Diketahui f’(x) = 2ax + (a – 1 ). Jika f(1) = 3 dan f(2) = 0, nilai a adalah … Terima kasih

f'(x) = 2ax + (a -1)

f(x)= ∫ 2ax + (a -1) dx
f(x)= a x² + (a-1) x + c

f(1) = 3
a + a - 1 + c = 3
2a + c = 4

f(2) = 0
a (4) + (a-1)2 + c=  0
4a  + 2a - 2 + c = 0
6a - c = 2

2a + c = 4
6a - c = 2
................(+)
8a = 6
a = 6/8 = 3/4