Question

Rolle's Theorem holds for the function x³ + bx² + cx, 1 ≤ x ≤ 2 at the point `4/3`, the value of b and c are

Options

• b = 8, c = – 5

• b = – 5, c = 8

• b = 5, c = – 8

• b = – 5, c = – 8

Answer 1

b = – 5, c = 8

Explanation:

Here, f(1) = f(2) and f'`(4/3)` = 0

⇒ 1 + b + c = 8 + 4b + 2c

⇒ – 7 = 3b + c  .......(i)

And `3(4/3)^2 + 2b(4/3) + c` = 0

⇒ `16/3 + (8b)/3 + (3c)/3` = 0

⇒ 8b + 3c = – 16  ......(ii)

Solving (i) and (ii), we get

b = – 5, c = 8.