ISI2015-MMA-30

Question

Suppose that a function $f$ defined on $\mathbb{R} ^2$ satisfies the following conditions:

$$\begin{array} &f(x+t,y) & = & f(x,y)+ty, \\ f(x,t+y) & = & f(x,y)+ tx \text{ and } \\ f(0,0) & = & K, \text{ a constant.} \end{array}$$

Then for all $x,y \in \mathbb{R}, \:f(x,y)$ is equal to

• A. $K(x+y)$
• B. $K-xy$
• C. $K+xy$
• D. none of the above

Answer 1

Simple check from the option. Only option (C) will satisfy all the conditions

Answer 2

Replacing  $y= 0$ in $f(x,t+y) = f(x,y) + tx $,

$f(x,t) = f(x,0) + tx$

i.e. $f(x,y) = f(x,0) + xy$     (equation $(1)$)

Replacing  $x=0,y= 0$ in $f(x+t,y) = f(x,y) + ty $,

$f(t,0)=f(0,0) = K$

i.e. $f(x,0)=K$

putting $f(x,0)=K$ in equation $(1),$

$f(x,y) = K + xy$