题目

Problem

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

The curve $C$ has equation

\begin{align*} y=\frac2x-k \end{align*}

where $k$ is a positive constant.

(a) Sketch the graph of $C$ .

Show on your sketch

the coordinates of any points of intersection of $C$ with the coordinate axes
the equation of the horizontal asymptote to $C$

stating each in terms of $k$ .

(3)

The line $l$ has equation $y=-kx-6$ .

Given that $l$ intersects $C$ at $2$ distinct points,

(b) find the range of possible values of $k$ .

(5)

解答

(a)

解法一

思路

展开

$y=\frac2x-k$ 是 $y=\frac2x$ 向下平移 $k$ 。因此水平渐近线是 $y=-k$ 。与 $x$ 轴交点令 $y=0$ ；没有 $y$ 轴交点，因为 $x=0$ 不在定义域内。

答题过程

展开

The horizontal asymptote is

\begin{align*} y=-k. \end{align*}

For the $x$ -intercept, set $y=0$ :

\begin{align*} \frac2x-k=&\,0\\ \frac2x=&\,k\\ x=&\,\frac2k. \end{align*}

So the curve intersects the $x$ -axis at

\begin{align*} \left(\frac2k,0\right). \end{align*}

There is no $y$ -intercept because $x=0$ is not allowed.

A completed sketch is:

(b)

解法一

思路

展开

两条图像有两个不同交点，等价于联立后关于 $x$ 的二次方程有两个不同实根。因此判别式要大于 $0$ 。同时题目给出 $k>0$ 。

答题过程

展开

At intersections,

\begin{align*} -kx-6=&\,\frac2x-k. \end{align*}

Multiply by $x$ :

\begin{align*} -kx^2-6x=&\,2-kx. \end{align*}

Bring all terms to one side:

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

For two distinct intersections,

\begin{align*} \Delta>0. \end{align*}

\begin{align*} (6-k)^2-4(k)(2)&>0\\ k^2-12k+36-8k&>0\\ k^2-20k+36&>0. \end{align*}

Factorise:

\begin{align*} k^2-20k+36=&\,(k-2)(k-18). \end{align*}

Since the quadratic is positive outside its roots,

\begin{align*} k<2\quad\text{or}\quad k>18. \end{align*}

Given $k>0$ ,

\begin{align*} 0<k<2\quad\text{or}\quad k>18. \end{align*}

IAL 2025 Jan Q7

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程