IAL 2025 June A Q1 | 國際數學站

Figure 1 In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.

Figure 1 shows a sketch of the curve $C$ with equation

\begin{align*} y = \frac{10x}{x - 6} \qquad x \neq 6 \end{align*}

and the line $l$ with equation $y = 2x + 12$

(a) Use algebra to determine the $x$ coordinates of the points of intersection of $C$ and $l$

(2)

Determine the range of values of $x$ for which

(i) $2x + 12 > \dfrac{10x}{x - 6}$

(ii) $\left|2x + 12\right| > \left|\dfrac{10x}{x - 6}\right|$

(4)

解答

(a)

To find the $x$ -coordinates of the points of intersection of $C$ and $l$ , we set their equations equal to each other:

\begin{align*} \frac{10x}{x-6} = &\,2x+12\\[4mm] 10x = &\,(2x+12)(x-6)\\[4mm] 10x = &\,2x^2 - 12x + 12x - 72\\[4mm] 2x^2 - 10x - 72 = &\,0\\[4mm] x^2 - 5x - 36 = &\,0\\[4mm] (x+4)(x-9) = &\,0 \end{align*}

Hence, the $x$ -coordinates are $x = -4$ and $x = 9$ .

(b)(i)

From the sketch and the points of intersection found in (a), the line $y=2x+12$ is above the curve $y=\frac{10x}{x-6}$ between $x=-4$ and the asymptote $x=6$ , and also for $x>9$ . Thus, the range of values is:

\begin{align*} -4 < x < 6 \quad \text{or} \quad x > 9 \end{align*}

(b)(ii)

We need to solve $|2x+12| > \left|\frac{10x}{x-6}\right|$ . First, find the points where $2x+12 = -\frac{10x}{x-6}$ :

\begin{align*} -\frac{10x}{x-6} = &\,2x+12\\[4mm] -10x = &\,2x^2 - 72\\[4mm] 2x^2 + 10x - 72 = &\,0\\[4mm] x^2 + 5x - 36 = &\,0\\[4mm] (x+9)(x-4) = &\,0 \end{align*}

The critical values for the reflected parts are $x = -9$ and $x = 4$ . By considering the shape of the modulus graphs or testing regions, the required range of values is:

\begin{align*} x < -9, \quad -4 < x < 4, \quad \text{or} \quad x > 9 \end{align*}