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A real valued function f(x) satisfies the functional equaiton f(x-y) = f(x)f(y) – f(a-x) f(a+y), where a is a given constant and f(0) =1, f(2a – x) is equal to
a)f(-x)
b)f(a) + f(a – x)
c)f(x)
d) – f(x)

Meghendra Agrawal , 7 Years ago
Grade
anser 1 Answers
Arun
Dear Meghendra
 
 
Ans:
f(x-y) = f(x)f(y)-f(a-x)f(a+y)
y = 0
f(x-0) = f(x)f(0)-f(a-x)f(a+0)
f(0) = 1
f(x) = f(x)-f(a-x)f(a)
f(a-x)f(a) = 0
\Rightarrow f(a) = 0
f(2a-x) = f(a-(x-a)) = f(a)f(x-a)-f(a-a)f(a+x-a)f(2a-x) = f(a)f(x-a)-f(0)f(x)
f(a) = 0, f(0) = 1
f(2a-x) = -f(x)
 
Hope it helps
 
In case of any query please feel free to ask
 
 
Regards
Arun (askIITians forum expert)
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Last Activity: 7 Years ago
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