Inequalities: Rational & \(\cos x\) restricted range — TMJC 2025 H2 Math Prelim Paper 2
What this question tests
Question
(a) Without using a calculator, solve the inequality \[ \frac{x+5}{-2x^2+5x+3} < 1. \]
(b) Hence, solve exactly the inequality \[ \frac{\cos x + 5}{-2\cos^2 x + 5\cos x + 3} < 1 \quad\text{for } 0 \le x \le \pi. \]
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(a) \[\begin{aligned} \frac{x+5}{-2x^2+5x+3} &< 1\\ \frac{x+5}{-2x^2+5x+3} - 1 &< 0\\ \frac{x+5-(-2x^2+5x+3)}{-2x^2+5x+3} &< 0\\ \frac{2x^2 - 4x + 2}{-2x^2+5x+3} &< 0\\ \frac{2(x-1)^2}{(3-x)(2x+1)} &< 0, \quad x\ne -\tfrac{1}{2},\; x\ne 3. \end{aligned}\] The numerator \(2(x-1)^2 \ge 0\) for all \(x\), and equals \(0\) at \(x=1\) (not part of the strict inequality). The sign of the quotient is governed by the sign of the denominator \((3-x)(2x+1)\):
The expression is negative on \(\left(-\infty, -\tfrac{1}{2}\right)\cup(3,\infty)\).
(b) Replace \(x\) by \(\cos x\): the inequality becomes \(\cos x < -\tfrac{1}{2}\) or \(\cos x > 3\). Since \(|\cos x|\leq 1\), reject \(\cos x > 3\). So we need \(\cos x < -\tfrac{1}{2}\).
For \(0 \le x \le \pi\), \(\cos x < -\tfrac{1}{2}\) when \(\tfrac{2\pi}{3} < x \le \pi\).