Question

Let $ \mathbb{R} $ denote the set of all real numbers. Define the function $ f: \mathbb{R} \to \mathbb{R} $ by $$ f(x) = \begin{cases} 2 - 2x^2 - x^2 \sin\left(\frac{1}{x}\right), & \text{if } x \ne 0, \\ 2, & \text{if } x = 0. \end{cases} $$ Then which one of the following statements is TRUE?

• A. The function \( f \) is NOT differentiable at \( x = 0 \)
• B. There is a positive real number \( \delta \), such that \( f \) is a decreasing function on the interval \( (0, \delta) \)
• C. For any positive real number \( \delta \), the function \( f \) is NOT an increasing function on the interval \( (-\delta, 0) \)
• D. \( x = 0 \) is a point of local minima of \( f \)

Hint:

When dealing with piecewise functions involving oscillations like \( \sin(1/x) \), check differentiability by evaluating the derivative limits and use bounding to analyze continuity or extrema.

Answer 1

The Correct Option is C

Step 1: Analyze continuity at \( x = 0 \) \[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \left[2 - 2x^2 - x^2 \sin\left(\frac{1}{x}\right)\right] \] Since \( -1 \le \sin\left(\frac{1}{x}\right) \le 1 \), \[ - x^2 \le x^2 \sin\left(\frac{1}{x}\right) \le x^2 \Rightarrow -x^2 \le -x^2 \sin\left(\frac{1}{x}\right) \le x^2 \] \[ -2x^2 - x^2 \le f(x) - 2 \le -2x^2 + x^2 \Rightarrow -3x^2 \le f(x) - 2 \le -x^2 \Rightarrow \lim_{x \to 0} f(x) = 2 \] So, \( f \) is continuous at \( x = 0 \).
Step 2: Check differentiability at \( x = 0 \) Let’s analyze: \[ f(x) = 2 - 2x^2 - x^2 \sin\left(\frac{1}{x}\right) \Rightarrow f'(x) = -4x - 2x \sin\left(\frac{1}{x}\right) + \cos\left(\frac{1}{x}\right) \] Here, \( \cos\left(\frac{1}{x}\right) \) oscillates as \( x \to 0 \) and does not converge. Hence, \( f \) is not differentiable at \( x = 0 \). 
Step 3: Examine local extrema We check the behavior of \( f(x) \) around 0: \[ f(x) = 2 - 2x^2 - x^2 \sin\left(\frac{1}{x}\right) \le 2 - x^2<2 = f(0) \Rightarrow f(x)<f(0) \text{ near } x = 0 \] So, \( x = 0 \) is actually a local maximum, not a local minimum. 
Step 4: Monotonicity Due to the oscillatory term \( \sin\left(\frac{1}{x}\right) \), the function is not monotonic in any neighborhood around 0. This rules out both strict increasing and decreasing nature in any interval \( (-\delta, 0) \) or \( (0, \delta) \).
Thus, the true statement is:
(C) For any positive real number \( \delta \), the function \( f \) is NOT an increasing function on the interval \( (-\delta, 0) \)