She wrote: No solution (the expression is always ≥ 0). A trick question. But she didn't fall for it.
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) ).
As she walked out, she thought: That wasn't a test. That was a rite of passage. core pure -as year 1- unit test 5 algebra and functions
Elena stared at the clock on the wall of Exam Hall 4. 9:02 AM. She had 58 minutes left.
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 ). She wrote: No solution (the expression is always ≥ 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.
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. hit her like a cold splash of water
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 ).