If the Rolle's theorem holds for the function f x = 2 x 3 + a x 2 + b x in the interval − 1 , 1 for the point c = 1 2 , then the value of 2 a + b is:

Step-by-step solution

Given that the function holds Rolle's theorem in the interval − 1 , 1 ⇒  f - 1 = f 1 ⇒ − 2 + a − b = 2 + a + b ⇒ b = - 2 Now, f ' x = 6 x 2 + 2 a x + b ∴     f ' 1 2 = 0 ⇒ 6 × 1 4 + 2 a × 1 2 + b = 0 ⇒     a = 1 2 ⇒  2 a + b = - 1