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Questions  

 If f(x)=x2sinπx2,|x|<1x|x|,|x|1, then f(x) is 

a
an even function
b
an odd function
c
a periodic function
d
none of these

detailed solution

Correct option is B

f(−x)=(−x)2sin⁡π(−x)2,|−x|<1(−x)|−x|,|−x|≥1=−x2sin⁡πx2,|x|<1−x|x|,|x|≥1=−f(x)

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