Mathematics 9709/31 — October/November 2018

Walkthrough

The inequality contains absolute values, so we cannot simply multiply out without considering cases. A standard and efficient way is to square both sides. This is valid because both $2|2x-a|$ and $|x+3a|$ are non-negative for every value of $x$. Squaring gives $4(2x-a)^2 < (x+3a)^2$.

Next, expand both sides. On the left, $4(2x-a)^2 = 16x^2 - 16ax + 4a^2$. On the right, $(x+3a)^2 = x^2 + 6ax + 9a^2$. Subtract the right side from the left to obtain $15x^2 - 22ax - 5a^2 < 0$. This step is often the place where sign errors occur, so make sure each term is collected carefully.

We now have a quadratic inequality in $x$, with $a$ treated as a positive constant. Factorise $15x^2 - 22ax - 5a^2$ as $(3x - 5a)(5x + a)$. The roots are $x = \frac{5}{3}a$ and $x = -\frac{1}{5}a$. Since $a>0$, $-\frac{1}{5}a < \frac{5}{3}a$, so these roots are ordered left to right.

For a quadratic inequality with positive leading coefficient, the expression is negative between its roots. Hence the required interval is $-\frac{1}{5}a < x < \frac{5}{3}a$. Because the original inequality is strict, the endpoints are not included.

Key Takeaways

This problem tests the method of squaring to remove modulus signs and then solving a quadratic inequality. It also requires treating a parameter $a$ as a positive constant and using its sign to order the critical values. The final interval is between the two critical values, not outside them.

Common Mistakes

• Forgetting to square the coefficient $2$: writing $(2x-a)^2 < (x+3a)^2$ instead of $4(2x-a)^2 < (x+3a)^2$.
• Incorrect expansion, especially the middle term in $16x^2 - 16ax + 4a^2$.
• Sign mistake when taking $x^2 + 6ax + 9a^2$ to the left, giving the wrong quadratic.
• Choosing the outside intervals instead of the inside interval.
• Writing $\leq$ instead of $<$; the mark scheme explicitly says do not condone $\leq$ for $<$ in the final answer.

Things to Be Careful About

The inequality is strict, so the critical values must not be included. Also, remember $a$ is positive; if $a$ could be negative, the order of the critical values would change. Since $a>0$, we can unambiguously state $-\frac{1}{5}a < x < \frac{5}{3}a$.