Inequalities: Modulus & rational inequalities — NYJC 2025 H2 Math Prelim Paper 2
Nanyang Junior College2025 PrelimPaper 2●●○ Standard6 marks
What this question tests
Modulus & rational inequalities.
Question
- Find the set of values of \(x\) for which \(3|x-3|\leq|1-2x|\).
- Without using a calculator, solve \[ \frac{x+18}{x^2+5x-14}\geq -1. \]
Show full worked solution▾
(a)
Method 1 (GC) From GC, the set required is \(\{x\in\mathbb{R}:2\leq x\leq 8\}\).
Method 2 (Algebraic) \(3|x-3|\leq|1-2x|\)
Squaring both sides (both sides non-negative): \[\begin{aligned} 9(x-3)^2 &\leq (1-2x)^2\\ 9x^2-54x+81 &\leq 1-4x+4x^2\\ 5x^2-50x+80 &\leq 0\\ x^2-10x+16 &\leq 0\\ (x-2)(x-8) &\leq 0 \end{aligned}\]
(b)
\(\dfrac{x+18}{x^2+5x-14}\geq -1\) \[\begin{aligned} \frac{x+18+(x^2+5x-14)}{(x+7)(x-2)} &\geq 0 \qquad x\neq -7\text{ or }2\\ \frac{x^2+6x+4}{(x+7)(x-2)} &\geq 0\\ \frac{(x+3)^2-5}{(x+7)(x-2)} &\geq 0\\ \frac{(x+3-\sqrt{5})(x+3+\sqrt{5})}{(x+7)(x-2)} &\geq 0 \end{aligned}\]Using a sign diagram with critical values \(-7,\;-3-\sqrt{5},\;-3+\sqrt{5},\;2\):
Answer: % (a) \(\{x\in\mathbb{R}:2\leq x\leq 8\}\) (b) \(x<-7\) or \(-3-\sqrt{5}\leq x\leq -3+\sqrt{5}\) or \(x>2\)%