Question

Let \( f(x, y) = \begin{cases \frac{xy}{(x^2 + y^2)^{\alpha}}, & (x, y) \neq (0,0)
0, & (x, y) = (0,0) \end{cases} \) Then which one of the following is TRUE for \( f \) at the point \( (0, 0) \)?}

• A. For \( \alpha = \frac{1}{2} \), \( f \) is continuous but not differentiable
• B. For \( \alpha = 1 \), \( f \) is continuous and differentiable
• C. For \( \alpha = \frac{1}{4} \), \( f \) is continuous and differentiable
• D. For \( \alpha = \frac{3}{4} \), \( f \) is neither continuous nor differentiable

Hint:

For functions involving powers of \( x^2 + y^2 \), the value of \( \alpha \) is critical in determining the continuity and differentiability of the function at the origin.

Answer 1

The Correct Option is C

Step 1: Analyzing the function at the origin.
We are given a piecewise function and need to check its behavior at the point \( (0,0) \). The function involves \( (x^2 + y^2)^{\alpha} \), which suggests that the value of \( \alpha \) will affect the continuity and differentiability of the function.

Step 2: Checking continuity.
For \( \alpha = \frac{1}{2} \), we check whether the limit of \( f(x, y) \) as \( (x, y) \to (0, 0) \) exists and equals 0. It turns out that \( f \) is continuous at \( (0, 0) \), but not differentiable at this point because the limit does not behave smoothly in all directions.

Step 3: Conclusion.
The correct answer is (A) For \( \alpha = \frac{1}{2} \), \( f \) is continuous but not differentiable.