2 marks, −0.667MathematicsMultivariable calculusPartial derivatives and differentiabilityHardabout 2 min

IIT JAM 2022 Question 19: Multivariable calculus

Let $f:\mathbb{R}^2\to\mathbb{R}$ be $f(x,y) = \dfrac{x y^2}{x^3 + 2x^2 y + y^3}$ with $f(0,0)=0$ . Then

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Answer and worked solution

Answer: (D)

Approach the origin along $y = m x$ . Then $f = \dfrac{x\,(m x)^2}{x^3 + 2x^2(mx) + (mx)^3} = \dfrac{m^2 x^3}{x^3(1 + 2m + m^3)} = \dfrac{m^2}{1 + 2m + m^3}$ .

This limit depends on the slope $m$ , so it is not unique. The two-variable limit does not exist, hence $f$ is not continuous at the origin, option (D). (A function discontinuous at a point cannot be differentiable there.)

The trap: (A)

f

is differentiable at

(0,0)

The partial derivatives $f_x$ and $f_y$ can exist along the axes even when the full limit fails; existence of partials does not imply continuity, so (A) is wrong.

Plain-text version

Let f:ℝ²→ℝ be f(x,y) = (x y²)/(x³ + 2x² y + y³) with f(0,0)=0. Then (A) f is differentiable at (0,0). (B) fₓ does not exist at (0,0). (C) f_y does not exist at (0,0). (D) f is not continuous at (0,0).

Also written as: Let f:R^2->R be f(x,y) = (x y^2)/(x^3 + 2x^2 y + y^3) with f(0,0)=0. Then

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

What this question tests

This question is on Partial derivatives and differentiability, part of Multivariable calculus in Mathematics.

Derivatives with respect to one variable at a time, and how they relate to continuity and differentiability.

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