If Rolle's theorem holds for the function $ f(x)=2 x^{3}+b x^{2}+c x, x \in[-1,1] $, at the point $ x=\frac{1}{2} $, then $ 2 \mathrm{~b}+\mathrm{c} $ equals:

Question

If Rolle's theorem holds for the function $ f(x)=2 x^{3}+b x^{2}+c x, x \in[-1,1] $, at the point $ x=\frac{1}{2} $, then $ 2 \mathrm{~b}+\mathrm{c} $ equals:, $ 2 $, $ 1 $, $ -1 $, $ -3 $

MARKS App · Accepted Answer

If Rolle's theorem is satisfied in the interval [-1, 1], then f - 1 = f ( 1 ) - 2 + b - c = 2 + b + c c = - 2 also f ′ x = 6 x 2 + 2 b x + c Also if f ′ 1 2 = 0 them 6 1 4 + 2 b 1 2 + c = 0 3 2 + b + c = 0 ∵ c = - 2 , b = 1 2 ∴ 2 b + c = 2 1 2 + ( - 2 ) = 1 - 2 = - 1 .

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