Inequalities: Rational & \(|x|\) substitution — ACJC 2025 H2 Math Prelim Paper 1
Anglo-Chinese Junior College2025 PrelimPaper 1●●○ Standard4 marks
What this question tests
Rational & \(|x|\) substitution.
Question
- Without using a calculator, solve the inequality \[ \frac{x}{x+2} \geq \frac{2}{2-x}\,. \]
- Hence, solve the inequality \[ \frac{|x|}{|x|+2} \geq \frac{2}{2-|x|}\,. \]
Show full worked solution▾
(a)
\[\begin{aligned} \frac{x}{x+2} &\geq \frac{2}{2-x}\\[4pt] \frac{x}{x+2} - \frac{2}{2-x} &\geq 0\\[4pt] \frac{x(2-x) - 2(x+2)}{(x+2)(2-x)} &\geq 0\\[4pt] \frac{-x^2 - 4}{(x+2)(2-x)} &\geq 0\\[4pt] \frac{x^2+4}{(x+2)(x-2)} &\geq 0 \end{aligned}\]Since \(x^2+4>0\) for all real \(x\), consider \((x+2)(x-2)\geq 0\), i.e. \((x-2)(x+2)\geq 0\).
(b)
Replace \(x\) with \(|x|\): from (a), \(|x|<-2\) has no solutions, so we need \(|x|>2\), giving \(x<-2\) or \(x>2\).
Answer: (a) \(x<-2\) or \(x>2\) (b) \(x<-2\) or \(x>2\)