Mathematics 9709/33 — May/June 2022

Walkthrough

The inequality contains absolute values, and the most reliable algebraic way to remove them is to square both sides. Because $|3x+a|$ and $|2x+3a|$ are never negative, squaring does not reverse the inequality. Notice the $2$ on the left must also be squared: $2|3x+a|$ becomes $4(3x+a)^2$.

Expanding gives a quadratic inequality in $x$ with coefficients containing the parameter $a$:

Treat $a$ as a fixed positive number throughout. To find where this quadratic is negative, first solve the equation

The quadratic formula gives the two roots. Since $a>0$, $\sqrt{784a^2}=28a$, so the roots are $\frac{a}{4}$ and $-\frac{5a}{8}$.

Finally, because the coefficient of $x^2$ is positive, the quadratic lies below zero between its two roots. Therefore the required interval is

Strict inequality is essential: at the endpoints the expression equals zero and no longer satisfies the original strict inequality.

Key Takeaways

This question combines the modulus inequality method with quadratic inequalities involving a parameter. The key skills are:

• Squaring a modulus inequality to remove the absolute value signs while preserving the inequality.
• Remembering to square the multiplier $2$ as well as the absolute value.
• Expanding and collecting a quadratic with a parameter.
• Solving the associated quadratic equation and using the sign of the leading coefficient to choose the inequality region.
• Expressing the final solution as a strict interval when the original inequality is strict.

Common Mistakes

• Forgetting to square the $2$ and writing $2(3x+a)^2$ instead of $4(3x+a)^2$.
• Using $\leq$ instead of $<$ in the final answer; the mark scheme states that $\leq$ must not be condoned.
• Stopping after solving the quadratic equation and not converting it into the correct inequality interval.
• Solving the quadratic equation for $a$ instead of for $x$; the mark scheme awards M0 for this.
• Substituting a numerical value for $a$ instead of keeping the answer in terms of $a$; the mark scheme limits this to at most 2 marks.
• Taking $\sqrt{784a^2}$ as $-28a$ without justification; since $a>0$, the positive root is correct here.

Things to Be Careful About

• The original inequality is strict, so the endpoints are excluded: $x = -\frac{5a}{8}$ and $x = \frac{a}{4}$ are not solutions.
• Since $a$ is positive, the smaller number is written first and $-\frac{5a}{8} < \frac{a}{4}$.
• Squaring both sides is valid because both sides are non-negative; this preserves equivalence for absolute values.
• Keep the method visible in the mark scheme: state the non-modular quadratic inequality, solve the quadratic, then state the final strict interval.