Core Pure -as Year 1- Unit Test 5 Algebra And Functions Apr 2026

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.

She wrote the final answer: ( \sqrt{x^2+3} ), domain ( [0, \infty) ). core pure -as year 1- unit test 5 algebra and functions

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 ). She flipped back

One down.

The invigilator called time.

Never. A square of a real number is always ( \geq 0 ). The only time it equals zero is at the roots. So no real ( x ) satisfies ( p(x) < 0 ). Prove that ( \sqrt{3} ) is irrational

She turned the page.