Question

Let \( f, g : \mathbb{R} \to \mathbb{R} \) be two functions defined by \[ f(x) = \begin{cases} x |x| \sin \frac{1}{x} & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \] and \[ g(x) = \begin{cases} x^2 \sin \frac{1}{x} + x \cos \frac{1}{x} & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \] Then, which one of the following is TRUE?

• A. \( f \) is differentiable at \( x = 0 \), and \( g \) is NOT differentiable at \( x = 0 \)
• B. \( f \) is NOT differentiable at \( x = 0 \), and \( g \) is differentiable at \( x = 0 \)
• C. \( f \) is differentiable at \( x = 0 \), and \( g \) is differentiable at \( x = 0 \)
• D. \( f \) is NOT differentiable at \( x = 0 \), and \( g \) is NOT differentiable at \( x = 0 \)

Hint:

For differentiability at \( x = 0 \), check the limit of the difference quotient. If it does not exist or if the function oscillates, the function is not differentiable.

Answer 1

The Correct Option is A

Step 1: Analyze the function \( f(x) \).
For \( f(x) \), we need to check if it is differentiable at \( x = 0 \). First, let's check the limit for \( f'(0) \): \[ f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} \] Since \( f(0) = 0 \), we have \[ f'(0) = \lim_{h \to 0} \frac{h |h| \sin \frac{1}{h}}{h} \] The term \( \sin \frac{1}{h} \) oscillates between -1 and 1, so the limit does not exist. Hence, \( f \) is not differentiable at \( x = 0 \). 
Step 2: Analyze the function \( g(x) \).
For \( g(x) \), we also check the limit for the derivative at \( x = 0 \): \[ g'(0) = \lim_{h \to 0} \frac{g(h) - g(0)}{h} \] Since \( g(0) = 0 \), we have \[ g'(0) = \lim_{h \to 0} \frac{h^2 \sin \frac{1}{h} + h \cos \frac{1}{h}}{h} \] This simplifies to \[ g'(0) = \lim_{h \to 0} \left( h \sin \frac{1}{h} + \cos \frac{1}{h} \right) \] Since \( \cos \frac{1}{h} \) oscillates between -1 and 1, the limit exists and is 0. Hence, \( g \) is differentiable at \( x = 0 \). 
Final Answer: \[ \boxed{f \text{ is NOT differentiable at } x = 0, \, g \text{ is differentiable at } x = 0} \]