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Let n be a natural and let 'a' be a real...

Let n be a natural and let 'a' be a real number. The number of zeros of `x^(2n+1)-(2n+1)x+a=0` in the interval `[-1,1]` is -

A

2 if a gt 0

B

2 if a lt 0

C

At most one for every value of a

D

At least three for every value of a

Text Solution

Verified by Experts

The correct Answer is:
C

`f(x)=x^(2n+1)-(2n+1)x+a`
`f'(x)=(2n+1)x^(2n)-(2n+1)`
`=(2n+1)(x^(2n)-1)le0` when x `in[-1,1]`
f(x) is strictly decreasing in `[-1,1]`
f(x) cut x axis at most one point in given interval
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