Question

Let \[ f(x, y) = \frac{xz y}{x^2 + y^2 + z^2}, \quad (x, y) \neq (0, 0). \] Then \[ \frac{\partial f}{\partial x} \text{ and } f \text{ are bounded and unbounded.} \]

• A. \( \frac{\partial f}{\partial x} \) and \( f \) are bounded
• B. \( \frac{\partial f}{\partial x} \) is bounded and \( f \) is unbounded
• C. \( \frac{\partial f}{\partial x} \) is unbounded and \( f \) is bounded
• D. \( \frac{\partial f}{\partial x} \) and \( f \) are unbounded

Hint:

For partial derivatives, check the behavior of the function at boundary points or near singularities to determine if it is bounded or unbounded.

Answer 1

The Correct Option is B

Step 1: Analyze the function \( f(x, y) \).
The function \( f(x, y) = \frac{xzy}{x^2 + y^2 + z^2} \) can be analyzed for its behavior as \( x, y, z \to 0 \). We observe that the function is unbounded as \( x, y \to 0 \).
Step 2: Analyze the derivative \( \frac{\partial f}{\partial x} \).
The partial derivative of \( f(x, y) \) with respect to \( x \) is bounded because the function behaves like a smooth function for small values of \( x \). Thus, \( \frac{\partial f}{\partial x} \) is bounded, while \( f \) is unbounded as \( x, y \to 0 \).
Step 3: Conclusion.
Thus, the correct answer is (B).