Let IR denote the set of all real numbers. Define the function f: IR → R by f(x)=\ [[2 - 2x ^ 2 - x ^ 2 * sin(1/x)], [2]] if x≠0, x = 0 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 8, such that f is a decreasing function on the interval (0, δ) For any positive real number 8, the function f is NOT an increasing function on the interval (-8, 0) x = 0 is a point of local minima of f

1. Setting up the function

For $x\neq0$ we can write

$$\begin{aligned} f(x)&=2-2x^2-x^2\sin\!\left(\tfrac{1}{x}\right)\\[4pt] &=2-x^2\bigl(2+\sin\!\left(\tfrac{1}{x}\right)\bigr). \end{aligned}$$

Since $\sin(\theta)$ is always between $-1$ and $1$, the bracket $\bigl(2+\sin(1/x)\bigr)$ is always between $1$ and $3$ (i.e. always positive).

---

2. Checking differentiability at $x=0$

Recall the definition of the derivative at $0$:

$$f'(0)=\lim_{h\to0}\frac{f(h)-f(0)}{h}.$$

Here

$$\frac{f(h)-f(0)}{h}=\frac{-h^2\bigl(2+\sin(1/h)\bigr)}{h} =-h\bigl(2+\sin(1/h)\bigr).$$

Because $|2+\sin(1/h)|\le3$, the whole expression is squeezed between $-3|h|$ and $3|h|$, which tends to $0$ as $h\to0$. Hence

$$f'(0)=0\quad\text{(the limit exists).}$$

So $f$ is differentiable at $0$; option A is wrong.

---

3. Derivative for $x\neq0$

For $x\neq0$ differentiate term by term:

$$\begin{aligned} f'(x)&=\frac{d}{dx}\bigl[2-2x^2-x^2\sin(1/x)\bigr]\\[4pt] &=-4x-\Bigl[2x\sin\!\left(\tfrac{1}{x}\right)-\cos\!\left(\tfrac{1}{x}\right)\Bigr]\\[4pt] &=\cos\!\left(\tfrac{1}{x}\right)-4x-2x\sin\!\left(\tfrac{1}{x}\right). \end{aligned}$$

Close to $0$, the two terms containing $x$ vanish (they are $O(x)$), so the sign of $f'(x)$ is essentially the sign of $\cos(1/x)$, which oscillates between $+1$ and $-1$ infinitely often when $x$ approaches $0$.

---

4. Consequences of the oscillating sign

• Because $\cos(1/x)$ keeps changing sign, no matter how tiny an interval $(0,\delta)$ you choose, $f'(x)$ will sometimes be positive and sometimes negative inside it. Hence we can never get purely decreasing behaviour on $(0,\delta)$. So option B is false.
• The same oscillation happens on the left: in every $(-\delta,0)$ the derivative changes sign, so $f$ cannot be purely increasing there. Thus option C is true.

---

5. Local maximum or minimum at 0?

Remember we rewrote

$$f(x)=2-x^2(2+\sin(1/x)).$$

Since the bracket is always positive, $x\neq0$ gives

$$f(x)<2=f(0).$$

Therefore all nearby points lie below the point $(0,2)$, making $x=0$ a point of local maximum, not minimum. Option D is false.

Hence,

$$\boxed{\text{Only option C is correct.}}$$

What is the question?

You have a hill‐shaped curve drawn by the rule

$$f(x)=\begin{cases} 2-2x^2-x^2\sin\!\left(\tfrac{1}{x}\right) & \text{if }x\neq0\\[4pt] 2 & \text{if }x=0\end{cases}$$

You must pick which sentence about this curve is right.

How do we judge the sentences?

1. Smoothness at 0 (differentiable or not) – check if the slope exists.
2. Always going down just to the right of 0 – see if the slope stays negative.
3. Always going up just to the left of 0 – see if the slope stays positive (we’ll test the opposite).
4. Is 0 the lowest point nearby? – compare the height at 0 with the heights of nearby points.

By using only the ideas of slopes (derivatives), sign of slope, and comparing heights, we can decide which line is true.

Step-by-step solution

1. Derivative at 0

   $$\frac{f(h)-f(0)}{h}=\frac{-h^2\bigl(2+\sin(1/h)\bigr)}{h}=-h\bigl(2+\sin(1/h)\bigr)\to0.$$

   So $f'(0)$ exists $\Rightarrow$ option A is false.

2. Derivative for $x\neq0$

   $$f'(x)=\cos\!\left(\tfrac{1}{x}\right)-4x-2x\sin\!\left(\tfrac{1}{x}\right).$$

   Near $0$, the dominant term is $\cos(1/x)$ which oscillates between $-1$ and $1$. Hence $f'(x)$ changes sign infinitely often in every neighbourhood of $0$.

   • Cannot be strictly decreasing on $(0,\delta)\;\forall\delta$ $\Rightarrow$ option B false.
   • Cannot be strictly increasing on $(-\delta,0)$ for any $\delta$ $\Rightarrow$ option C true.

3. Local extrema at 0

   $$f(x)=2-x^2\bigl(2+\sin(1/x)\bigr)<2=f(0)\quad(\forall x\neq0~\text{near }0).$$

   Hence $0$ is a local maximum, not a minimum $\Rightarrow$ option D false.

Therefore, the correct statement is option C.