Question

Let f : \(\R^2 → \R\) be the function defined by
\(f(x,y) = \begin{cases}   x^2\sin\frac{1}{x}+y^2\cos y, & x \ne 0 \\   0, & x=0. \end{cases}\)
Then which one of the following statements is NOT true ?

• A. f is continuous at (0, 0)
• B. The partial derivative of f with respect to x is not continuous at (0, 0)
• C. The partial derivative of f with respect to y is continuous at (0, 0)
• D. f is not differentiable at (0, 0)

Answer 1

The Correct Option is D

To determine which statement about the function \(f(x, y)\) is not true, let's analyze each option systematically.

The function is given by:

\(f(x,y) = \begin{cases} x^2\sin\frac{1}{x} + y^2\cos y, & x \neq 0 \\ 0, & x = 0 \end{cases}\)

1. Checking Continuity at (0, 0):

For continuity at \( (0, 0) \), the limit of \( f(x, y) \) as \( (x, y) \to (0, 0) \) must equal \( f(0, 0) = 0 \).

As \( x \to 0 \), \( x^2 \sin \frac{1}{x} \to 0 \) because \(|x^2 \sin \frac{1}{x}| \leq x^2\) and tends to 0.

As \( y \to 0 \), \( y^2 \cos y \to 0 \) because \(y^2 \cos y \approx y^2\) tends to 0.

Thus, \(f(x, y) \to 0\) as \( (x, y) \to (0, 0) \), indicating continuity at the point.

1. Partial Derivative of f with respect to x at (0, 0):

The partial derivative with respect to \( x \) is:

1. \(\frac{\partial f}{\partial x} = \begin{cases} 2x \sin \frac{1}{x} - \cos \frac{1}{x}, & x \neq 0 \\ 0, & x=0 \end{cases}\)

As \( x \to 0 \), \( \cos \frac{1}{x} \) oscillates between -1 and 1. Therefore, \(\frac{\partial f}{\partial x}\) is not continuous at \( (0, 0) \).

1. Partial Derivative of f with respect to y at (0, 0):

The partial derivative with respect to \( y \) is:

1. \(\frac{\partial f}{\partial y} = \begin{cases} -y^2 \sin y + 2y \cos y, & y \neq 0 \\ 0, & y=0 \end{cases}\)

As \( y \to 0 \), both terms \( -y^2 \sin y \) and \( 2y \cos y \) tend to 0. Hence, \(\frac{\partial f}{\partial y}\) is continuous at \( (0, 0) \).

1. Differentiability of f at (0, 0):

A function \( f \) is differentiable at a point if it is linear approximation at that point. Here, since \( \frac{\partial f}{\partial x} \) is not continuous at \( (0, 0) \), \( f \) is not differentiable at \( (0, 0) \).

Based on the above explanation, the statement which is NOT true is: f is not differentiable at (0, 0).