Question

Let \(f:\mathbb{R}\setminus\{0\}\to \mathbb{R}\) and \(g:\mathbb{R}\setminus\{0\}\to \mathbb{R}\) be two bounded and differentiable functions. Define \(F:\mathbb{R}^2\to \mathbb{R}\) by

• A. \(P\) is correct and \(Q\) is NOT correct
• B. \(P\) is NOT correct and \(Q\) is correct
• C. Both \(P\) and \(Q\) are correct
• D. Neither \(P\) nor \(Q\) is correct

Hint:

Boundedness helps in proving continuity near the origin, but it does not always guarantee existence of partial derivatives at boundary-type points.

Answer 1

The Correct Option is A

Step 1: Check continuity at \((0,0)\).
Since \(f\) and \(g\) are bounded, there exist constants \(M>0\) and \(N>0\) such that
\[ |f(y)|\leq M \] and
\[ |g(x)|\leq N \] For \(xy\neq 0\),
\[ F(x,y)=xf(y)+y^2g(x) \] So,
\[ |F(x,y)| \leq |x||f(y)|+y^2|g(x)| \] \[ |F(x,y)| \leq M|x|+Ny^2 \] As \((x,y)\to (0,0)\),
\[ M|x|+Ny^2 \to 0 \] Hence,
\[ F(x,y)\to 0 \] Also,
\[ F(0,0)=0 \] Therefore, \(F\) is continuous at \((0,0)\).
Thus, statement \(P\) is correct.

Step 2: Check \(\frac{\partial F}{\partial x}\) at \((1,0)\).
By definition,
\[ \frac{\partial F}{\partial x}(1,0) = \lim_{h\to 0} \frac{F(1+h,0)-F(1,0)}{h} \] Since \(y=0\), we have \(xy=0\), so
\[ F(1+h,0)=0 \] and
\[ F(1,0)=0 \] Therefore,
\[ \frac{\partial F}{\partial x}(1,0) = \lim_{h\to 0}\frac{0-0}{h}=0 \] So, \(\frac{\partial F}{\partial x}(1,0)\) exists.

Step 3: Check \(\frac{\partial F}{\partial y}\) at \((1,0)\).
By definition,
\[ \frac{\partial F}{\partial y}(1,0) = \lim_{k\to 0} \frac{F(1,k)-F(1,0)}{k} \] For \(k\neq 0\),
\[ F(1,k)=f(k)+k^2g(1) \] and
\[ F(1,0)=0 \] Thus,
\[ \frac{F(1,k)-F(1,0)}{k} = \frac{f(k)+k^2g(1)}{k} \] \[ = \frac{f(k)}{k}+kg(1) \] Now, \(f\) is only given to be bounded and differentiable on \(\mathbb{R}\setminus\{0\}\).
This does not guarantee that
\[ \lim_{k\to 0}\frac{f(k)}{k} \] exists.
For example, if \(f(k)=1\), then
\[ \frac{f(k)}{k}=\frac{1}{k} \] which does not have a finite limit as \(k\to 0\).
Therefore, \(\frac{\partial F}{\partial y}(1,0)\) need not exist.
Thus, statement \(Q\) is NOT correct.

Step 4: Final conclusion.
Hence,
\[ \boxed{(A)} \]