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

Elena stared at the clock on the wall of Exam Hall 4. 9:02 AM. She had 58 minutes left.

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 killer. The one that separated the A from the B. The function ( p(x) = x^4 - 8x^2 + 16 ). Find all real roots. Hence solve the inequality ( p(x) < 0 ). She factorised: let ( u = x^2 ). Then ( u^2 - 8u + 16 = (u-4)^2 ). So ( p(x) = (x^2 - 4)^2 = (x-2)^2 (x+2)^2 ).

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

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

One down.

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. Elena stared at the clock on the wall of Exam Hall 4

Unit Test 5 wasn't just about algebra. It was about precision. About checking every assumption. About remembering that a square can never be negative.

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.

Elena set her pen on the desk. Her palms were damp, but her mind was clear. She had faced the domain restrictions, the partial fraction decomposition, the inverse function trap, the composite’s hidden conditions, and the elegant emptiness of the squared inequality. Roots: ( x = 2 ) and (

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 ).

brought the first real resistance. The function ( g(x) = \frac{3x+1}{x-2} ), ( x \neq 2 ). Find ( g^{-1}(x) ) and state its domain. She swapped ( x ) and ( y ): ( x = \frac{3y+1}{y-2} ). Cross-multiplied: ( x(y-2) = 3y+1 ). ( xy - 2x = 3y + 1 ). Grouped terms: ( xy - 3y = 2x + 1 ). Factored: ( y(x-3) = 2x+1 ). So ( g^{-1}(x) = \frac{2x+1}{x-3} ).