Question 3

SCQHARD

Let $\mathbb{R}$ denote the set of all real numbers. Define the function $f: \mathbb{R} \to \mathbb{R}$ by

$f(x) = \begin{cases} 2 - 2x^2 - x^2 \sin \frac{1}{x} & \text{if } x \neq 0, \\ 2 & \text{if } x = 0. \end{cases}$ 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 $f$

Detailed Solution

First, check differentiability at $x=0$ : $f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{(2 - 2h^2 - h^2 \sin \frac{1}{h}) - 2}{h} = \lim_{h \to 0} (-2h - h \sin \frac{1}{h}) = 0$ .

So, $f$ is differentiable at $x=0$ , making (A) false.

Now find $f'(x)$ for $x \neq 0$ : $f'(x) = -4x - [2x \sin \frac{1}{x} + x^2 \cos \frac{1}{x} (-\frac{1}{x^2})] = -4x - 2x \sin \frac{1}{x} + \cos \frac{1}{x}$ .

As $x \to 0$ , the terms $-4x$ and $-2x \sin \frac{1}{x}$ approach $0$ , but $\cos \frac{1}{x}$ oscillates between $-1$ and $1$ infinitely often in any neighborhood of $0$ .

Thus, $f'(x)$ changes sign infinitely often in any interval $(-\delta, 0)$ or $(0, \delta)$ . This implies $f$ is not monotonic (neither strictly increasing nor strictly decreasing) on any such interval. Hence, (B) is false and (C) is true.

Finally, for $x \neq 0$ , $f(x) - f(0) = -2x^2 - x^2 \sin \frac{1}{x} = -x^2 (2 + \sin \frac{1}{x})$ .

Since $|\sin \frac{1}{x}| \le 1$ , we have $2 + \sin \frac{1}{x} \ge 1$ . Thus $f(x) - f(0) < 0$ for all $x \neq 0$ , meaning $x=0$ is a point of local maxima. (D) is false.

Question 3

Detailed Solution

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