If the function f (x) = 2ax2-x-1 has exactly one zero point in (0,1), then the value range of a is () A. (1,+∞)B. (-∞,-1)C. (-1,1)D. [0,1)

If the function f (x) = 2ax2-x-1 has exactly one zero point in (0,1), then the value range of a is () A. (1,+∞)B. (-∞,-1)C. (-1,1)D. [0,1)

When △ = 0, a = - 18, there is a zero point x = - 2, which is not on (0, 1), so it does not hold. The function f (x) = 2ax2-x-1 has exactly a zero point in (0, 1), that is, f (0) f (1) < 0, that is - 1 × (2a-1) < 0. The solution is a > 1, so a is selected