Section A: Pure Mathematics [40 marks]

1. Sketch the linear graph $y = 3x$.

Draw a straight line passing through the origin $(0, 0)$ with a positive gradient. Label this line as $y = 3x$.

2. Sketch the modulus curve $y = |x^2 - a^2|$ and identify key features.

The underlying graph $y = x^2 - a^2$ is an upward-opening parabola with $x$-intercepts at $x = -a$ and $x = a$, and a minimum turning point at $(0, -a^2)$.
Applying the modulus reflects the portion below the $x$-axis upwards.
The final curve $y = |x^2 - a^2|$ consists of the original parabola for $x \le -a$ and $x \ge a$, and an inverted hump for $-a < x < a$ with a local maximum at $(0, a^2)$.
Sketch this on the same axes, marking the $x$-intercepts $(-a, 0)$ and $(a, 0)$, and the $y$-intercept $(0, a^2)$.

1. Remove the modulus by determining the sign of $x^2 - a^2$ in the given domain.

For $0 < x < a$, we know that $x^2 < a^2$, which means $x^2 - a^2 < 0$.
Therefore, the modulus expression simplifies to:
$|x^2 - a^2| = -(x^2 - a^2) = a^2 - x^2$

2. Set up and solve the resulting quadratic equation.

Substitute the simplified expression into the given equation:
$a^2 - x^2 = 3x$
Rearranging into standard quadratic form:
$x^2 + 3x - a^2 = 0$
Using the quadratic formula with coefficients $A=1$, $B=3$, and $C=-a^2$:
$x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-a^2)}}{2}$
$x = \frac{-3 \pm \sqrt{9 + 4a^2}}{2}$

3. Select the valid root by checking against the domain $0 < x < a$.

Since the domain requires $x > 0$, and $\sqrt{9 + 4a^2} > \sqrt{9} = 3$, the negative root $x = \frac{-3 - \sqrt{9 + 4a^2}}{2}$ is strictly negative and must be rejected.
We also verify the positive root satisfies $x < a$:
For $a > 0$, $\sqrt{9 + 4a^2} < \sqrt{9 + 12a + 4a^2} = \sqrt{(3 + 2a)^2} = 3 + 2a$.
Therefore, $x = \frac{-3 + \sqrt{9 + 4a^2}}{2} < \frac{-3 + (3 + 2a)}{2} = a$.
Thus, the only valid solution is the positive root.