A) \[-f(x)\]
B) \[f(x)\]
C) \[f(a)+f(a-x)\]
D) \[f(-x)\]
Correct Answer: A
Solution :
[a] \[f(2a-x)=f(a-(x-a))=f(a)f(x-a)-f(0)\] \[f(x)=f(a)f(x-a)-f(x)=-f(x)\] \[[\because \,\,\,x=0,y=0,f(0)={{f}^{2}}(0)-{{f}^{2}}(a)\] \[\Rightarrow {{f}^{2}}(a)=0\] \[\Rightarrow f(a)=0]\Rightarrow f(2a-x)=-f(x)\]You need to login to perform this action.
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