2017/Prelim/HCI/P1/Q3

1. Express the given quadratic expression in completed square form.

$\begin{aligned} 3x - x^2 - 4 &= -(x^2 - 3x) - 4 \\ &= -\left( x^2 - 3x + \left(\frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^2 \right) - 4 \\ &= -\left( x - \frac{3}{2} \right)^2 + \frac{9}{4} - 4 \\ &= -\left( x - \frac{3}{2} \right)^2 - \frac{7}{4} \end{aligned}$

2. Deduce that the expression is always negative.

Since $\left( x - \frac{3}{2} \right)^2 \ge 0$ for all real values of $x$,
$-\left( x - \frac{3}{2} \right)^2 \le 0$.

Therefore, $-\left( x - \frac{3}{2} \right)^2 - \frac{7}{4} \le -\frac{7}{4} < 0$.

Hence, $3x - x^2 - 4$ is always negative for all real values of $x$.

1. Use the result from part (i) to simplify the rational inequality.

From part (i), we know that $3x - x^2 - 4 < 0$ for all real values of $x$.
Dividing both sides of the given inequality by this strictly negative expression reverses the inequality sign:
$\frac{(x - 1)^2}{x^2 - 2x - 5} \ge 0$

2. Identify the condition where the numerator satisfies the inequality.

For the rational expression to be $\ge 0$, the inequality is satisfied if the numerator is exactly $0$ (provided the denominator is non-zero).
$(x - 1)^2 = 0 \implies x = 1$
At $x = 1$, the denominator is $1^2 - 2(1) - 5 = -6 \neq 0$.
Thus, $x = 1$ is a valid solution.

3. Determine the roots of the denominator to find critical values.

For $x \neq 1$, the numerator $(x - 1)^2$ is strictly positive. Therefore, for the entire fraction to be strictly positive (to satisfy $\ge 0$), the denominator must also be strictly positive:
$x^2 - 2x - 5 > 0$
We find the critical values by solving $x^2 - 2x - 5 = 0$ using the quadratic formula:
$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-5)}}{2(1)}$
$x = \frac{2 \pm \sqrt{4 + 20}}{2} = \frac{2 \pm \sqrt{24}}{2}$
$x = \frac{2 \pm 2\sqrt{6}}{2} = 1 \pm \sqrt{6}$

4. Solve the quadratic inequality for the denominator and state the combined final solution.

Since the coefficient of $x^2$ is positive ($1 > 0$), the parabola $y = x^2 - 2x - 5$ opens upwards.
Hence, the regions where $x^2 - 2x - 5 > 0$ are:
$x < 1 - \sqrt{6} \text{ or } x > 1 + \sqrt{6}$
Combining this with the solution from the numerator, the complete exact solution is:
$x < 1 - \sqrt{6} \text{ or } x = 1 \text{ or } x > 1 + \sqrt{6}$