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Let f: R rarr R be a function defined as...

Let `f: R rarr R` be a function defined as `f(x)=[(x+1)^2]^(1/3)+[(x-1)^2]^(1/3)+x/(2^x-1)+x/2+1`

A

Even

B

Odd

C

Neither even nor odd

D

Even and odd both

Text Solution

Verified by Experts

The correct Answer is:
A
`f(x)=[(x+1)^(2)]^(1//3) +[(x-1)^(2)]^(1//3)+(x)/(2^(x)-1)+(x)/(2)+1`
`f(-x)=[(-x+1)^(2)]^(1//3) +[(-x-1)^(2)]^(1//3)+ ((-x))/(2^(-x)-1)+((-x))/(2)+1`
`f(x)-f(-x)=[(x+1)^(2)]^(1//3)+[(x-1)^(2)]^(1//3)-[(x-1)^(2)]^(1//3)=(x)/(2^(x)-1) +(x.2^(x))/(1-2^(x))+x`
`-[(x+1)^(2)]^(1//3) +(x)/(2^(x)-1) +(x)/(2^(x)-1)+(x)/(2) +1 +(x)/(2^(-x)-1) +(x)/(2)-1`
`=x[(1)/(2^(x)-1) -(2^(x))/(2^(x)-1)]+x`
`=x[(1-2^(x))/(2^(x)-1)]+x`
`= -x[(2^(x)-1)/(2^(x)-1)] +x=0`
`:. f(x)-f(-x)=0`. Hence function is even.
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