If Rolle's theorem holds for the function f x = 2 x 3 + b x 2 + c x , x ∈ - 1 , 1 at the point x = 1 2 , then 2 b + c is equal to

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If Rolle's theorem holds for the function f x = 2 x 3 + b x 2 + c x , x &#8712; - 1 , 1 at the point x = 1 2 , then 2 b + c is equal to

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Correct answer: C. - 1

Since, Rolle's theorem is satisfied in [ - 1 , &#160; 1 ] and corresponding to x = 1 2 we have f ( &#8722; 1 ) = f ( 1 ) & f ' 1 2 = 0 So, &#8722; 2 + b &#8722; c = 2 + b + c &#8658; c = &#8722; 2 Further f ' x = 6 x 2 + 2 b x + c Put x = 1 2 So f ' 1 2 = 3 2 + b + c = 0 &#160; &#160; &#8658; b = 1 2 Hence, 2 b + c = - 1

If Rolle's theorem holds for the function f x = 2 x 3 + b x 2 + c x , x ∈ - 1 , 1 at the point x = 1 2 , then 2 b + c is equal to

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If Rolle&#39;s theorem holds for the function f x = 2 x 3 + b x 2 + c x , x &#8712; - 1 , 1 at the point x = 1 2 , then 2 b + c is equal to

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Correct answer

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More from Application of Derivatives

If Rolle's theorem holds for the function f x = 2 x 3 + b x 2 + c x , x ∈ - 1 , 1 at the point x = 1 2 , then 2 b + c is equal to