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

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.

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

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. Unit Test 5 wasn't just about algebra

She turned the page.

And for the first time, she felt like a real mathematician. About remembering that a square can never be negative

The answer formed: ( \frac{1}{x-1} - \frac{1}{x+2} + \frac{5}{x-3} ). Clean. Elegant.

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} ).