2024/25 AHSMC Part 1 Problem 1Problem 1 of 16Let f(x)=4x2−4x+1f(x) = 4x^2 - 4x + 1f(x)=4x2−4x+1 and suppose that f(a)=0f(a) = 0f(a)=0 and f(0)=bf(0) = bf(0)=b for some real numbers aaa and bbb. Then the value of a+ba+ba+b is:Answer choicesA−12-\dfrac{1}{2}−211B12\dfrac{1}{2}212C32\dfrac{3}{2}233D3334Enone of these5Submit answerShow solutionSolutionWe have f(x)=(2x−1)2f(x) = (2x-1)^2f(x)=(2x−1)2. Then a=12a = \dfrac{1}{2}a=21 and b=1b = 1b=1, so a+b=32a+b = \dfrac{3}{2}a+b=23. The answer is (C).University of AlbertaCopy problem