2021/Prelim/HCI/P1/Q3

1. Rearrange the inequality to compare with zero and factorise the denominators.

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

2. Combine the fractions over a common denominator and simplify.

$\frac{2(2x^2 + 3x) - (2x - 1)}{2(2x - 1)(x + 1)} \le 0$
$\frac{4x^2 + 6x - 2x + 1}{2(2x - 1)(x + 1)} \le 0$
$\frac{4x^2 + 4x + 1}{2(2x - 1)(x + 1)} \le 0$

3. Factorise the numerator into a perfect square.

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

4. Analyse the signs of the factors to deduce the solution.

Since $(2x + 1)^2 \ge 0$ for all real $x$, the fraction is $\le 0$ when the denominator is strictly negative, or when the numerator is exactly zero.
The numerator is zero when $x = -\frac{1}{2}$.
The denominator is negative when $(2x - 1)(x + 1) < 0$, which gives $-1 < x < \frac{1}{2}$.
Since $x = -\frac{1}{2}$ is included in this interval, the overall solution is:
$-1 < x < \frac{1}{2}$

1. Solve the inequality from part (i) concisely to obtain the valid range of values.

$\text{Let the variable in part (i) be } y:$
$\frac{2y^2 + 3y}{(2y - 1)(y + 1)} - \frac{1}{2(y + 1)} \le 0$
$\frac{4y^2 + 4y + 1}{2(2y - 1)(y + 1)} \le 0 \implies \frac{(2y + 1)^2}{2(2y - 1)(y + 1)} \le 0$
$\text{Since } (2y + 1)^2 \ge 0 \text{, we require } (2y - 1)(y + 1) < 0 \text{ or } (2y + 1)^2 = 0$
$\implies -1 < y < \frac{1}{2}$

2. Substitute $\cos x$ into the solution of part (i).

$\text{Comparing the given inequality with part (i), we substitute } y = \cos x:$
$-1 < \cos x < \frac{1}{2}$

3. Solve the trigonometric inequalities for $x$ in the domain $-\pi \le x \le \pi$.

$\text{For } \cos x < \frac{1}{2} \text{ in the domain } -\pi \le x \le \pi:$
$x \in \left[-\pi, -\frac{\pi}{3}\right) \cup \left(\frac{\pi}{3}, \pi\right]$
$\text{For } \cos x > -1 \text{, since } \cos x \ge -1 \text{ for all } x \text{, we just exclude values where } \cos x = -1:$
$x \neq -\pi \text{ and } x \neq \pi$
$\text{Taking the intersection of these conditions:}$
$-\pi < x < -\frac{\pi}{3} \text{ or } \frac{\pi}{3} < x < \pi$