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IAL 2019 May Q6

A Level / Edexcel / P1

IAL 2019 May Paper · Question 6

题目

Problem

The line with equation y=4x+cy=4x+c, where cc is a constant, meets the curve with equation

y=x(x−3)\begin{align*} y=x(x-3) \end{align*}

at only one point.

(a) Find the value of cc.

(4)

(b) Hence find the coordinates of the point of intersection.

(3)

解答

(a)

解法一

思路

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直线和曲线只有一个交点，所以联立后得到的二次方程有一个重复根，判别式等于 00。

答题过程

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At the point of intersection,

4x+c= x(x−3)4x+c= x2−3xx2−7x−c= 0.\begin{align*} 4x+c=&\,x(x-3)\\ 4x+c=&\,x^2-3x\\ x^2-7x-c=&\,0. \end{align*}

For exactly one point of intersection,

(−7)2−4(1)(−c)= 049+4c= 0c= −494.\begin{align*} (-7)^2-4(1)(-c)=&\,0\\ 49+4c=&\,0\\ c=&\,-\frac{49}{4}. \end{align*}

解法二

思路

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只有一个交点说明直线是曲线的切线。切线斜率是 44，所以让曲线导数等于 44，先求切点，再求 cc。

答题过程

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For the curve,

y=x(x−3)=x2−3x,\begin{align*} y=x(x-3)=x^2-3x, \end{align*}

so

dydx=2x−3.\begin{align*} \frac{dy}{dx}=2x-3. \end{align*}

Since the tangent has gradient 44,

2x−3= 4x= 72.\begin{align*} 2x-3=&\,4\\ x=&\,\frac72. \end{align*}

At this point,

y= 72(72−3)= 72⋅12= 74.\begin{align*} y=&\,\frac72\left(\frac72-3\right)\\ =&\,\frac72\cdot\frac12\\ =&\,\frac74. \end{align*}

Substitute into y=4x+cy=4x+c:

74= 4(72)+c74= 14+cc= −494.\begin{align*} \frac74=&\,4\left(\frac72\right)+c\\ \frac74=&\,14+c\\ c=&\,-\frac{49}{4}. \end{align*}

(b)

解法一

思路

展开

把 c=−494c=-\frac{49}{4} 代入联立得到的二次方程。它应该有一个重复根。

答题过程

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Using

x2−7x−c=0\begin{align*} x^2-7x-c=0 \end{align*}

and c=−494c=-\frac{49}{4},

x2−7x+494= 0(x−72)2= 0.\begin{align*} x^2-7x+\frac{49}{4}=&\,0\\ \left(x-\frac72\right)^2=&\,0. \end{align*}

Thus

x=72.\begin{align*} x=\frac72. \end{align*}

Then

y=4(72)−494=74.\begin{align*} y=4\left(\frac72\right)-\frac{49}{4} =\frac74. \end{align*}

Therefore the point of intersection is

(72,74).\begin{align*} \left(\frac72,\frac74\right). \end{align*}