IAL 2024 Jan Q1 | 國際數學站

Using algebra, solve the inequality

\begin{align*} \frac{1}{x + 2} > 2x + 3 \end{align*}

(5)

解答

Given the inequality:

\begin{align*} \frac{1}{x+2} > &\,2x+3 \end{align*}

Rearrange the inequality to bring all terms to one side:

\begin{align*} \frac{1}{x+2} - (2x+3) > &\,0\\[4mm] \frac{1 - (2x+3)(x+2)}{x+2} > &\,0\\[4mm] \frac{1 - (2x^2 + 7x + 6)}{x+2} > &\,0\\[4mm] \frac{-2x^2 - 7x - 5}{x+2} > &\,0 \end{align*}

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

\begin{align*} \frac{2x^2 + 7x + 5}{x+2} < &\,0\\[4mm] \frac{(2x+5)(x+1)}{x+2} < &\,0 \end{align*}

The critical values are $x = -2$ , $x = -\frac{5}{2}$ , and $x = -1$ . Arranging these in order, we have $-\frac{5}{2}$ , $-2$ , and $-1$ .

We test the sign of $\frac{(2x+5)(x+1)}{x+2}$ in the intervals defined by the critical values:

For $x < -\frac{5}{2}$ , the expression is $\frac{(-)(-)}{(-)}$ , which is negative.
For $-\frac{5}{2} < x < -2$ , the expression is $\frac{(+)(-)}{(-)}$ , which is positive.
For $-2 < x < -1$ , the expression is $\frac{(+)(-)}{(+)}$ , which is negative.
For $x > -1$ , the expression is $\frac{(+)(+)}{(+)}$ , which is positive.

We need the expression to be strictly less than zero. Thus, the correct range of values is:

\begin{align*} x < -\frac{5}{2} \quad \text{or} \quad -2 < x < -1 \end{align*}