1.

A function f such thatf'(a)=f''(a)=….=f^(2n)(a)=0 , and fhas a local maximum value b at x=a ,if f (x) is

Answer»

`(x-a)^(2n-2)+B`
`b-1-(x+a)^(2n+1)`
`b-(x-a)^(2n+a)`
`(x-a)^(2n+2)+b `

SOLUTION :It is given that
`f(a) = f'(a)= ….= f^(2n) (a)=0`
`rArrx=a` is root of f(x) of order (2n+1) or more.
Also ,it is given that f(a)=b. Therefore
`f(x)=b pm (x-a)^(2n+2)`
`If f(x)=b -(x-a)^(2n+2)` then
`f(x)=- 2(n-2)(x-a)^(2n+1)`
Clearly f'(x) chages its sign from positive to negative in the neighbourhood of x=a
Therefore , f(x) attatins a LOCAL maximum at x=a .
Hence ,option (c ) is correct


Discussion

No Comment Found

Related InterviewSolutions