MI Promo 9758/2020/PU1/Q2 (modified (ii))

1. Rearrange the inequality so that one side is zero and combine the terms into a single fraction.

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

2. Expand and simplify the numerator.

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

3. Multiply by $-1$ to make the leading coefficient positive, flip the inequality sign, and factorise the quadratic numerator.

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

4. Identify critical values and use a sign diagram to determine the solution ranges, taking care to exclude the denominator's zero.

Critical values are $x = -2$, $x = 1$, and $x = \frac{3}{2}$.

Using a sign test or number line:

• For $x < -2$, expression is negative.
• For $-2 < x < 1$, expression is positive.
• For $1 < x < \frac{3}{2}$, expression is negative.
• For $x > \frac{3}{2}$, expression is positive.

Since we require the expression to be $\le 0$ and $x - 1 \neq 0$ (so $x \neq 1$), the solution is:
$x \le -2 \text{ or } 1 < x \le \frac{3}{2}$

1. Concisely solve the inequality from part (i) to establish the prerequisite result.

$\frac{3}{x - 1} \ge 2x + 3$
$\frac{3 - (2x+3)(x-1)}{x-1} \ge 0$
$\frac{3 - (2x^2 + x - 3)}{x - 1} \ge 0$
$\frac{-2x^2 - x + 6}{x - 1} \ge 0$
$\frac{2x^2 + x - 6}{x - 1} \le 0$
$\frac{(2x - 3)(x + 2)}{x - 1} \le 0$
Testing intervals on a number line gives the solution:
$x \le -2 \quad \text{or} \quad 1 < x \le \frac{3}{2}$

2. Substitute $u = e^x$ into the inequality in (ii) and relate it to the inequality in (i).

Let $u = e^x$. The inequality in (ii) becomes:
$\frac{3}{(u - 1)(2u + 3)} \ge 1$
Since $u = e^x > 0$ for all real $x$, we know that $2u + 3 > 3 > 0$.
Multiplying both sides by the positive expression $(2u + 3)$ preserves the inequality direction:
$\frac{3}{u - 1} \ge 2u + 3$

3. Apply the solution from part (i) to find the valid range for $u$.

This inequality is identical to the one solved in (i). Using the result, we have:
$u \le -2 \quad \text{or} \quad 1 < u \le \frac{3}{2}$
Since $u = e^x > 0$, the condition $u \le -2$ is rejected.
Thus, we must have:
$1 < u \le \frac{3}{2}$

4. Solve for $x$ exactly by taking the natural logarithm.

$1 < e^x \le \frac{3}{2}$
Taking the natural logarithm ($\ln$) on all parts:
$\ln 1 < x \le \ln \frac{3}{2}$
$0 < x \le \ln \frac{3}{2}$