Question

CSATHardPrelims 2026Maths

For $\frac{1}{3} < x < y < 2$ , which of the following statements is/are always correct? I. $x + \frac{1}{x} < y + \frac{1}{y}$ II. $\frac{\sqrt{1 + y^2}}{y} < \frac{\sqrt{1 + x^2}}{x}$ Select the answer using the code given below.

Explanation

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 $(0, \infty)$ .