Obtain all function satisfying
(i) `f (x + y) = f (x) + f (y)` `(x` `in R,` `y in R)`
(ii) `f (x + y) = f (x).f (y)` `(x` `in R,` `y in R)`
(iii)`f (xy) = f (x) + f (y)` `(x gt 0, y gt 0)`
(iv) `f (xy) = f (x). f (y)` `(x gt 0, y gt 0)`

Explanation

(i) Differentiate the given equation with respect to x keeping y constant and then with respect to y keeping x constant.
`f ′(x) = f ′(x + y) = f ′(y)`
`implies f ′(x) = c`
`implies f (x) = cx + k` ('k' is integration constant)
Put `x = 0, f (0) = k = 0` (from original equation `f (0 + 0) = f (0) = f (0))`
`implies f (0) = 0`
Ans: `f (x) = c x`
(ii) `f ′(x + y) = f ′(x).f (y)` (As in (i))
`f ′(x + y) = f (x).f ′(y)`
`implies f ′(x).f(y) = f (x).f ′(y)`
`implies (f'(x))/(f(x)) = (f'(y))/(f(y)) implies (f'(x))/(f(x)) = c`
`implies (d f(x))/(f(x)) = c dx`
`implies` ln `f (x) = c x + k (Q f (0) = (f (0)^2)`
ln `f (0) = c.0 + k`
ln `1 = k`
`implies k = 0` Ans: `f (x) = e^(c x)`
(iii) `f (xy) = f (x) + f (y)` (As in (i))
`{(f'(xy).y=f'(x)),( f'(xy).x=f'(y)):} implies f ′(xy) = (f'(xy))/y = (f'(y))/x`
`implies x f ′(x) = "constant" = c implies d f (x) = c/x dx`
`implies f (x) = c` ln `x + k`
`f (1) = c` ln `1 = k`
`k = f (1) = 0 (f(1) = 2 f (1))` Ans: `f (x) = c` ln `x`
(iv)` yf'(xy) ⇒ f ′(xy) = =
⇒ = = c
⇒ = dx ⇒ ln f (x) = c ln x + k Ans: f (x) = xc