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

One down.

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

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

She turned the page.

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

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 ).

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. She flipped back

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

The invigilator called time.

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 ).

But the domain of ( h \circ k ) is ( { x \in \text{dom}(k) \mid k(x) \in \text{dom}(h) } ). ( x \geq 0 ) and ( x^2 - 1 \geq -4 ) — which is always true. So the domain is simply ( x \geq 0 ).