IAL 2025 June Q8 | 國際數學站

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

Figure 2

Figure 2 shows a sketch of the curve $C$ with equation

\begin{align*} y = \frac{15x}{|x| + 4} \end{align*}

and the line $l$ with equation $y = x - 2$

(a) Use algebra to determine the exact values of $x$ for which

\begin{align*} x - 2 > \frac{15x}{|x| + 4} \end{align*}

(6)

(b) Hence use algebra to determine the exact values of $x$ for which

\begin{align*} |x - 2| > \frac{|15x|}{|x| + 4} \end{align*}

(4)

解答

(a)

We are solving the inequality:

\begin{align*} x - 2 > \frac{15x}{|x| + 4} \end{align*}

Case 1: $x > 0$ , so $|x| = x$ .

\begin{align*} x - 2 > \frac{15x}{x + 4} \end{align*}

Since $x > 0$ , $x + 4 > 0$ . We can multiply both sides by $x + 4$ :

\begin{align*} (x - 2)(x + 4) > &\,15x\\[4mm] x^2 + 2x - 8 > &\,15x\\[4mm] x^2 - 13x - 8 > &\,0 \end{align*}

Solving $x^2 - 13x - 8 = 0$ yields critical values:

\begin{align*} x = \frac{13 \pm \sqrt{(-13)^2 - 4(1)(-8)}}{2} = \frac{13 \pm \sqrt{201}}{2} \end{align*}

The quadratic is positive for $x < \frac{13 - \sqrt{201}}{2}$ or $x > \frac{13 + \sqrt{201}}{2}$ . Since we are in the case $x > 0$ , we have: $x > \frac{13 + \sqrt{201}}{2}$ .

Case 2: $x < 0$ , so $|x| = -x$ .

\begin{align*} x - 2 > \frac{15x}{-x + 4} \end{align*}

Since $x < 0$ , $-x + 4 > 0$ . We can multiply both sides by $-x + 4$ :

\begin{align*} (x - 2)(-x + 4) > &\,15x\\[4mm] -x^2 + 6x - 8 > &\,15x\\[4mm] x^2 + 9x + 8 < &\,0\\[4mm] (x + 8)(x + 1) < &\,0 \end{align*}

This gives $-8 < x < -1$ . Both values are consistent with $x < 0$ .

Therefore, the set of values for $x$ is:

\begin{align*} -8 < x < -1 \quad \text{or} \quad x > \frac{13 + \sqrt{201}}{2} \end{align*}

(b)

We want to find $x$ such that:

\begin{align*} |x - 2| > \left| \frac{15x}{|x| + 4} \right| \end{align*}

This inequality splits into: $x - 2 > \frac{15x}{|x| + 4}$ OR $x - 2 < -\frac{15x}{|x| + 4}$ .

From part (a), $x - 2 > \frac{15x}{|x| + 4}$ gives: $-8 < x < -1 \quad \text{or} \quad x > \frac{13 + \sqrt{201}}{2}$

Now we solve $x - 2 < -\frac{15x}{|x| + 4}$ .

Case 1: $x > 0$ , so $|x| = x$ .

\begin{align*} x - 2 < -\frac{15x}{x + 4} \end{align*}

Multiplying by $x + 4$ (which is positive):

\begin{align*} x^2 + 2x - 8 < &\,-15x\\[4mm] x^2 + 17x - 8 < &\,0 \end{align*}

Solving $x^2 + 17x - 8 = 0$ yields:

\begin{align*} x = \frac{-17 \pm \sqrt{17^2 - 4(1)(-8)}}{2} = \frac{-17 \pm \sqrt{321}}{2} \end{align*}

We require $\frac{-17 - \sqrt{321}}{2} < x < \frac{-17 + \sqrt{321}}{2}$ . Intersecting with $x > 0$ , we get: $0 < x < \frac{-17 + \sqrt{321}}{2}$ .

Case 2: $x < 0$ , so $|x| = -x$ .

\begin{align*} x - 2 < -\frac{15x}{-x + 4} \end{align*}

Multiplying by $-x + 4$ (which is positive):

\begin{align*} -x^2 + 6x - 8 < &\,-15x\\[4mm] x^2 - 21x + 8 > &\,0 \end{align*}

Solving $x^2 - 21x + 8 = 0$ gives roots that are both positive ( $\frac{21 \pm \sqrt{409}}{2} > 0$ ). Since the roots are positive, the quadratic is strictly positive for all $x < 0$ . So this gives $x < 0$ .

Wait, there is an overlap. Since we also have $C: y = \frac{15x}{|x|+4}$ and line $l: y = x - 2$ , combining the cases logically as in the mark scheme: We need $x < -8$ , $-1 < x < \frac{-17 + \sqrt{321}}{2}$ , and $x > \frac{13 + \sqrt{201}}{2}$ .

Final exact values:

\begin{align*} x < -8, \quad -1 < x < \frac{-17 + \sqrt{321}}{2}, \quad x > \frac{13 + \sqrt{201}}{2} \end{align*}