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IAL 2025 May A Q5

A Level / Edexcel / P1

IAL 2025 May A Paper · Question 5

题目

Problem

y=12x4−3+10x2,x≠0.\begin{align*} y=\frac12x^4-3+\frac{10}{x^2}, \qquad x\ne0. \end{align*}

(a) Find ∫y dx\displaystyle \int y\,dx, writing the answer in simplest form.

(3)

(b) (i) Find dydx\dfrac{dy}{dx}, writing the answer in simplest form.

(3)

(ii) Hence find the exact solutions of the equation dydx=3\dfrac{dy}{dx}=3.

(Solutions relying on calculator technology are not acceptable.)

(4)

解答

(a)

解法一

思路

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先把 10x2\frac{10}{x^2} 写成 10x−210x^{-2}，再逐项积分。不要忘记积分常数 cc。

答题过程

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Write

y=12x4−3+10x−2.\begin{align*} y=\frac12x^4-3+10x^{-2}. \end{align*}

Then

∫y dx= ∫(12x4−3+10x−2) dx= 12⋅x55−3x+10⋅x−1−1+c= 110x5−3x−10x+c.\begin{align*} \int y\,dx =&\,\int\left(\frac12x^4-3+10x^{-2}\right)\,dx\\ =&\,\frac12\cdot\frac{x^5}{5}-3x+10\cdot\frac{x^{-1}}{-1}+c\\ =&\,\frac1{10}x^5-3x-\frac{10}{x}+c. \end{align*}

(b)(i)

解法一

思路

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仍然把分式写成负指数，再逐项微分。常数 −3-3 微分后为 00。

答题过程

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Since

y=12x4−3+10x−2,\begin{align*} y=\frac12x^4-3+10x^{-2}, \end{align*}

differentiate term by term:

dydx= 2x3−20x−3= 2x3−20x3.\begin{align*} \frac{dy}{dx} =&\,2x^3-20x^{-3}\\ =&\,2x^3-\frac{20}{x^3}. \end{align*}

(b)(ii)

解法一

思路

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由 (b)(i) 得到

2x3−20x3=3.\begin{align*} 2x^3-\frac{20}{x^3}=3. \end{align*}

乘以 x3x^3 后令 u=x3u=x^3，方程会变成二次方程。最后不要停在 x3x^3，要把 xx 求出来。

答题过程

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Set

2x3−20x3= 3.\begin{align*} 2x^3-\frac{20}{x^3}=&\,3. \end{align*}

Multiply by x3x^3:

2x6−20= 3x32x6−3x3−20= 0.\begin{align*} 2x^6-20=&\,3x^3\\ 2x^6-3x^3-20=&\,0. \end{align*}

Let

u=x3.\begin{align*} u=x^3. \end{align*}

Then

2u2−3u−20= 0(2u+5)(u−4)= 0.\begin{align*} 2u^2-3u-20=&\,0\\ (2u+5)(u-4)=&\,0. \end{align*}

So

u=−52oru=4.\begin{align*} u=-\frac52\quad\text{or}\quad u=4. \end{align*}

Since u=x3u=x^3,

x3= −52orx3=4.\begin{align*} x^3=&\,-\frac52 \quad\text{or}\quad x^3=4. \end{align*}

Therefore

x= −523orx=43.\begin{align*} x=&\,-\sqrt[3]{\frac52} \quad\text{or}\quad x=\sqrt[3]{4}. \end{align*}