Part CCSIR NET December 2024partials-exist-without-differentiability

Partials exist without differentiability

For a positive real number $a, a$ denotes the positive square root of a. Consider the function $f : R^{2} \to R$ defined by f(x, y) = (x/|x| $) x^{2} + y^{2}$ for x ≠ 0, and f(x, y) = 0 for x = 0. Which of the following statements are true?

A.f is continuous at (0, 0).✓
B.The partial derivatives $\partial f / \partial x$ and $\partial f / \partial y$ exist at (0, 0).✓
C.f is differentiable at (0, 0).
D.f is not differentiable at (0, 0).✓

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward.

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Partials exist without differentiability

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