2023 H2 Maths Paper 2, Question 1

Approach: both sides of $3|x - 2| < |5 - 2x|$ are non-negative, so squaring removes the modulus signs without changing the inequality’s direction — this avoids splitting into cases on the sign of $x - 2$ and $5 - 2x$.

Squaring both sides,

Solving $5x^{2} - 16x + 11 = 0$ by the quadratic formula (discriminant $\text{=}( - 16)^{2} - 4(5)(11) = 36$),

Since the coefficient of $x^{2}$ is positive, $5x^{2} - 16x + 11 < 0$ between the roots.

Approach: combine the given expression into a single fraction first — it factorises — then reuse that factorisation to solve the inequality, rather than expanding everything again from scratch.

Factorising the numerator (discriminant $\text{=}( - 11)^{2} - 4(3)(10) = 1$) and the denominator,

so

Hence, solving $\frac{(x - 2)(3x - 5)}{(x - 5)(x + 1)} > - 3$: moving $- 3$ to the left and combining into a single fraction,

Solving $6x^{2} - 23x - 5 = 0$ by the quadratic formula (discriminant $\text{=}( - 23)^{2} - 4(6)( - 5) = 649$, not a perfect square),

Since $\sqrt{649} \approx 25.5$, these roots are approximately $- 0.206$ and $4.04$. With the four critical values $x = - 1,\ \frac{23 - \sqrt{649}}{12},\ \frac{23 + \sqrt{649}}{12},\ 5$ in increasing order, a sign chart gives $\frac{6x^{2} - 23x - 5}{(x - 5)(x + 1)} > 0$ on $( - \infty, - 1)$, on $\left( \frac{23 - \sqrt{649}}{12},\frac{23 + \sqrt{649}}{12} \right)$, and on $(5,\infty)$, with $x = - 1$ and $x = 5$ excluded since the original expression is undefined there.