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

Question

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

TestHub · Accepted Answer

∵ f ( 1 ) = f ( 3 ) ⇒ ⇒ and ⇒ ​ a + b + 11 − 6 = 27 a + 9 b + 33 − 6 13 a + 4 b = − 11 f ′ ( x ) = 3 a x 2 + 2 b x + 11 f ′ ( 2 + 3 ​ 1 ​ ) = 3 a ( 2 + 3 ​ 1 ​ ) 2 + 2 b ( 2 + 3 ​ 1 ​ ) + 11 = 0 ​ ⇒ 3 a ( 4 + 3 1 ​ + 3 ​ 4 ​ ) + 2 b ( 2 + 3 ​ 1 ​ ) + 11 = 0 From eqs. (i) and (ii), we get a = 1 , b = − 6 .

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