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IAL 2024 May R Q4

A Level / Edexcel / P1

IAL 2024 May (R) Paper · Question 4

题目

Problem

The curve CC has equation

y=2x+3x−4,x≠0.\begin{align*} y=\frac{2}{x}+3x-4,\qquad x\ne0. \end{align*}

The straight line ll has equation

y=kx+2\begin{align*} y=kx+2 \end{align*}

where kk is a constant.

(a) Show that ll meets CC when

(k−3)x2+6x−2=0.\begin{align*} (k-3)x^2+6x-2=0. \end{align*}
(2)

(b) Hence find the value of kk for which ll is a tangent to CC.

(3)

解答

(a)

解法一

思路

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直线和曲线相交时 yy 值相等。把 kx+2kx+2 和 2x+3x−4\frac2x+3x-4 相等，再乘以 xx 去分母。

答题过程

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At intersections,

kx+2=2x+3x−4.\begin{align*} kx+2=\frac2x+3x-4. \end{align*}

Bring all terms to one side:

kx−3x+6−2x=0.\begin{align*} kx-3x+6-\frac2x=0. \end{align*}

Multiply by xx:

(k−3)x2+6x−2=0.\begin{align*} (k-3)x^2+6x-2=0. \end{align*}

This is the required result.

(b)

解法一

思路

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若直线与曲线相切，那么交点方程只有一个实根，所以判别式等于 00。

答题过程

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For tangency, the quadratic

(k−3)x2+6x−2=0\begin{align*} (k-3)x^2+6x-2=0 \end{align*}

has equal roots. Therefore its discriminant is zero:

b2−4ac= 062−4(k−3)(−2)= 036+8(k−3)= 08k+12= 0k= −32.\begin{align*} b^2-4ac=&\,0\\ 6^2-4(k-3)(-2)=&\,0\\ 36+8(k-3)=&\,0\\ 8k+12=&\,0\\ k=&\,-\frac32. \end{align*}

解法二

思路

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也可以用切线斜率。曲线的导数等于直线斜率 kk，同时切点也必须满足 (a) 的交点方程。把 k=3−2x2k=3-\frac2{x^2} 代回去即可。

答题过程

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For the curve

y=2x+3x−4,\begin{align*} y=\frac2x+3x-4, \end{align*}

the gradient is

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

At a point of tangency, the gradient of the curve is the gradient of the line, so

k=3−2x2.\begin{align*} k=3-\frac{2}{x^2}. \end{align*}

Substitute this into the equation from part (a):

(k−3)x2+6x−2= 0(3−2x2−3)x2+6x−2= 0−2+6x−2= 06x= 4x= 23.\begin{align*} (k-3)x^2+6x-2=&\,0\\ \left(3-\frac{2}{x^2}-3\right)x^2+6x-2=&\,0\\ -2+6x-2=&\,0\\ 6x=&\,4\\ x=&\,\frac23. \end{align*}

Then

k= 3−2(2/3)2= 3−24/9= 3−92= −32.\begin{align*} k =&\,3-\frac{2}{(2/3)^2}\\ =&\,3-\frac{2}{4/9}\\ =&\,3-\frac92\\ =&\,-\frac32. \end{align*}