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 ).
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 ).
She wrote: No solution (the expression is always ≥ 0). A trick question. But she didn't fall for it. core pure -as year 1- unit test 5 algebra and functions
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 ).
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) ). was the killer
The invigilator called time.
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 ). Find all real roots
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} ).
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 ).
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.