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IAL 2024 Oct Q6

A Level / Edexcel / P1

IAL 2024 Oct Paper · Question 6

题目

Problem

In this question you must show all stages of your working.

Solutions relying on calculator technology are not acceptable.

(a) Sketch the curve CC with equation

y=12−x,x≠2.\begin{align*} y=\frac{1}{2-x},\qquad x\ne2. \end{align*}

State on your sketch

  • the equation of the vertical asymptote
  • the coordinates of the intersection of CC with the yy-axis
(3)

The straight line ll has equation y=kx−4y=kx-4, where kk is a constant.

Given that ll cuts CC at least once,

(b) (i) show that

k2−5k+4≥0\begin{align*} k^2-5k+4\ge0 \end{align*}

(ii) find the range of possible values for kk.

(6)

解答

(a)

解法一

思路

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y=12−xy=\dfrac1{2-x} 是倒数型图像。分母为 00 的地方是竖直渐近线，所以 x=2x=2。令 x=0x=0 可以求出 yy 轴截距。

答题过程

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The vertical asymptote is where the denominator is zero:

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

The intersection with the yy-axis occurs when x=0x=0:

y=12−0=12.\begin{align*} y=\frac{1}{2-0}=\frac12. \end{align*}

So the sketch should show a reciprocal-type curve with

x=2\begin{align*} x=2 \end{align*}

as the vertical asymptote and

(0,12)\begin{align*} \left(0,\frac12\right) \end{align*}

as the yy-intercept.

(b)(i)

解法一

思路

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直线和曲线相交时，它们的 yy 值相等。整理后会得到关于 xx 的二次方程。题目说至少相交一次，所以这个二次方程至少有一个实根，即判别式 Δ≥0\Delta\ge0。

答题过程

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At intersections,

kx−4=12−x.\begin{align*} kx-4=\frac{1}{2-x}. \end{align*}

Multiply by 2−x2-x:

(kx−4)(2−x)= 12kx−kx2−8+4x= 1.\begin{align*} (kx-4)(2-x)=&\,1\\ 2kx-kx^2-8+4x=&\,1. \end{align*}

Rearrange:

−kx2+(2k+4)x−9= 0kx2+(−2k−4)x+9= 0.\begin{align*} -kx^2+(2k+4)x-9=&\,0\\ kx^2+(-2k-4)x+9=&\,0. \end{align*}

For the line to cut the curve at least once, this quadratic must have at least one real root. Hence

b2−4ac≥0(−2k−4)2−4(k)(9)≥0.\begin{align*} b^2-4ac&\ge0\\ (-2k-4)^2-4(k)(9)&\ge0. \end{align*}

Expand and simplify:

4k2+16k+16−36k≥04k2−20k+16≥04(k2−5k+4)≥0.\begin{align*} 4k^2+16k+16-36k&\ge0\\ 4k^2-20k+16&\ge0\\ 4(k^2-5k+4)&\ge0. \end{align*}

Therefore

k2−5k+4≥0.\begin{align*} k^2-5k+4\ge0. \end{align*}

This is the required result.

(b)(ii)

解法一

思路

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解二次不等式。先因式分解找到临界值 11 和 44，再根据开口向上判断外侧区间满足 ≥0\ge0。

答题过程

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From part (b)(i),

k2−5k+4≥0(k−1)(k−4)≥0.\begin{align*} k^2-5k+4&\ge0\\ (k-1)(k-4)&\ge0. \end{align*}

The critical values are

k=1,k=4.\begin{align*} k=1,\qquad k=4. \end{align*}

Since the quadratic opens upwards,

(k−1)(k−4)≥0\begin{align*} (k-1)(k-4)\ge0 \end{align*}

outside the interval between the roots.

Therefore

k≤1ork≥4.\begin{align*} k\le1\quad\text{or}\quad k\ge4. \end{align*}