题目

Problem

(a) Sketch the curve with equation

\begin{align*} y=-\frac{k}{x},\qquad k>0,\quad x\ne0. \end{align*}

(2)

(b) On a separate diagram, sketch the curve with equation

\begin{align*} y=-\frac{k}{x}+k,\qquad k>0,\quad x\ne0, \end{align*}

stating the coordinates of the point of intersection with the $x$ -axis and, in terms of $k$ , the equation of the horizontal asymptote.

(3)

\begin{align*} y=-\frac{k}{x}+k,\qquad k>0,\quad x\ne0, \end{align*}

does not touch or intersect the line with equation $y=3x+4$ .

(5)

解答

(a)

解法一

思路

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因为 $k>0$ ， $y=-\frac{k}{x}$ 是 negative reciprocal 图像，分支在第二、第四象限。

答题过程

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The sketch should have two branches in quadrants 2 and 4, with asymptotes

\begin{align*} x=0,\qquad y=0. \end{align*}

(b)

解法一

思路

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相比 $y=-\frac{k}{x}$ ，图像向上平移 $k$ 。所以水平渐近线从 $y=0$ 变为 $y=k$ 。求 $x$ 轴交点时令 $y=0$ 。

答题过程

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The horizontal asymptote is

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

For the $x$ -intercept,

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

Since $k>0$ ,

\begin{align*} x=1. \end{align*}

So the point of intersection with the $x$ -axis is

\begin{align*} (1,0). \end{align*}

(c)

解法一

思路

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不相交也不相切，表示联立两方程后没有实数解。把方程整理成关于 $x$ 的二次方程，要求判别式小于 $0$ 。

答题过程

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Set the curve equal to the line:

\begin{align*} 3x+4=&\,-\frac{k}{x}+k. \end{align*}

Multiply by $x$ :

\begin{align*} 3x^2+4x=&\,-k+kx\\ 3x^2+(4-k)x+k=&\,0. \end{align*}

For no intersection and no touching, this quadratic must have no real roots:

\begin{align*} b^2-4ac&<0\\ (4-k)^2-4(3)(k)&<0\\ k^2-8k+16-12k&<0\\ k^2-20k+16&<0. \end{align*}

Find the critical values:

\begin{align*} k =&\,\frac{20\pm\sqrt{20^2-4(1)(16)}}{2}\\ =&\,\frac{20\pm\sqrt{336}}2\\ =&\,10\pm2\sqrt{21}. \end{align*}

Since the quadratic in $k$ opens upwards,

\begin{align*} 10-2\sqrt{21}<k<10+2\sqrt{21}. \end{align*}

IAL 2021 Jan Q6

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程