If 4a + 2b + c = 0, then the equation 3ax2 + 2bx + c = 0 has at least

1.	If 4a + 2b + c = 0, then the equation 3ax2 + 2bx + c = 0 has at least one real root lying in the interval.A. (0, 1)B. (1, 2)C. (0, 2)D. None of these
Answer» Correct answer is C. 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 \(\because\) 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).

If 4a + 2b + c = 0, then the equation 3ax2 + 2bx + c = 0 has at least one real root lying in the interval.A. (0, 1)B. (1, 2)C. (0, 2)D. None of these

Answer»

Correct answer is C.

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

\(\because\) 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).