Question

Let \( f: \mathbb{R}^2 \to \mathbb{R} \) be given by \[ f(x, y) = \begin{cases} \sqrt{x^2 + y^2} \sin\left( \frac{y^2}{x} \right) & \text{if } x \neq 0, \\ 0 & \text{if } x = 0. \end{cases} \] Consider the following statements: P: \( f \) is continuous at \( (0, 0) \) but \( f \) is NOT differentiable at \( (0, 0) \). 
Q: The directional derivative \( D_u f(0, 0) \) of \( f \) at \( (0, 0) \) exists in the direction of every unit vector \( u \in \mathbb{R}^2 \). Then:

• A. both P and Q are TRUE
• B. P is TRUE and Q is FALSE
• C. P is FALSE and Q is TRUE
• D. both P and Q are FALSE

Hint:

To check if a function is differentiable, ensure that the partial derivatives exist and are continuous. A function can be continuous but not differentiable.

Answer 1

The Correct Option is A

Step 1: Analyze statement P.
The function \( f(x, y) \) is continuous at \( (0, 0) \) since the limit of \( f(x, y) \) as \( (x, y) \to (0, 0) \) is 0. However, the function is not differentiable at \( (0, 0) \) because the partial derivatives at \( (0, 0) \) do not exist. Thus, statement P is TRUE.

Step 2: Analyze statement Q.
The directional derivative of \( f \) at \( (0, 0) \) in any direction exists because \( f(x, y) \) approaches 0 as \( (x, y) \to (0, 0) \) from any direction. Therefore, statement Q is TRUE.

Final Answer: (A) both P and Q are TRUE