Question

please answer the question ....​

Answer 1

Answer:

7

Step-by-step explanation:

Given :

$\displaystyle \sf f ( x ) = x^3-6x^2+ax+b$

Now :

Diff. w.r.t. 'x' :

f' ( x ) = 3 x² - 12 x + a + 0

= > f' ( x ) = 3 x² - 12 x + a  

Since it satisfies Rolles theorem at [ 1 , 3 ]

= > f ( 1 ) = f ( 3 )

= > 1 - 6 + a + b = 27 - 54 + 3 a + b

= > > 1 - 6 + a = 27 - 54 + 3 a

= > 2 a = 27 - 5

= > a = 22 / 1

= > a = 11

We are asked to find value of ' a - 4 '

= > a - 4 = 11 - 4

= > a - 4 = 7

Hence we get required answer.