Questions
a
-1
b
0
c
1
d
none of these
detailed solution
Correct option is B
Let fx+1y+fx−1y=2f(x)f1y∀x,y∈R Given f(0)=0 Putting x=0,y=1x, we get f(x)+f(−x)=2f(0)f(x)⇒ f(x)+f(−x)=0⇒ f(x)=−f(−x) Putting x=1,y=−1, we get ⇒ f(2)+f(0)=2[f(1)]2⇒f(2)=2[f(1)]2 Putting x=−1,y=−1, we get ⇒f(−2)=2f(−1)f(−1)⇒ −f(2)=2(f(1))2⇒f(2)=−f(2)⇒f(2)=0⇒f(1)=0∴ f(1)=f(2)=0∴ f(1)+f(2)=0Talk to our academic expert!
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