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Question

Let f(x+y)=f(x)f(y) x,y R. If f(5)=2 and f(0)=3, then the value of f(5) is

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Solution

Given, f(5)=2 and f(0)=3
f(x+y)=f(x)f(y) ...(1)
Put x=5 and y=0, we get
f(5)=f(5)f(0)
f(0)=1

Differentiating (1) w.r.t x
f(x+y)(1+y)=f(x)f(y)+f(x)f(y)y
Put x=5 and y=0, we get
f(5)(1+y)=f(5)f(0)+f(5)f(0)y
Since, f(0)=1, f(5)=2 and f(0)=3
f(5)+yf(5)=f(5)+23y
yf(5)=6y
f(5)=6

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