标分式。

答题过程

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Differentiating with respect to xx,

dydx=11+x2x21+x2.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,\frac{1}{\sqrt{1+x^2}}\\ &\,-\frac{x^{-2}}{\sqrt{1+x^{-2}}}. \end{align*}

Since x>0x>0,

1+x2=1+x2x.\sqrt{1+x^{-2}} =\frac{\sqrt{1+x^2}}{x}.

Therefore,

dydx=11+x21x1+x2=x1x1+x2,\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,\frac{1}{\sqrt{1+x^2}} -\frac{1}{x\sqrt{1+x^2}}\\ =&\,\boxed{ \frac{x-1}{x\sqrt{1+x^2}}}, \end{align*}

as required.

(b)

解法一

思路

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“Hence” 要直接令 (a) 的导数为零。由于定义域内分母恒正,只需令分子为零得到 x=1x=1;代回原式后,使用 arsinhu=ln(u+u2+1)\operatorname{arsinh}u=\ln(u+\sqrt{u^2+1}),再合并对数。

答题过程

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From part (a),

x1x1+x2=0.\frac{x-1}{x\sqrt{1+x^2}}=0.

Since x>0x>0, the denominator is positive. Hence

x1=0,x-1=0,

so x=1x=1. Substituting into the original expression,

y=2arsinh1=2ln(1+2)=ln((1+2)2)=ln(3+22).\begin{align*} y =&\,2\operatorname{arsinh}1\\ =&\,2\ln(1+\sqrt2)\\ =&\,\ln\big((1+\sqrt2)^2\big)\\ =&\,\boxed{\ln(3+2\sqrt2)}. \end{align*}

解法二

思路

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官方评分资料还接受由 sinh(y/2)=1\sinh(y/2)=1 转成指数方程。令 u=ey/2>0u=e^{y/2}>0 后得到二次方程,舍去负根,再把结果写成最简自然对数。

答题过程

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As in the first method, part (a) gives x=1x=1, and hence

y=2arsinh1.y=2\operatorname{arsinh}1.

Therefore,

sinh(y2)=1.\sinh\bigg(\frac y2\bigg)=1.

Using the exponential definition of sinh\sinh,

ey/2ey/22=1.\frac{e^{y/2}-e^{-y/2}}{2}=1.

Let u=ey/2u=e^{y/2}, where u>0u>0. Multiplying by 2u2u gives

u22u1=0.u^2-2u-1=0.

Thus

u=1±2.u=1\pm\sqrt2.

Since u>0u>0, only u=1+2u=1+\sqrt2 is valid. Hence

y2=ln(1+2),y=2ln(1+2)=ln(3+22).\begin{align*} \frac y2=&\,\ln(1+\sqrt2),\\ y=&\,2\ln(1+\sqrt2)\\ =&\,\boxed{\ln(3+2\sqrt2)}. \end{align*}