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Part CCSIR NET December 2024oddness-makes-the-restriction-linear-not-just-positively-homogeneous

Oddness makes the restriction linear not just positively homogeneous

Let g be a real-valued continuous function on the set {(x,y)∈R2(x, y) \in \mathbb{R}^{2} | x2+y2=1x^{2} + y^{2} = 1} such that g(0, 1) = g(1, 0) = 0 and g(−x, −y) = −g(x, y). Define f:R2→Rf : \mathbb{R}^{2} \to \mathbb{R} by f(x,y)=x2+y2⋅g(x/x2+y2,y/x2+y2)f(x, y) = \sqrt{x^{2} + y^{2}}\cdot{}g(x/\sqrt{x^{2} + y^{2}}, y/\sqrt{x^{2} + y^{2}}) if (x, y) ≠ (0, 0), and f(x, y) = 0 if (x, y) = (0, 0). For each (a,b)∈R2(a, b) \in \mathbb{R}^{2}, define h(a,b):R→Rh_(a,b) : \mathbb{R} \to \mathbb{R} by h_(a,b)(t) = f(ta, tb). Which of the following statements are necessarily true?

  1. A.The function h_(a,b) is differentiable on R\mathbb{R} for each (a,b)∈R2(a, b) \in \mathbb{R}^{2}.✓ Correct
  2. B.There exists (a,b)∈R2(a, b) \in \mathbb{R}^{2} such that h_(a,b) is not differentiable at t = 0.
  3. C.The function f is differentiable at the point (0, 0).
  4. D.If the function f is differentiable at (0, 0) then g is identically zero.✓ Correct

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, and why this one tests hypothesis dropped.

See pricing

50 are analysed free — try those first.

The trap it tests

Hypothesis dropped

A theorem applied without checking one of its preconditions.

Drill statements like this

Related counterexample: If all partial derivatives exist at a point then f is continuous there

More on this topic

The chapter behind this: Differentiability in several variables — free to read

From Functions of Several Variables › Partial derivatives, differentiability, chain rule

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

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