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

Options

A. `(x-a)^(2n-2)+b`
B. `b-1-(x+a)^(2n+1)`
C. `b-(x-a)^(2n+2)`
D. `(x-a)^(2n+2)+b `

Answer

C

Explanation

It is given that
`f(a) = f'(a)= ….= f^(2n) (a)=0`
`rArr x=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