She wrote: No solution (the expression is always ≥ 0). A trick question. But she didn't fall for it.

On her desk lay . The front cover was deceptively calm, featuring only the exam board’s logo and the instruction: Attempt all questions. Use algebraic methods unless otherwise stated.

She turned the page.

But the domain of ( h \circ k ) is ( { x \in \text{dom}(k) \mid k(x) \in \text{dom}(h) } ). ( x \geq 0 ) and ( x^2 - 1 \geq -4 ) — which is always true. So the domain is simply ( x \geq 0 ).

was a curveball—a partial fractions problem disguised as a rational function. Express ( \frac{5x^2 + 4x - 11}{(x-1)(x+2)(x-3)} ) in partial fractions. Her pen flew. She set up the identity: ( 5x^2 + 4x - 11 \equiv A(x+2)(x-3) + B(x-1)(x-3) + C(x-1)(x+2) ). She chose the cover-up rule for speed: ( x=1 ) gave ( A = 1 ). ( x=-2 ) gave ( B = -1 ). ( x=3 ) gave ( C = 5 ).

She flipped back. Question 6 (not mentioned yet) was a proof by contradiction involving a rational root of a cubic. She had left it till last. Prove that ( \sqrt{3} ) is irrational. She wrote: Assume ( \sqrt{3} = \frac{a}{b} ) in lowest terms. Then ( 3b^2 = a^2 ). So 3 divides ( a^2 ), so 3 divides ( a ). Let ( a = 3k ). Then ( 3b^2 = 9k^2 ) → ( b^2 = 3k^2 ). So 3 divides ( b^2 ), so 3 divides ( b ). Contradiction — ( a ) and ( b ) have a common factor 3, not lowest terms. Hence ( \sqrt{3} ) is irrational.

One down.

The invigilator called time.