Mathematics 9709/32 — October/November 2012

Walkthrough

The modulus sign makes the inequality piecewise, but squaring both sides is a clean way to remove it because $|u|^2 = u^2$. Since both $3|x-1|$ and $|2x+1|$ are non-negative, squaring preserves the direction of the inequality. Expand the squares to get a quadratic inequality. Then bring all terms to one side and factorise. The roots of the quadratic are the critical values where the two sides are equal. Because the quadratic has a positive $x^2$ coefficient, its graph is a U-shape, so it is negative between the roots. Therefore the solution set is the open interval between $\frac{2}{5}$ and $4$.

Key Takeaways

• Squaring is a valid way to solve modulus inequalities because both sides are non-negative.
• The critical values come from solving the corresponding equality.
• A quadratic inequality with a positive leading coefficient is satisfied between its roots when the inequality is $<0$.
• Strict inequality means the endpoints are not included.

Common Mistakes

• Forgetting to square the coefficient $3$: writing $3(x-1)^2$ instead of $(3(x-1))^2$. This gives wrong critical values.
• Using $\leq$ instead of $<$; the mark scheme explicitly says do not condone $\leq$.
• Incorrect sign of the quadratic: mixing up less than zero and greater than zero intervals.
• Not checking the direction of inequality after squaring. Here it is preserved, but students should remember that squaring is only valid for non-negative quantities.

Things to Be Careful About

• The critical values are $x = \frac{2}{5}$ and $x = 4$, not the roots of the original linear equations without considering coefficients.
• Since the original inequality is strict, the answer must be open intervals: $\frac{2}{5} < x < 4$, not $\frac{2}{5} \leq x \leq 4$.
• When squaring, both sides are non-negative, so no sign reversal occurs. If one side could be negative, squaring would require more care.