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IAL 2020 Oct Q8

A Level / Edexcel / P1

IAL 2020 Oct Paper · Question 8

题目

Problem

The curve CC has equation

y=(x−2)(x−4)2.\begin{align*} y=(x-2)(x-4)^2. \end{align*}

(a) Show that

dydx=3x2−20x+32.\begin{align*} \frac{dy}{dx}=3x^2-20x+32. \end{align*}
(4)

The line l1l_1 is the tangent to CC at the point where x=6x=6.

(b) Find an equation for l1l_1, giving your answer in the form y=mx+cy=mx+c, where mm and cc are constants to be found.

(4)

The line l2l_2 is another tangent to CC.

Given that l2l_2 is parallel to l1l_1 and that l2l_2 touches CC at the point where x=αx=\alpha,

(c) find the value of α\alpha.

(3)

解答

(a)

解法一

思路

展开

先展开成三次多项式，再逐项求导。Show that 题要把展开和求导步骤写清楚。

答题过程

展开 y= (x−2)(x−4)2= (x−2)(x2−8x+16)= x3−8x2+16x−2x2+16x−32= x3−10x2+32x−32.\begin{align*} y=&\,(x-2)(x-4)^2\\ =&\,(x-2)(x^2-8x+16)\\ =&\,x^3-8x^2+16x-2x^2+16x-32\\ =&\,x^3-10x^2+32x-32. \end{align*}

Therefore

dydx= 3x2−20x+32.\begin{align*} \frac{dy}{dx} =&\,3x^2-20x+32. \end{align*}

解法二

思路

展开

也可以直接用乘积法则。这样不用先完全展开原函数，但最后仍要整理成题目要求的形式。

答题过程

展开 y= (x−2)(x−4)2.\begin{align*} y=&\,(x-2)(x-4)^2. \end{align*}

Using the product rule,

dydx= (x−2)⋅2(x−4)+(x−4)2= 2(x−2)(x−4)+(x−4)2= (x−4){2(x−2)+(x−4)}= (x−4)(3x−8)= 3x2−20x+32.\begin{align*} \frac{dy}{dx} =&\,(x-2)\cdot 2(x-4)+(x-4)^2\\ =&\,2(x-2)(x-4)+(x-4)^2\\ =&\,(x-4)\{2(x-2)+(x-4)\}\\ =&\,(x-4)(3x-8)\\ =&\,3x^2-20x+32. \end{align*}

(b)

解法一

思路

展开

切点的 xx 坐标是 66。先代入原方程求 yy 坐标，再代入导函数求切线斜率，最后用点斜式写直线。

答题过程

展开

At x=6x=6,

y=(6−2)(6−4)2=16.\begin{align*} y=(6-2)(6-4)^2=16. \end{align*}

The gradient is

dydx∣x=6= 3(6)2−20(6)+32= 20.\begin{align*} \frac{dy}{dx}\bigg|_{x=6} =&\,3(6)^2-20(6)+32\\ =&\,20. \end{align*}

So the tangent is

y−16= 20(x−6)y= 20x−104.\begin{align*} y-16=&\,20(x-6)\\ y=&\,20x-104. \end{align*}

(c)

解法一

思路

展开

平行切线有相同斜率，所以在 x=αx=\alpha 处的导数也等于 2020。解出两个可能的 xx 后，排除已经属于 l1l_1 的 x=6x=6。

答题过程

展开

Since l2l_2 is parallel to l1l_1,

3α2−20α+32= 203α2−20α+12= 0(3α−2)(α−6)= 0.\begin{align*} 3\alpha^2-20\alpha+32=&\,20\\ 3\alpha^2-20\alpha+12=&\,0\\ (3\alpha-2)(\alpha-6)=&\,0. \end{align*}

Thus

α=23orα=6.\begin{align*} \alpha=\frac23 \quad\text{or}\quad \alpha=6. \end{align*}

Since l2l_2 is another tangent, α≠6\alpha\ne6. Therefore

α=23.\begin{align*} \alpha=\frac23. \end{align*}