Given that,
- $f( 0,y) = y+1 \quad \longrightarrow (1)$
- $f(x+1,0) = f( x, 1) \quad \longrightarrow (2)$
- $f( x+1, y+1) = f\left( x,f( x+1,y) \right) \quad \longrightarrow (3)$
Now, put $x = 0,y = 1,$ in equation $(3)$ we get,
- $f(1,2) = f(0,f(1,1)) \quad \longrightarrow (4)$
And, put $x = 0,y = 0$ in equation $(3)$ we get,
- $f(1,1) = f(0,f(1,0)) \quad \longrightarrow (5)$
And, put $x = 0$ in equation $(2)$ we get,
- $f(1,0) = f( 0, 1) \quad \longrightarrow (6)$
And, put $y = 1,$ in equation $(1)$ we get,
Now, put $f(0,1) = 2,$ in equation $(6),$ we get,
Now, Put $f(0,1) = 2,$ in equation $(5),$ we get,
- $f(1,1) = f(0,2) \quad \longrightarrow (7)$
Put $y = 2$ in equation $(1)$ we get, $f(0,2) = 2+1 = 3$
Now, put $f(1,1) = 3,$ in equation $(4),$ we get,
- $f(1,2) = f(0,3) \quad \longrightarrow (8)$
Put $y = 3$ in equation $(1)$ we get, $f(0,3) = 3+1 = 4$
$\therefore \boxed{f(1,2) = 4}$
Correct Answer $:4$