MI Promo 9758/2020/PU1/Q2（修改自第 (ii) 问）

1. Rearrange the inequality to compare with zero and express it as a single algebraic fraction.

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

2. Expand, simplify and factorise the numerator. Adjust the inequality sign when reversing the signs to make the leading coefficient positive.

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

3. Identify critical values and use a sign test (number line) to determine the solution intervals, ensuring the denominator is strictly non-zero.

Critical values are $x = -2$, $x = 1$, and $x = \frac{3}{2}$.
Since $x - 1$ is in the denominator, we must have $x \neq 1$.

Testing the intervals:

• For $x \le -2$: $\frac{(-)(-)}{(-)} \le 0$ (Valid)
• For $-2 < x < 1$: $\frac{(-)(+)}{(-)} > 0$ (Invalid)
• For $1 < x \le \frac{3}{2}$: $\frac{(-)(+)}{(+)} \le 0$ (Valid)
• For $x > \frac{3}{2}$: $\frac{(+)(+)}{(+)} > 0$ (Invalid)

Solution: $x \le -2 \text{ or } 1 < x \le \frac{3}{2}$

1. Reduce the given inequality to the structure of part (i)

Since $e^x>0$ for all real $x$, the factor $2e^x+3 \ge 3 > 0$ is always positive. Multiplying both sides of $\frac{3}{(e^x-1)(2e^x+3)} \ge 1$ by the positive quantity $(2e^x+3)$ leaves the inequality sign unchanged: $\frac{3}{e^x-1} \ge 2e^x+3.$ This is exactly the inequality in part (i) with $x$ replaced by $e^x$.

2. Apply the part (i) solution to $e^x$

From part (i), $\dfrac{3}{X-1} \ge 2X+3$ has solution $X \le -2$ or $1 < X \le \dfrac{3}{2}$. With $X = e^x$: $e^x \le -2 \quad\text{or}\quad 1 < e^x \le \tfrac{3}{2}.$ The branch $e^x \le -2$ has no solution because $e^x>0$.

3. Solve the remaining exponential inequality exactly

From $1 < e^x \le \tfrac{3}{2}$, taking $\ln$ (an increasing function) of all three parts: $\ln 1 < x \le \ln\tfrac{3}{2} \;\Rightarrow\; 0 < x \le \ln\tfrac{3}{2}.$