Question

Consider the function \(f:\mathbb{R}^2\to\mathbb{R}\) defined by
\[ f(x,y)= \begin{cases} \dfrac{x^3+y^3}{\sqrt{x^2+2y^2}} & (x,y)\neq (0,0) \\ 0 & (x,y)=(0,0). \end{cases} \]
Which of the following statements is/are correct?

• A. \(f\) is continuous at \((0,0)\)
• B. Partial derivatives \(f_x\) and \(f_y\) exist at \((0,0)\) and \(f_x(0,0)=0\), \(f_y(0,0)=0\)
• C. \(f\) is not differentiable at \((0,0)\)
• D. \(f\) is differentiable

Hint:

Bound |f(x,y)| by x^2 + y^2/sqrt(2) using sqrt(x^2+2y^2) >= |x| and >= sqrt(2)|y|; this single bound settles continuity, the partial derivatives, and differentiability at once.

Answer 1

The Correct Option is A, B, D

Step 1: Bound.
Since \(\sqrt{x^2+2y^2}\ge|x|\) and \(\ge\sqrt2|y|\), \[ |f(x,y)|\le x^2+\frac{y^2}{\sqrt2}. \]

Step 2: (A) continuity.
Bound \(\to0\), so continuous. TRUE.

Step 3: (B) partials.
Along axes, \(f_x(0,0)=0\), \(f_y(0,0)=0\). TRUE.

Step 4: Differentiability.
\[ \left|\frac{f(h,k)}{\sqrt{h^2+k^2}}\right|\le\sqrt{h^2+k^2}\to0. \] So differentiable at origin. (C) FALSE.

Step 5: (D) everywhere.
Away from origin, quotient of smooth functions with nonzero denominator, differentiable. TRUE.

Final Answer: \[ \boxed{\text{A, B, D}} \]