Answer: B (II only)

Statement I — Not always true.

Consider $f(t) = t + \tfrac{1}{t}$. Its derivative is $f'(t) = 1 - \tfrac{1}{t^2}$, which is negative on $(0,1)$ and positive on $(1,\infty)$. So $f$ decreases on $(\tfrac{1}{3}, 1)$ and increases on $(1, 2)$ — not monotonic on the whole interval.

Counterexample: $x = \tfrac{1}{2}, y = \tfrac{3}{4}$. Then $x + \tfrac{1}{x} = 2.5$ but $y + \tfrac{1}{y} = 0.75 + 1.333\ldots \approx 2.083$. So $x + \tfrac{1}{x} > y + \tfrac{1}{y}$, contradicting I.

Statement II — Always true.

Rewrite:

$\frac{\sqrt{1+t^2}}{t} = \sqrt{\frac{1+t^2}{t^2}} = \sqrt{\frac{1}{t^2} + 1}$

For $t > 0$, as $t$ increases, $\tfrac{1}{t^2}$ decreases, so $\sqrt{\tfrac{1}{t^2}+1}$ decreases. Thus $g(t) = \tfrac{\sqrt{1+t^2}}{t}$ is strictly decreasing on $(0, \infty)$.

Since $x < y$ (both positive), $g(x) > g(y)$, i.e.

$\frac{\sqrt{1+y^2}}{y} < \frac{\sqrt{1+x^2}}{x}. \checkmark$