Discussion
Let \(f\) be a function from R2 to R given by :
\[ f(x, y) = \begin{cases} x & \text{if } x = y \\ 1 & \text{if } x \neq y \end{cases} \]
Then, which of the following statements is true ?
S1 : Both partial derivatives of \(f\) at (0, 0) exist and are equal
S2 : \(f\) is differentiable at (0, 0)
Answer: C
Neither statement holds, so the answer is Neither \(S_{1}\) nor \(S_{2}\).
At the origin \(f(0, 0) = 0\) because \(x = y\) there. Along the \(x\)-axis, every point \((h, 0)\) with \(h \neq 0\) has \(x \neq y\), so \(f(h, 0) = 1\). The difference quotient is
\[ \frac{f(h, 0) - f(0, 0)}{h} = \frac{1}{h}, \]
which has no limit as \(h \to 0\). So \(f_{x}(0, 0)\) does not exist, and by the same computation along the \(y\)-axis neither does \(f_{y}(0, 0)\). \(S_{1}\) is false.
\(f\) is not even continuous at the origin: points arbitrarily close to (0, 0) off the diagonal have value 1 while \(f(0, 0) = 0\). A function that is not continuous at a point cannot be differentiable there, so \(S_{2}\) is false as well. That rules out every option that asserts \(S_{1}\) or \(S_{2}\).
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