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Question

Is the function f(x)=[(x+1)2]1/3+[(x1)2]1/3 odd, even or neither?
  1. Even
  2. Odd
  3. Neither even nor odd
  4. Even for some values of x

A
Even
B
Neither even nor odd
C
Odd
D
Even for some values of x
Solution
Verified by Toppr

Given, f(x)=[(x+1)2]1/3+[(x1)2]1/3
Now f(x)=[(x+1)2]1/3+[(x1)2]1/3=[(x1)2]1/3+[(x+1)2]1/3=[(x+1)2]1/3+[(x1)2]1/3
Clearly f(x)=f(x), Hence f(x) is an even function.

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