Core Pure -as Year 1- Unit Test 5 Algebra And Functions May 2026

She wrote the final answer: ( \sqrt{x^2+3} ), domain ( [0, \infty) ).

She turned the page.

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} ). 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 ). She wrote the final answer: ( \sqrt{x^2+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 ). The function ( g(x) = \frac{3x+1}{x-2} ), ( x \neq 2 )