search
Log In
0 votes
179 views

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

  1. $K(x+y)$
  2. $K-xy$
  3. $K+xy$
  4. none of the above
in Calculus
recategorized by
179 views

2 Answers

0 votes
Simple check from the option. Only option (C) will satisfy all the conditions
0 votes
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 $(1),$

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

Related questions

1 vote
1 answer
1
129 views
If $f(x)$ is a real valued function such that $2f(x)+3f(-x)=15-4x,$ for every $x \in \mathbb{R}$, then $f(2)$ is $-15$ $22$ $11$ $0$
asked Sep 23, 2019 in Calculus Arjun 129 views
0 votes
1 answer
2
168 views
If $f(x) = \dfrac{\sqrt{3}\sin x}{2+\cos x}$, then the range of $f(x)$ is the interval $[-1, \sqrt{3}/2]$ the interval $[- \sqrt{3}/2, 1]$ the interval $[-1, 1]$ none of the above
asked Sep 23, 2019 in Calculus Arjun 168 views
1 vote
1 answer
3
125 views
For non-negative integers $m$, $n$ define a function as follows $f(m,n) = \begin{cases} n+1 & \text{ if } m=0 \\ f(m-1, 1) & \text{ if } m \neq 0, n=0 \\ f(m-1, f(m,n-1)) & \text{ if } m \neq 0, n \neq 0 \end{cases}$ Then the value of $f(1,1)$ is $4$ $3$ $2$ $1$
asked Sep 23, 2019 in Calculus Arjun 125 views
0 votes
1 answer
4
215 views
The limit $\:\:\:\underset{n \to \infty}{\lim} \Sigma_{k=1}^n \begin{vmatrix} e^{\frac{2 \pi i k }{n}} – e^{\frac{2 \pi i (k-1) }{n}} \end{vmatrix}\:\:\:$ is $2$ $2e$ $2 \pi$ $2i$
asked Sep 23, 2019 in Calculus Arjun 215 views
...