题目

Problem

The curve $C$ has equation

\begin{align*} y=\frac{1}{2-x}. \end{align*}

(a) Sketch $C$ , showing the coordinates of any points of intersection with the coordinate axes and giving the equations of the asymptotes.

(3)

The line with equation $y=4x+k$ , where $k$ is a constant, meets $C$ at two distinct points.

(b) Show that

\begin{align*} k^2+16k+48>0. \end{align*}

(4)

解答

(a)

解法一

思路

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分母为零给出竖直渐近线。分母越来越大时，函数值趋近于 $0$ ，所以水平渐近线是 $y=0$ 。代入 $x=0$ 可得 $y$ 轴截距。

答题过程

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The vertical asymptote is

\begin{align*} 2-x=0 \quad\Rightarrow\quad x=2. \end{align*}

The horizontal asymptote is

\begin{align*} y=0. \end{align*}

At $x=0$ ,

\begin{align*} y=\frac12. \end{align*}

So the curve crosses the $y$ -axis at $\left(0,\frac12\right)$ and has no $x$ -intercept. The sketch should have one branch above the $x$ -axis for $x<2$ and one branch below the $x$ -axis for $x>2$ .

(b)

解法一

思路

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交点满足直线方程和曲线方程同时成立。把它们联立后得到关于 $x$ 的二次方程。若有两个不同交点，二次方程必须有两个不同实根，所以判别式要大于 $0$ 。

答题过程

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At the points of intersection,

\begin{align*} 4x+k=&\,\frac{1}{2-x}. \end{align*}

Multiplying by $2-x$ gives

\begin{align*} (4x+k)(2-x)=&\,1\\ 8x+2k-4x^2-kx=&\,1\\ 4x^2+(k-8)x+1-2k=&\,0. \end{align*}

For two distinct intersections, the discriminant must be positive:

\begin{align*} (k-8)^2-4(4)(1-2k)&>0\\ k^2-16k+64-16+32k&>0\\ k^2+16k+48&>0. \end{align*}

(c)

解法一

思路

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把二次不等式因式分解。因为 $k^2$ 的系数是正数，抛物线开口向上，所以取两个根外侧的范围。

答题过程

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\begin{align*} k^2+16k+48&>0\\ (k+12)(k+4)&>0. \end{align*}

The critical values are

\begin{align*} k=-12,\qquad k=-4. \end{align*}

Since the quadratic is positive outside these two values,

\begin{align*} k<-12 \quad\text{or}\quad k>-4. \end{align*}

IAL 2020 Oct 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

(c)

解法一

思路

答题过程