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Question

The function f(x)=sin[log(x+x2+1)] is
  1. even
  2. odd
  3. neither even nor odd
  4. periodic

A
even
B
odd
C
periodic
D
neither even nor odd
Solution
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f(x)=sin[log(x+1+x2)]

f(x)=sin[log(x+1+x2)]

f(x)=sinlog⎢ ⎢(1+x2x)(1+x2+x)(1+x2+x)⎥ ⎥

f(x)=sinlog⎢ ⎢1(x+1+x2)⎥ ⎥

f(x)=sin[log(x+1+x2)]

f(x)=sin[log(x+1+x2)]

f(x)=f(x)

f(x) is an odd function.

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