Mark the correct alternative in the following:
If 4a + 2b + c = 0, then the equation 3ax2 + 2bx + c = 0 has at least one real root lying in the interval.
Let f(x) = ax3 + bx2 + cx + d --------- (i)
f(0) = d
f(2) = a(2)3 + b(2)2 + c(2) + d
= 8a + 4b + 2c + d
= 2(4a + 2b + c) + d
4a + 2b + c = 0 {Given}
= 2 (0) + d
= 0 + d
= d
f is continuous in closed interval [0, 2] and f is derivable in the open interval (0, 2).
Also, f(0) = f(2)
As per Rolle’s Theorem,
f’(α) = 0 for 0 < α < 2
f’(x) = 3ax2 + 2bx + c
f’(α) = 3aα2 + 2b(α) + c
3aα2 + 2b(α) + c = 0
Hence equation (i) has at least one root in the interval (0, 2).
Thus, f’(x) must have one root in the interval (0, 2).
Hence, Option (C) is the answer.