[|f(x)|+1] is even where, [x]= greatest integer ≤x
f(x)−f(−x)2 is odd function
None of these
A
f(−x)+f(x)2is a constant function
B
[|f(x)|+1] is even where, [x]= greatest integer ≤x
C
f(x)−f(−x)2 is odd function
D
None of these
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Solution
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Given: f(−x)=−f(x) f(x)+f(−x)=0. Hence, f(x)+f(−x)2 is a constant function.
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