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IAL 2026 Jan Q7

A Level / Edexcel / P1

IAL 2026 Jan Paper · Question 7

题目

Problem

The curve CC has equation

y=23x3−8x2+43x−203\begin{align*} y=\frac23x^3-8x^2+43x-\frac{20}{3} \end{align*}

(a) Show that dydx\dfrac{\mathrm{d}y}{\mathrm{d}x} can be written in the form

p(x+q)2+r\begin{align*} p(x+q)^2+r \end{align*}

where pp, qq and rr are constants to be found.

(5)

(b) Hence state

(i) the minimum value of dydx\dfrac{\mathrm{d}y}{\mathrm{d}x}

(ii) the value of xx at which this minimum value occurs.

(2)

Given that SS is the point on CC at which the gradient is a minimum,

(c) find the equation of the tangent to CC at SS, giving your answer in the form y=mx+cy=mx+c, where mm and cc are constants.

(3)

解答

(a)

解法一

思路

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先求导，再把二次式配方。配方后最低值就会在下一问直接看出来。

答题过程

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Differentiate:

dydx= 2x2−16x+43\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\, 2x^2-16x+43 \end{align*}

Complete the square:

2x2−16x+43= 2(x2−8x)+43= 2[(x−4)2−16]+43= 2(x−4)2−32+43= 2(x−4)2+11\begin{align*} 2x^2-16x+43 =&\, 2(x^2-8x)+43\\[3mm] =&\, 2\left[(x-4)^2-16\right]+43\\[3mm] =&\, 2(x-4)^2-32+43\\[3mm] =&\, 2(x-4)^2+11 \end{align*}

Therefore

dydx=2(x−4)2+11\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} = 2(x-4)^2+11 \end{align*}

So

p=2,q=−4,r=11\begin{align*} p=2,\qquad q=-4,\qquad r=11 \end{align*}

(b)

解法一

思路

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平方项 2(x−4)22(x-4)^2 的最小值是 00，发生在 x=4x=4。所以导数的最小值就是剩下的常数 1111。

答题过程

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Since

dydx=2(x−4)2+11\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x}=2(x-4)^2+11 \end{align*}

and

2(x−4)2⩾0,\begin{align*} 2(x-4)^2\geqslant0, \end{align*}

the minimum value of dydx\dfrac{\mathrm{d}y}{\mathrm{d}x} is

11\begin{align*} 11 \end{align*}

This occurs when

x=4\begin{align*} x=4 \end{align*}

(c)

解法一

思路

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点 SS 的 xx 坐标是 44，切线斜率是最小梯度 1111。先把 x=4x=4 代回曲线求 yy 坐标，再写切线方程。

答题过程

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At SS, x=4x=4.

Find the yy coordinate:

y= 23(4)3−8(4)2+43(4)−203= 1283−128+172−203= 80\begin{align*} y =&\, \frac23(4)^3-8(4)^2+43(4)-\frac{20}{3}\\[3mm] =&\, \frac{128}{3}-128+172-\frac{20}{3}\\[3mm] =&\, 80 \end{align*}

The gradient of the tangent is 1111, so

y−80=11(x−4)\begin{align*} y-80=11(x-4) \end{align*}

Hence

y=11x+36\begin{align*} y=11x+36 \end{align*}