Is the function f(x)=[(x+1)2]1/3+[(x−1)2]1/3 odd, even or neither?
Even
Odd
Neither even nor odd
Even for some values of x
A
Even
B
Neither even nor odd
C
Odd
D
Even for some values of x
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Solution
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Given, f(x)=[(x+1)2]1/3+[(x−1)2]1/3 Now f(−x)=[(−x+1)2]1/3+[(−x−1)2]1/3=[(x−1)2]1/3+[(x+1)2]1/3=[(x+1)2]1/3+[(x−1)2]1/3 Clearly f(x)=f(−x), Hence f(x) is an even function.
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Maths byjus worksheet 11 and 12 JEE
Chapter relation and function 1
Sub topic Algebra of real function , equal or identical function , odd and even function