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Q.

If the function f satisfies the relation fx+y+fx-y=2fxfy∀x,y∈R   and   f0≠0, then

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a

f(x) is an even function

b

f(x) is an odd function

c

If f2=a,    then   f-2=a

d

If f4=b,    then   f-4=-b

answer is A.

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Detailed Solution

fx+y+fx-y=2fx.fyPut  x=0.   Then   fy+f-y=2f0fyPut  x=y=0.   Then   f0  +f0=2f0f0∴  f0=1as   f0≠0∴f-y=fy
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If the function f satisfies the relation fx+y+fx-y=2fxfy∀x,y∈R   and   f0≠0, then