For the function, $f(x) = x\cos\frac{1}{x}$, $x \geq 1$,

For the function, f(x) = xcos1/x, x ≥ 1, $f(x) = x\cos\dfrac{1}{x}, \qquad x \in [1, \infty)$ $\text{Which of (1), (2), (3), (4) are TRUE?}$ $1.\ f'(x), \ f''(x)$ $2.\ \lim_{x\to\infty} f'(x), \text{ monotonicity of } f'$ $3.\ \text{MVT: } f(x+2) - f(x) = 2f'(c)$ $f'(x) = \cos\dfrac{1}{x} + x\left(-\sin\dfrac{1}{x}\right)\left(-\dfrac{1}{x^2}\right) = \cos\dfrac{1}{x} + \dfrac{1}{x}\sin\dfrac{1}{x}$ $f''(x) = \dfrac{\sin(1/x)}{x^2} - \dfrac{\sin(1/x)}{x^2} - \dfrac{\cos(1/x)}{x^3} = -\dfrac{\cos(1/x)}{x^3}$ $x \geq 1 \Rightarrow \cos\dfrac{1}{x} > 0 \Rightarrow f''(x) < 0 \Rightarrow f' \text{ strictly decreasing} \Rightarrow (4) \text{ TRUE}$ $\lim_{x \to \infty} f'(x) = \cos 0 + 0 = 1 \Rightarrow (2) \text{ TRUE}$ $g(t) = \cos t + t\sin t, \quad g(0) = 1, \quad g'(t) = t\cos t > 0 \ (0 < t \leq 1)$ $\Rightarrow f'(x) = g\!\left(\dfrac{1}{x}\right) > 1 \quad \forall\, x \geq 1$ $\text{MVT on } [x, x+2]: \ f(x+2) - f(x) = 2f'(c), \ c \in (x, x+2)$ $f'(c) > 1 \Rightarrow f(x+2) - f(x) > 2 \ \forall x \Rightarrow (3) \text{ TRUE}, \ (1) \text{ FALSE}$ $\boxed{\text{TRUE: } (2), (3), (4)}$