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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 |
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Answer» `(x-a)^(2n-2)+B` `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 |
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