IAL 2020 Oct Q3 | 國際數學站

Use algebra to obtain the set of values of $x$ for which

\begin{align*} \left\vert \frac{x^2 + 3x + 10}{x + 2} \right\vert < 7 - x \end{align*}

(9)

解答

\begin{align*} &\,\frac{x^2 + 3x + 10}{x + 2} < 7 - x \\[4mm] \iff &\,\frac{x^2 + 3x + 10 - (7 - x)(x + 2)}{x + 2} < 0\\[4mm] \iff &\,\frac{x^2 + 3x + 10 - (-x^2 + 5x + 14)}{x + 2} < 0\\[4mm] \iff &\,\frac{2x^2 - 2x - 4}{x + 2} < 0\\[4mm] \iff &\,\frac{(x - 2)(x + 1)}{x + 2} < 0 \end{align*}

so $x < -2$ or $-1 < x < 2$ .

\begin{align*} &\,\frac{x^2 + 3x + 10}{x + 2} > -(7 - x) \\[4mm] \iff &\,\frac{x^2 + 3x + 10 - (x - 7)(x + 2)}{x + 2} > 0\\[4mm] \iff &\,\frac{x^2 + 3x + 10 - (x^2 - 5x - 14)}{x + 2} > 0\\[4mm] \iff &\,\frac{8x + 24}{x + 2} > 0\\[4mm] \iff &\,\frac{x + 3}{x + 2} > 0 \end{align*}

so $x < -3$ or $x > -2$ .

Hence the required set of values of $x$ is

x < -3 \quad \text{or} \quad -1 < x < 2.