IAL 2024 June Q5 | 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.

Use algebra to determine the values of $x$ for which

\begin{align*} \frac{x + 1}{(x - 3)(x + 2)} \leqslant 1 - \frac{2}{x - 3} \end{align*}

(6)

解答

Given the inequality:

\begin{align*} \frac{x+1}{(x-3)(x+2)} \leqslant &\,1 - \frac{2}{x-3} \end{align*}

Rearrange to bring all terms to one side:

\begin{align*} \frac{x+1}{(x-3)(x+2)} - 1 + \frac{2}{x-3} \leqslant &\,0 \end{align*}

Combine into a single fraction with common denominator $(x-3)(x+2)$ :

\begin{align*} \frac{x+1 - (x-3)(x+2) + 2(x+2)}{(x-3)(x+2)} \leqslant &\,0\\[4mm] \frac{x+1 - (x^2 - x - 6) + 2x + 4}{(x-3)(x+2)} \leqslant &\,0\\[4mm] \frac{x + 1 - x^2 + x + 6 + 2x + 4}{(x-3)(x+2)} \leqslant &\,0\\[4mm] \frac{-x^2 + 4x + 11}{(x-3)(x+2)} \leqslant &\,0 \end{align*}

Multiply the inequality by $-1$ , which flips the inequality sign:

\begin{align*} \frac{x^2 - 4x - 11}{(x-3)(x+2)} \geqslant &\,0 \end{align*}

Find the critical values. From the denominator, we have $x = 3$ and $x = -2$ . From the numerator, solve $x^2 - 4x - 11 = 0$ using the quadratic formula:

\begin{align*} x = &\,\frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-11)}}{2(1)}\\[4mm] = &\,\frac{4 \pm \sqrt{16 + 44}}{2}\\[4mm] = &\,\frac{4 \pm \sqrt{60}}{2}\\[4mm] = &\,\frac{4 \pm 2\sqrt{15}}{2}\\[4mm] = &\,2 \pm \sqrt{15} \end{align*}

The critical values in increasing order are: $-2$ , $2 - \sqrt{15}$ (approx. $-1.87$ ), $3$ , and $2 + \sqrt{15}$ (approx. $5.87$ ).

We test the sign of $\frac{(x - (2+\sqrt{15}))(x - (2-\sqrt{15}))}{(x-3)(x+2)}$ in the intervals defined by these critical values:

For $x > 2 + \sqrt{15}$ , the expression is positive.
For $3 < x < 2 + \sqrt{15}$ , the expression is negative.
For $2 - \sqrt{15} < x < 3$ , the expression is positive.
For $-2 < x < 2 - \sqrt{15}$ , the expression is negative.
For $x < -2$ , the expression is positive.

We need the expression to be $\geqslant 0$ . The numerator can be exactly zero, so we include $2 \pm \sqrt{15}$ , but the denominator cannot be zero, so $x \neq 3$ and $x \neq -2$ .

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

\begin{align*} x < -2 \quad \text{or} \quad 2 - \sqrt{15} \leqslant x < 3 \quad \text{or} \quad x \geqslant 2 + \sqrt{15} \end{align*}