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IAL 2019 May Q10

A Level / Edexcel / P1

IAL 2019 May Paper · Question 10

题目

Problem

A curve has equation y=f(x)y=f(x), where

f(x)=(x−4)(2x+1)2.\begin{align*} f(x)=(x-4)(2x+1)^2. \end{align*}

The curve touches the xx-axis at the point PP and crosses the xx-axis at the point QQ.

(a) State the coordinates of the point PP.

(1)

(b) Find f′(x)f'(x).

(4)

(c) Hence show that the equation of the tangent to the curve at the point where x=52x=\dfrac52 can be expressed in the form y=ky=k, where kk is a constant to be found.

(3)

The curve with equation y=f(x+a)y=f(x+a), where aa is a constant, passes through the origin OO.

(d) State the possible values of aa.

(2)

解答

(a)

解法一

思路

展开

(2x+1)2(2x+1)^2 是重复因式，所以曲线在对应的根处接触 xx 轴。

答题过程

展开 (2x+1)2=0⇒x=−12.\begin{align*} (2x+1)^2=0 \quad\Rightarrow\quad x=-\frac12. \end{align*}

Therefore

P=(−12,0).\begin{align*} P=\left(-\frac12,0\right). \end{align*}

(b)

解法一

思路

展开

先展开成三次多项式，再逐项求导。

答题过程

展开 f(x)= (x−4)(2x+1)2= (x−4)(4x2+4x+1)= 4x3+4x2+x−16x2−16x−4= 4x3−12x2−15x−4.\begin{align*} f(x) =&\,(x-4)(2x+1)^2\\ =&\,(x-4)(4x^2+4x+1)\\ =&\,4x^3+4x^2+x-16x^2-16x-4\\ =&\,4x^3-12x^2-15x-4. \end{align*}

Therefore

f′(x)=12x2−24x−15.\begin{align*} f'(x)=12x^2-24x-15. \end{align*}

解法二

思路

展开

也可以直接用乘积法则。因为有平方因式，乘积法则后还可以提出公因式整理。

答题过程

展开 f(x)=(x−4)(2x+1)2.\begin{align*} f(x)=(x-4)(2x+1)^2. \end{align*}

Using the product rule,

f′(x)= (2x+1)2+(x−4)⋅2(2x+1)⋅2= (2x+1)2+4(x−4)(2x+1)= (2x+1){(2x+1)+4(x−4)}= (2x+1)(6x−15)= 12x2−24x−15.\begin{align*} f'(x) =&\,(2x+1)^2+(x-4)\cdot2(2x+1)\cdot2\\ =&\,(2x+1)^2+4(x-4)(2x+1)\\ =&\,(2x+1)\{(2x+1)+4(x-4)\}\\ =&\,(2x+1)(6x-15)\\ =&\,12x^2-24x-15. \end{align*}

(c)

解法一

思路

展开

若切线形如 y=ky=k，说明切线是水平线，斜率为 00。先验证 f′(52)=0f'\left(\frac52\right)=0，再求对应的 yy 值。

答题过程

展开 f′(52)= 12(52)2−24(52)−15= 75−60−15= 0.\begin{align*} f'\left(\frac52\right) =&\,12\left(\frac52\right)^2-24\left(\frac52\right)-15\\ =&\,75-60-15\\ =&\,0. \end{align*}

So the tangent is horizontal.

Now find the yy coordinate:

f(52)= (52−4)(2⋅52+1)2= (−32)(6)2= −54.\begin{align*} f\left(\frac52\right) =&\,\left(\frac52-4\right)\left(2\cdot\frac52+1\right)^2\\ =&\,\left(-\frac32\right)(6)^2\\ =&\,-54. \end{align*}

Therefore the tangent is

y=−54.\begin{align*} y=-54. \end{align*}

(d)

解法一

思路

展开

曲线 y=f(x+a)y=f(x+a) 过原点，表示当 x=0x=0 时 f(a)=0f(a)=0。所以 aa 必须是 ff 的根。

答题过程

展开

For the curve to pass through the origin,

f(0+a)=0.\begin{align*} f(0+a)=0. \end{align*}

So

f(a)=0.\begin{align*} f(a)=0. \end{align*}

The roots of f(x)=(x−4)(2x+1)2f(x)=(x-4)(2x+1)^2 are

x=4,x=−12.\begin{align*} x=4,\qquad x=-\frac12. \end{align*}

Therefore

a=4ora=−12.\begin{align*} a=4 \quad\text{or}\quad a=-\frac12. \end{align*}