${[y] }$ denotes the greatest integer less than or equal to $ y$ , then choose the correct option.

Question

Define $f :\mathbb{R^2} \rightarrow \mathbb{R}$ by $f(x,y) = x^{[y]}$. where ${[y] }$ denotes the greatest integer less than or equal to $ y$ ..then

choose the correct option.

$1)$ $f$ is continuous on $\mathbb{ R^2} $

$2)$for every $ y \in \mathbb{ R}$ , $x \rightarrow f(x,y)$ is continuous on $\mathbb{R}$ \ $\{0\}$

$3)$ For every $x \in \mathbb{R}$ , $y \rightarrow f(x,y)$ is continuous on $\mathbb{R}$

$4)$ $f$ is continuous at no point of $\mathbb{R^2}$

My answer : as For option 1) is not correct because I can claim that $f$ is not continuous on $\mathbb{R^2}$ becuse if I take $f(x) = \frac {1}{x}$ which is not defined at $x =0$

For option $4)$ is also False if i take $f(x) = x$

I am in confusion about option $2)$ and option $3)$

Please help me,,,

Thanks in adavnce.

Any hints/solution will be appreciated...

Answer 1

Answers: (1) No. $f(0,y)$ is undefined on $0<y<1$ and $x=0$. Another argument: use solution to question (3). (2) yes, even though for $y=-1$, $f(x,-1)=1/x$. (3) no, $f(2,y)=4$ for $2<y<3$ but $f(2,y=3)=8$. (4) no, $f$ is continuous at point $(1,y)$ for any $y$.

Answer 2

Given that $f :\mathbb{R^2} \rightarrow \mathbb{R}$ is defined by $f(x,y) = x^{[y]}$.

For option $\bf{(1)}$, clearly $~f~$ is not defined at $(0,0)$ and therefore it is not continuous at $(0,0)$.
Hence $~f~$ is not continuous on $ \mathbb{R}^2$ and therefore option $(1)$ is not true.

For option $\bf{(2)}$, for $~y~$ fixed in $ \mathbb{R}$, $~~f(x,y)=\phi(x) = x^{[y]}~$ is a polynomial on $ \mathbb{R}$ \ $\{0\}$.
Hence $~\phi~$ is continuous on $ \mathbb{R}$ \ $\{0\}$ and therefore option $(2)$ is true.

For option $\bf{(3)}$, for $~x~$ fixed in $ \mathbb{R}$, $$f(x,y)=g(y) = x^{[y]}=\begin{cases} x^0=1~~~ ;~~~ 0\le y\lt 1 \\ x^1~~~~~~~~~~~;~~~ 1\leq y\lt 2 \\ x^2~~~~~~~~~~~;~~~ 2\leq y\lt3\\ \vdots~~~~~~~~~~~~~~~~~~~~~~\vdots \end{cases}$$ Clearly $~g~$ is not continuous at $ ~y\in\mathbb Z~$ and hence option $(3)$ is not true.

For option$\bf{(4)}$, for $~x\ne0~$ and $~y\notin\mathbb Z~$, from option $(3)$, $~f(x,y)~$ is continuous.
Therefore option $(4)$ is not true.