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IAL 2025 Oct Q6

A Level / Edexcel / P1

IAL 2025 Oct Paper · Question 6

题目

Problem

(a) Sketch the graph of the curve CC with equation

y=4kx−2k\begin{align*} y=\frac{4k}{x-2k} \end{align*}

where kk is a positive constant.

On your sketch show

  • the coordinates of any points where CC cuts the coordinate axes
  • the equation of the vertical asymptote to CC
(4)

The straight line ll has equation

y=6−2x\begin{align*} y=6-2x \end{align*}

Given that there is at least one point of intersection between ll and CC,

(b) find the range of possible values of kk.

(5)

解答

(a)

解法一

思路

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这是 reciprocal graph 的平移和伸缩。分母为 00 时给出竖直渐近线；当 x=0x=0 时求 yy 轴截距。由于分子 4k>04k>0，所以没有 xx 轴截距。

答题过程

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The vertical asymptote occurs when the denominator is zero:

x−2k= 0x= 2k.\begin{align*} x-2k=&\,0\\ x=&\,2k. \end{align*}

At the yy-axis, x=0x=0, so

y= 4k0−2k= −2.\begin{align*} y=&\,\frac{4k}{0-2k}\\ =&\,-2. \end{align*}

Thus the curve cuts the yy-axis at

(0,−2).\begin{align*} (0,-2). \end{align*}

There is no xx-axis intercept, because

4kx−2k=0\begin{align*} \frac{4k}{x-2k}=0 \end{align*}

is impossible when k>0k>0.

The sketch should have vertical asymptote x=2kx=2k and horizontal asymptote y=0y=0.

A completed sketch is:

(b)

解法一

思路

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交点满足两条曲线的 yy 值相等。整理成关于 xx 的二次方程后，“至少一个交点”表示这个二次方程有实根，所以判别式大于等于 00。最后别忘了题目给定 k>0k>0。

答题过程

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At intersections,

4kx−2k=6−2x.\begin{align*} \frac{4k}{x-2k}=6-2x. \end{align*}

Cross multiply:

4k= (6−2x)(x−2k)= 6x−12k−2x2+4kx.\begin{align*} 4k=&\,(6-2x)(x-2k)\\ =&\,6x-12k-2x^2+4kx. \end{align*}

Rearrange:

2x2−(4k+6)x+16k= 0.\begin{align*} 2x^2-(4k+6)x+16k=&\,0. \end{align*}

For at least one real intersection,

b2−4ac⩾0.\begin{align*} b^2-4ac\geqslant0. \end{align*}

So

(−(4k+6))2−4(2)(16k)⩾0(4k+6)2−128k⩾016k2+48k+36−128k⩾016k2−80k+36⩾04k2−20k+9⩾0.\begin{align*} \bigl(-(4k+6)\bigr)^2-4(2)(16k)&\geqslant0\\ (4k+6)^2-128k&\geqslant0\\ 16k^2+48k+36-128k&\geqslant0\\ 16k^2-80k+36&\geqslant0\\ 4k^2-20k+9&\geqslant0. \end{align*}

Factorise:

4k2−20k+9= (2k−1)(2k−9).\begin{align*} 4k^2-20k+9 =&\,(2k-1)(2k-9). \end{align*}

Hence

(2k−1)(2k−9)⩾0.\begin{align*} (2k-1)(2k-9)&\geqslant0. \end{align*}

Since this is positive or zero outside the critical values,

k⩽12ork⩾92.\begin{align*} k\leqslant\frac12 \quad\text{or}\quad k\geqslant\frac92. \end{align*}

Also k>0k>0, so

0<k⩽12ork⩾92.\begin{align*} 0<k\leqslant\frac12 \quad\text{or}\quad k\geqslant\frac92. \end{align*}