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.

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