f(x) = x2 -(a + 3)x + 4a f(0) = 0 - 0 + 4a f(0) = 4a f(0) = positive then for exactly one root between (0, 2) f(2) should be negative f(2) = (2)* - (a-3)2+4.3 f(2) = 4-2a-6+4a F(2) = 2a-2 = f(2) < 0 = (2a-2) < 0 = a < 1 So, a = (0,1) Case 2: When a < 0
Case 2: When a < o f(0) = 4a = negative Then for exactly one root between (0,2), f(2) should be greater than zero. So, f(2) > 0 = 2a-2 > 0 But here 'a' is already < 0 so No value of 'a' possible So, a ∈ (0,1)