Question

Let \( f : \mathbb{R} \to \mathbb{R} \) be defined by \[ f(x) = \begin{cases} x^4(2 + \sin \frac{1}{x}), & x \neq 0, \\ 0, & x = 0. \end{cases} \] Then which of the following statement(s) is (are) true? 
 

• A. \( f \) attains its minimum at 0
• B. \( f \) is monotone
• C. \( f \) is differentiable at 0
• D. \( f(x) > 2x^4 + x^3, \text{ for all } x > 0 \)

Hint:

For functions that oscillate, check the value at \( x = 0 \) and compare it with the behavior of the function at nearby points to determine if it attains a minimum there.

Answer 1

The Correct Option is A, C

Given: $$f(x) = \begin{cases} x^4\left(2 + \sin\frac{1}{x}\right), & x \neq 0 \ 0, & x = 0 \end{cases}$$

Option (A): $f$ attains its minimum at 0

For $x \neq 0$: $$f(x) = x^4\left(2 + \sin\frac{1}{x}\right)$$

Since $-1 \leq \sin\frac{1}{x} \leq 1$, we have: $$1 \leq 2 + \sin\frac{1}{x} \leq 3$$

Therefore, for $x \neq 0$: $$x^4 \leq f(x) \leq 3x^4$$

This means $f(x) \geq x^4 > 0$ for all $x \neq 0$.

At $x = 0$: $f(0) = 0$

Since $f(x) > 0$ for all $x \neq 0$ and $f(0) = 0$, the function attains its minimum value of 0 at $x = 0$.

Option (A) is TRUE 

Option (B): $f$ is monotone

Let's check if $f$ is monotone by examining its behavior near 0.

Consider small positive values: For $x = \frac{2}{(2n+1)\pi}$ where $\sin\frac{1}{x} = \sin\left(\frac{(2n+1)\pi}{2}\right) = \pm 1$ alternates.

The function oscillates due to the $\sin\frac{1}{x}$ term, so $f$ is not monotone.

Option (B) is FALSE 

Option (C): $f$ is differentiable at 0

To check differentiability at 0, we compute: $$f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{h^4\left(2 + \sin\frac{1}{h}\right)}{h}$$

$$= \lim_{h \to 0} h^3\left(2 + \sin\frac{1}{h}\right)$$

Since $\left|2 + \sin\frac{1}{h}\right| \leq 3$ and $h^3 \to 0$ as $h \to 0$: $$|f'(0)| = \left|h^3\left(2 + \sin\frac{1}{h}\right)\right| \leq 3|h|^3 \to 0$$

Therefore, $f'(0) = 0$ exists.

Option (C) is TRUE 

Option (D): $f(x) > 2x^4 + x^3$ for all $x > 0$

We need to check if: $$x^4\left(2 + \sin\frac{1}{x}\right) > 2x^4 + x^3$$

This simplifies to: $$x^4\sin\frac{1}{x} > x^3$$ $$x\sin\frac{1}{x} > 1$$

However, we know that $|\sin\frac{1}{x}| \leq 1$, so $|x\sin\frac{1}{x}| \leq |x|$.

For small $x > 0$, we have $x\sin\frac{1}{x} \leq x < 1$, which means the inequality doesn't hold for all $x > 0$.

Specifically, for $x = 0.5$: $x\sin\frac{1}{x} = 0.5\sin(2) \approx 0.5(0.909) \approx 0.455 < 1$.

Option (D) is FALSE 

Answer: (A) and (C)