2021/Prelim/HCI/P1/Q3

1. Rearrange the inequality to have zero on one side and factorise the denominators.

$\dfrac{2x^2 + 3x}{2x^2 + x - 1} - \dfrac{1}{2x + 2} \le 0$

$\dfrac{2x^2 + 3x}{(2x - 1)(x + 1)} - \dfrac{1}{2(x + 1)} \le 0$

2. Combine the terms into a single algebraic fraction and simplify the numerator.

$\dfrac{2(2x^2 + 3x) - (2x - 1)}{2(2x - 1)(x + 1)} \le 0$

$\dfrac{4x^2 + 6x - 2x + 1}{2(2x - 1)(x + 1)} \le 0$

$\dfrac{4x^2 + 4x + 1}{2(2x - 1)(x + 1)} \le 0$

$\dfrac{(2x + 1)^2}{2(2x - 1)(x + 1)} \le 0$

3. Identify the critical values of the rational expression.

The numerator gives a critical value when $(2x + 1)^2 = 0 \implies x = -\dfrac{1}{2}$.

The denominator gives critical values when $2x - 1 = 0 \implies x = \dfrac{1}{2}$ and when $x + 1 = 0 \implies x = -1$.

4. Perform a sign test to determine the interval(s) satisfying the inequality.

Since the numerator $(2x + 1)^2 \ge 0$ for all real $x$, the sign of the fraction is determined by the denominator $(2x - 1)(x + 1)$.

Testing the intervals separated by $x = -1, -\dfrac{1}{2}, \dfrac{1}{2}$:

• For $x < -1$: $(-) \times (-) > 0 \implies$ fraction $> 0$
• For $-1 < x < -\dfrac{1}{2}$: $(-) \times (+) < 0 \implies$ fraction $< 0$
• For $x = -\dfrac{1}{2}$: fraction $= 0$ (satisfies $\le 0$)
• For $-\dfrac{1}{2} < x < \dfrac{1}{2}$: $(-) \times (+) < 0 \implies$ fraction $< 0$
• For $x > \dfrac{1}{2}$: $(+) \times (+) > 0 \implies$ fraction $> 0$

Excluding the asymptotes $x = -1$ and $x = \dfrac{1}{2}$, the inequality is satisfied for:
$-1 < x < \dfrac{1}{2}$

1. Concisely solve the corresponding algebraic inequality from part (i) to obtain the valid range for the substitution.

First, we concisely recover the result of part (i) which is required here.\nLet $y$ be the variable in part (i):\n$\frac{2y^2 + 3y}{(2y-1)(y+1)} - \frac{1}{2(y+1)} \le 0$\n$\frac{2(2y^2 + 3y) - (2y-1)}{2(2y-1)(y+1)} \le 0$\n$\frac{4y^2 + 4y + 1}{2(2y-1)(y+1)} \le 0$\n$\frac{(2y+1)^2}{2(2y-1)(y+1)} \le 0$\nSince $(2y+1)^2 \ge 0$ for all real $y$, the fraction is $\le 0$ when the denominator is strictly negative (the numerator being zero at $y = -\frac{1}{2}$ is naturally included in this case):\n$2(2y-1)(y+1) < 0 \implies -1 < y < \frac{1}{2}$

2. Substitute the trigonometric function into the algebraic solution.

Using the answer from part (i), substitute $y = \cos x$:\n$-1 < \cos x < \frac{1}{2}$

3. Solve the trigonometric inequality over the domain $-\pi \le x \le \pi$.

For $\cos x < \frac{1}{2}$ in the domain $-\pi \le x \le \pi$, the reference angle is $\frac{\pi}{3}$. This gives the range:\n$-\pi \le x < -\frac{\pi}{3} \quad \text{or} \quad \frac{\pi}{3} < x \le \pi$\nHowever, the inequality also requires $\cos x > -1$, which means $\cos x \neq -1$. \nIn the given domain, $\cos x = -1$ occurs at $x = -\pi$ and $x = \pi$. \nExcluding these endpoints, we obtain the exact solution:\n$-\pi < x < -\frac{\pi}{3} \quad \text{or} \quad \frac{\pi}{3} < x < \pi$