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Standard XII
Maths
Question

Let f(x)=|x−2|+|x−3|+|x−4| and g(x)=f(x+1). Then
  1. g(x) is an even function
  2. g(x) is an odd function
  3. g(x) is periodic
  4. g(x) is neither even nor odd

A
g(x) is an odd function
B
g(x) is an even function
C
g(x) is periodic
D
g(x) is neither even nor odd
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Solution
Verified by Toppr

f(x)=|x−2|+|x−3|+|x−4|

f((x+1))=|x+1−2|+|x+1−3|+|x+1−4|

g(x)=|x−1|+|x−2|+|x−3|

g(−x)=|−x−1|+|−x−2|+|−x−3|=|x+1|+|x+2|+|x+3|≠g(x) ,also≠−g(x)

Hence it will neither be an odd or an even function.

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