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: Bound near the origin for $P$.
For $F(x,y)=xf(y)+y^2g(x)$ with $f,g$ bounded by $M,N$, we get $|F(x,y)|\le M|x|+Ny^2$.

Step 2: Take the limit.
As $(x,y)\to(0,0)$ both pieces go to $0$, and $F(0,0)=0$, so $F$ is continuous at the origin. $P$ is correct.

Step 3: Check $F_x$ at $(1,0)$ for $Q$.
Along $y=0$ we have $xy=0$ so $F\equiv0$, giving $F_x(1,0)=0$, fine.

Step 4: Check $F_y$ at $(1,0)$.
The difference quotient is $\frac{f(k)}{k}+kg(1)$. Since $f$ is only bounded, $\frac{f(k)}{k}$ may blow up (take $f\equiv1$, giving $1/k$). So $F_y(1,0)$ need not exist, and $Q$ is not correct.

Step 5: Conclude.
$P$ right, $Q$ wrong, option (A).
\[ \boxed{(A)} \]