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Questions › CUET-PG › CUET-PG 2023 Shift 1 › Q82
4 marks, −1Mathematics › Multivariable calculus › Homogeneous functions and Euler's theoremModerateabout 1 min

CUET-PG 2023 Shift 1 Question 82: Multivariable calculus

Suppose z=f(x,y)z = f(x, y)z=f(x,y) is homogeneous of degree kkk. Which of the following statements is true?

  1. ∂z∂x\frac{\partial z}{\partial x}∂x∂z​ is homogeneous of degree (k−2)(k-2)(k−2)
  2. ∂2z∂y∂x\frac{\partial^2 z}{\partial y \partial x}∂y∂x∂2z​ is homogeneous of degree (k−1)(k-1)(k−1)
  3. ∂2z∂y2\frac{\partial^2 z}{\partial y^2}∂y2∂2z​ is homogeneous of degree (k−2)(k-2)(k−2)
  4. ∂2z∂x∂y\frac{\partial^2 z}{\partial x \partial y}∂x∂y∂2z​ is homogeneous of degree (k−1)(k-1)(k−1)

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Plain-text version

Suppose z = f(x, y) is homogeneous of degree k. Which of the following statements is true? 1. (∂ z)/(∂ x) is homogeneous of degree (k-2) 2. (∂² z)/(∂ y ∂ x) is homogeneous of degree (k-1) 3. (∂² z)/(∂ y²) is homogeneous of degree (k-2) 4. (∂² z)/(∂ x ∂ y) is homogeneous of degree (k-1) (A) (∂ z)/(∂ x) is homogeneous of degree (k-2) (B) (∂² z)/(∂ y ∂ x) is homogeneous of degree (k-1) (C) (∂² z)/(∂ y²) is homogeneous of degree (k-2) (D) (∂² z)/(∂ x ∂ y) is homogeneous of degree (k-1)

Also written as: Suppose z = f(x, y) is homogeneous of degree k. Which of the following statements is true? 1. (d z)/(d x) is homogeneous of degree (k-2) 2. (d^2 z)/(d y d x) is homogeneous of degree (k-1) 3. (d^2 z)/(d y^2) is homogeneous of degree (k-2) 4. (d^2 z)/(d x d y) is homogeneous of degree (k-1)

Multivariable calculus has come up 32 times in 7 exams since 2007.

What this question tests

This question is on Homogeneous functions and Euler's theorem, part of Multivariable calculus in Mathematics.

Functions that scale in a fixed way when all inputs scale, and the identity this gives.

TestStop has 12 past-paper questions on this subtopic; the nearest in year are below.

  • IIT JAM 2022 · Q38Which of the following functions is/are homogeneous?
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Question from the CUET-PG 2023 Shift 1 paper. Solutions are TestStop's own. Not affiliated with or endorsed by any examination authority.