UPSC Prelims 2026 CSAT, question 21

Solution.

Answer: (b) II only

1. Statement I is incorrect: on $(\frac{1}{3}, 2)$, $f(t) = t + \frac{1}{t}$ decreases on $(\frac{1}{3}, 1)$ and increases on $(1, 2)$; for example, $x = \frac{1}{2}$ and $y = 1$ give $x + \frac{1}{x} = 2.5 > y + \frac{1}{y} = 2$.
2. Statement II is correct: for any $t > 0$, $\frac{\sqrt{1+t^2}}{t} = \sqrt{\frac{1+t^2}{t^2}} = \sqrt{\frac{1}{t^2} + 1}$, which is a strictly decreasing function of $t$ on $(0, \infty)$.
3. Because $0 < \frac{1}{3} < x < y < 2$, we always have $x^2 < y^2 \Rightarrow \frac{1}{y^2} + 1 < \frac{1}{x^2} + 1 \Rightarrow \frac{\sqrt{1+y^2}}{y} < \frac{\sqrt{1+x^2}}{x}$.

Trap: Do not drop the square root in Statement II: without the radical, $\frac{1+t^2}{t} = t + \frac{1}{t}$ would match Statement I and fail on $(1, 2)$, whereas $\frac{\sqrt{1+t^2}}{t}$ decreases across the entire interval.