题目

Problem

(a) Given that $k$ is a positive constant such that $0<k<4$ sketch, on separate axes, the graphs of

(i) $y=(2x-k)(x+4)^2$

(ii) $y=\dfrac{k}{x^2}$

showing the coordinates of any points where the graphs cross or meet the coordinate axes, leaving coordinates in terms of $k$ , where appropriate.

(5)

(b) State, with a reason, the number of roots of the equation

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

(1)

解答

(a)(i)

解法一

思路

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这是一个正首项系数的三次图像。因式 $(x+4)^2$ 给出 $x=-4$ 的重根，所以图像在 $(-4,0)$ 处接触 $x$ 轴；因式 $2x-k$ 给出 $x=\frac{k}{2}$ 的单根，所以图像在这里穿过 $x$ 轴。

答题过程

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For

\begin{align*} y=(2x-k)(x+4)^2, \end{align*}

the $x$ -intercepts are found from

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

\begin{align*} x=\frac{k}{2} \quad\text{or}\quad x=-4. \end{align*}

The graph crosses the $x$ -axis at

\begin{align*} \left(\frac{k}{2},0\right), \end{align*}

and meets the $x$ -axis at

\begin{align*} (-4,0). \end{align*}

The $y$ -intercept is found by setting $x=0$ :

\begin{align*} y=&\,(2(0)-k)(0+4)^2\\ =&\,-16k. \end{align*}

So the $y$ -intercept is

\begin{align*} (0,-16k). \end{align*}

The sketch should be a positive cubic, touching the $x$ -axis at $(-4,0)$ and crossing the $x$ -axis at $\left(\frac{k}{2},0\right)$ .

(a)(ii)

解法一

思路

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因为 $k>0$ 且 $x^2>0$ ，所以 $y=\frac{k}{x^2}$ 永远为正。图像在第一、第二象限，靠近但不接触两条坐标轴。

答题过程

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For

\begin{align*} y=\frac{k}{x^2}, \end{align*}

there are no intercepts with the coordinate axes.

The graph has two branches, one in quadrant I and one in quadrant II, with asymptotes

\begin{align*} x=0 \quad\text{and}\quad y=0. \end{align*}

(b)

解法一

思路

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方程的根数就是两张图像的交点个数。根据图像位置，三次图像与 $y=\frac{k}{x^2}$ 只相交一次。

答题过程

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The number of roots is the number of intersections of the two graphs.

From the sketches, the graphs intersect once only.

Therefore the equation has

\begin{align*} 1 \end{align*}

root.

IAL 2022 Oct Q6

题目

解答

(a)(i)

解法一

思路

答题过程

(a)(ii)

解法一

思路

答题过程

(b)

解法一

思路

答题过程