Inequalities: Rational then \(|x+1|\) substitution — TJC 2025 H2 Math Prelim Paper 1
Temasek Junior College2025 PrelimPaper 1●●○ Standard7 marks
What this question tests
Rational then \(|x+1|\) substitution.
Question
- Without using a calculator, solve the inequality \(\dfrac{2x - 25}{x^2 - 2x - 3} \le 2\).
- Hence solve the inequality \(\dfrac{2|x+1| - 25}{(x+1)^2 - 2|x+1| - 3} \le 2\).
Show full worked solution▾
(a) \[ \frac{2x-25}{x^2-2x-3} \leq 2 \implies \frac{2x-25 - 2(x^2-2x-3)}{x^2-2x-3} \leq 0 \implies \frac{-2x^2+6x-19}{x^2-2x-3} \leq 0 \] \[ \implies \frac{-2\!\left(x-\tfrac{3}{2}\right)^2 - \tfrac{29}{2}}{x^2-2x-3} \leq 0 \quad\cdots(1) \]
Since the numerator \(-2\!\left(x-\tfrac{3}{2}\right)^2 - \tfrac{29}{2} < 0\) for all real \(x\), we require \(x^2 - 2x - 3 > 0\):
\((x-3)(x+1) > 0 \implies x < -1\) or \(x > 3\).
(b) Replace \(x\) by \(|x+1|\) in part (a):
From (a): \(|x+1| < -1\) (no solution) or \(|x+1| > 3\).
\(|x+1| > 3 \implies x+1 > 3\) or \(x+1 < -3 \implies x > 2\) or \(x < -4\).
Answer: (a) \(x < -1\) or \(x > 3\) (b) \(x < -4\) or \(x > 2\)