Suppose `f: RvecR^+` be a differentiable function such that `3f(x+y)=f(x)f(y)AAx ,y in R` with `f(1)=6.` Then the value of `f(2)` is `6` b. `9` c. `12` d. `15`

Options

A. 6
B. 9
C. 12
D. 15

Answer

C

Explanation

`3f(x+y)=f(x)f(y)" (i)"`
`f'(x)=underset(hrarr0)(lim)(f(x+h)-f(x))/(h)`
`=underset(hrarr0)(lim)((f(x).f(h))/(3)-f(x))/(h)" (using f rule)"`
`=underset(hrarr0)(lim)(f(x).f(h)-3f(x))/(3h)`
`rArr" "f'(x)=underset(hrarr0)(lim)(f(x))/(3).([f(h)-3])/(h)`
`"Put "x=0, y=1`
`therefore" "3f(1)=f(0)f(1)`
`therefore" "f(0)=3`
`therefore" "f'(x)=(f(x))/(3)underset(hrarr0)(lim)(f(h)-f(0))/(h)`
`rArr" "f'(x)=(f(x))/(3)f'(0),`
`rArr" "3.(f'(x))/(f(x))=f'(0)=k("say")`
`rArr" Integrating, we get "3log(f(x))=kx+C,`
`"Using "f(0)=3 rArr 3log 3=C`
`rArr" "3log(f(x))=kx+3log3`
`rArr" "3log((f(x))/(3))=kx,`
Using `f(1)=6 rArr 3log2=k`
`rArr" "f(x)=3.2^(x)`
`rArr" "f(2)=12`