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Part CCSIR NET June 2025the-second-order-euler-constant-is-k-times-k-minus-1-not-k-squared

The second order euler constant is k times k minus 1 not k squared

Let f:R2→Rf : \mathbb{R}^{2} \to \mathbb{R} be a twice continuously differentiable non-zero function such that f(tx1_{1}, tx2)=t3f(x1,x2)_{2}) = t^{3}f(x_{1}, x_{2}) for all t > 0 and (x1,x2)∈R2(x_{1}, x_{2}) \in \mathbb{R}^{2}. Which of the following statements are necessarily true?

  1. A.3∂f/∂x1(1,1)+3∂f/∂x2(1,1)=f(1,1)3 \partial{}f/\partial{}x_{1}(1, 1) + 3 \partial{}f/\partial{}x_{2}(1, 1) = f(1, 1)
  2. B.∂f/∂x1(1,−1)−∂f/∂x2(1,−1)=3f(1,−1)\partial{}f/\partial{}x_{1}(1, -1) - \partial{}f/\partial{}x_{2}(1, -1) = 3f(1, -1)✓ Correct
  3. C.x12∂2f/∂x12(x1,x2)+x22∂2f/∂x22(x1,x2)+2x1x2∂2f/∂x1∂x2(x1,x2)=6f(x1,x2)x_{1}^{2}\partial^{2}f/\partial{}x_{1}^{2}(x_{1}, x_{2}) + x_{2}^{2}\partial^{2}f/\partial{}x_{2}^{2}(x_{1}, x_{2}) + 2x_{1}x_{2} \partial^{2}f/\partial{}x_{1}\partial{}x_{2}(x_{1}, x_{2}) = 6f(x_{1}, x_{2})✓ Correct
  4. D.x12∂2f/∂x12(x1,x2)+x22∂2f/∂x22(x1,x2)+2x1x2∂2f/∂x1∂x2(x1,x2)=9f(x1,x2)x_{1}^{2}\partial^{2}f/\partial{}x_{1}^{2}(x_{1}, x_{2}) + x_{2}^{2}\partial^{2}f/\partial{}x_{2}^{2}(x_{1}, x_{2}) + 2x_{1}x_{2} \partial^{2}f/\partial{}x_{1}\partial{}x_{2}(x_{1}, x_{2}) = 9f(x_{1}, x_{2})

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 execution slip.

See pricing

50 are analysed free — try those first.

The trap it tests

Execution slip

The idea was right. The computation was not.

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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