Core Pure -as Year 1- Unit Test 5 Algebra And Functions 【ULTIMATE | ROUNDUP】

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.

As she walked out, she thought: That wasn't a test. That was a rite of passage.

One down.

Elena stared at the clock on the wall of Exam Hall 4. 9:02 AM. She had 58 minutes left. core pure -as year 1- unit test 5 algebra and functions

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.

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

Domain of the inverse = range of the original. The original had a horizontal asymptote at ( y=3 ) and a vertical asymptote at ( x=2 ). So the range of ( g ) is all real numbers except 3. Therefore, domain of ( g^{-1} ): ( x \in \mathbb{R}, x \neq 3 ). She flipped back

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 ).

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

She wrote the final answer: ( \sqrt{x^2+3} ), domain ( [0, \infty) ). Prove that ( \sqrt{3} ) is irrational

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.

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 ).

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.