FunctionHard
Question
If Roll′s theorem holds for the function f(x) = 2x3 + bx2 + cx, x ∈ [−1, 1], at the point x = 1/2 then 2b + c equals :
Options
A.1
B.2
C.- 1
D.- 3
Solution
f(1) = f(-1)
2 + b + c = - 2 + b - c ⇒ c = - 2
f′(x) = 6x2 + 2bx + c
= 6
+ c
=
+ b = c = 0 ⇒ b = 
Now 2b + c = 1 - 2 = - 1
2 + b + c = - 2 + b - c ⇒ c = - 2
f′(x) = 6x2 + 2bx + c
= 6
+ c=
+ b = c = 0 ⇒ b = Now 2b + c = 1 - 2 = - 1
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