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

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

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. core pure -as year 1- unit test 5 algebra and functions

Elena stared at the clock on the wall of Exam Hall 4. 9:02 AM. She had 58 minutes left. She wrote: No solution (the expression is always ≥ 0)

And for the first time, she felt like a real mathematician. She flipped back

The invigilator called time.

Roots: ( x = 2 ) and ( x = -2 ), both repeated (multiplicity 2). The inequality ( p(x) < 0 ) asked: when is a square less than zero?

was the function composition trap. Given ( h(x) = \sqrt{x+4} ) for ( x \geq -4 ), and ( k(x) = x^2 - 1 ) for ( x \geq 0 ). Find ( h(k(x)) ) and state its domain. She composed carefully: ( h(k(x)) = \sqrt{(x^2 - 1) + 4} = \sqrt{x^2 + 3} ). Wait, she thought. That’s defined for all real ( x ), but ( k ) only takes ( x \geq 0 ). And ( k(x) ) gives outputs ( \geq -1 ), but ( h ) requires inputs ( \geq -4 ). That’s fine.