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IAL 2024 Jan Q7

A Level / Edexcel / P1

IAL 2024 Jan Paper · Question 7

题目

Problem

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

y=4xk,\begin{align*} y=\frac{4}{x-k}, \end{align*}

where kk is a positive constant.

Show on your sketch

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

Given that the straight line with equation y=9xy=9-x does not cross or touch CC,

(b) find the range of values of kk.

(5)

解答

(a)

解法一

思路

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这是 reciprocal graph。分母为 00 的位置给出竖直渐近线 x=kx=k。由于 k>0k>0,这条渐近线在 yy 轴右侧。令 x=0x=0 可得 yy 轴截距;但分子恒为 44,所以不会有 xx 轴截距。

答题过程

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The vertical asymptote occurs when the denominator is zero:

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

The yy-intercept is found by setting x=0x=0:

y=40k=4k.\begin{align*} y=\frac{4}{0-k}=-\frac4k. \end{align*}

So the curve cuts the yy-axis at

(0,4k).\begin{align*} \left(0,-\frac4k\right). \end{align*}

There is no xx-intercept, since

4xk=0\begin{align*} \frac{4}{x-k}=0 \end{align*}

has no solution.

The sketch should show a decreasing reciprocal curve with vertical asymptote x=kx=k, a right-hand branch above the xx-axis, and a left-hand branch passing through (0,4k)\left(0,-\frac4k\right).

(b)

解法一

思路

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直线不与曲线相交或相切,意思是联立后没有实根。联立得到关于 xx 的二次方程,再令判别式小于 00

答题过程

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At an intersection,

4xk=9x.\begin{align*} \frac{4}{x-k}=9-x. \end{align*}

Multiplying by xkx-k,

4=(9x)(xk)=9x9kx2+kx.\begin{align*} 4=&\,(9-x)(x-k)\\ =&\,9x-9k-x^2+kx. \end{align*}

Hence

x2(9+k)x+9k+4=0.\begin{align*} x^2-(9+k)x+9k+4=0. \end{align*}

For the line not to cross or touch the curve, this quadratic must have no real roots, so

Δ<0[(9+k)]24(1)(9k+4)<0(9+k)236k16<0k2+18k+8136k16<0k218k+65<0(k13)(k5)<0.\begin{align*} \Delta&<0\\ [-(9+k)]^2-4(1)(9k+4)&<0\\ (9+k)^2-36k-16&<0\\ k^2+18k+81-36k-16&<0\\ k^2-18k+65&<0\\ (k-13)(k-5)&<0. \end{align*}

Therefore

5<k<13.\begin{align*} 5<k<13. \end{align*}