Elena stared at the clock on the wall of Exam Hall 4. 9:02 AM. She had 58 minutes left.
hit her like a cold splash of water. Given that ( f(x) = 2x^3 + 3x^2 - 8x + 3 ), show that ( (x-1) ) is a factor, and hence fully factorise ( f(x) ). Elena took a breath. Polynomials. I can do this. She scribbled the substitution: ( f(1) = 2 + 3 - 8 + 3 = 0 ). Yes. Then came the algebraic long division, the careful subtraction of terms, the descent into the quadratic. ( (x-1)(2x^2 + 5x - 3) ). Then the final break: ( (x-1)(2x-1)(x+3) ). core pure -as year 1- unit test 5 algebra and functions
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. Elena stared at the clock on the wall of Exam Hall 4
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} ). hit her like a cold splash of water
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.