Question is as below :

Asked by spuneet23 | 20th Sep, 2010, 05:16: PM

Expert Answer:

Dear Student,
 
Firstly let y = 0,
then we have
 
f(x+0) + f(x-0) = 2f(x).f(0)
 
=> 2f(x) = 2f(x).f(0)
 
=> f(0) = 1                --------------- (1)
 
 
Now, putting x = 0, we get
 
f(0+y) + f(0-y) = 2f(0).f(y)
 
=> f(y) + f(-y) = 2f(y)              (as f(0) = 1 from (1))
 
=> f(-y) = 2f(y) - f(y)
 
=> f(-y) = f(y)
 
or equivalently, f(x) = f(-x)
 
Hence f(x) is an even function.
 
Therefore, option b is correct.
 
 
Regards Topperlearning.

Answered by  | 20th Sep, 2010, 09:22: PM

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