If the function f ( x ) = a x 3 + b x 2 + 11 x − 6 satisfies the conditions of Rolle's theorem in [ 1 , 3 ] and f ′ ( 2 + 3 1 ) = 0 , then a + b =

Question

If the function f ( x ) = a x 3 + b x 2 + 11 x − 6 satisfies the conditions of Rolle's theorem in [ 1 , 3 ] and f ′ ( 2 + 3 ​ 1 ​ ) = 0 , then a + b =, -5, -3, 4, 7

MARKS App · Accepted Answer

Given, f ( x ) = a x 3 + b x 2 + 11 x − 6 satisfies the Rolle's theorem in [ 1 , 3 ] . So, $ f ( 1 ) = 0 and f ( 3 ) = 0 im g src = " h ttp s : // c d n − q u es t i o n − p oo l . g e t ma r k s . a pp / p y q / a p e ​ am ce t / Ky n 5 m E o vv I p h c CS 7 hb c M 6 U A S 1 g BF V R W w u s ZO D 5 V e p J 0. or i g ina l . f u ll s i ze . p n g " > b r > \begin{array}{lrlrl}	ext { and } & & 27 a+9 b+27 & =0 \ \Rightarrow & & 9 a+3 b+9 & =0\end{array} im g src = " h ttp s : // c d n − q u es t i o n − p oo l . g e t ma r k s . a pp / p y q / a p e ​ am ce t / f 9 t A − j v Mp A f 2 p L 9 A Z 9 KHe E o A k U M U e 7 U d 4 f 8 cc a f B L k M . or i g ina l . f u ll s i ze . p n g " > b r > F ro m Eq . ( i ) , w e g e t a+b=-5$

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