The correct option is B - II only.

[as per provisional answerkey]

Solution

We are given the range: $1 / 3 < x < y < 2$ . We need to check if the statements are always correct.

Evaluating Statement I: $x + 1 / x < y + 1 / y$
Let $f (t) = t + 1 / t$ . To see if $f (x) < f (y)$ when $x < y$ , we check if the function is increasing.
The derivative $f^{'} (t) = 1 - 1 / t^{2}$ .
- If $t > 1$ , $f^{'} (t) > 0$ (Function is increasing).
- If $t < 1$ , $f^{'} (t) < 0$ (Function is decreasing).
Since our range (1/3 to 2) includes values both below and above 1, the inequality will not always hold.
Example: Let $x = 1 / 2$ and $y = 1$ .
$x + 1 / x = 0.5 + 2 = 2.5$
$y + 1 / y = 1 + 1 = 2$
Here, $2.5 < 2$ is False. Thus, Statement I is not always correct.

Evaluating Statement II: $\sqrt{1 + y^{2}} / y < \sqrt{1 + x^{2}} / x$
Let $g (t) = \sqrt{1 + t^{2}} / t$ . We can rewrite this as $g (t) = \sqrt{(1 + t^{2}) / t^{2}} = \sqrt{1 / t^{2} + 1}$ .
As t increases:
1. t^2 increases.
2. $1 / t^{2}$ decreases.
3. $(1 / t^{2} + 1)$ decreases.
4. $\sqrt{1 / t^{2} + 1}$ decreases.
Since g(t) is a strictly decreasing function for all $t > 0$ , if $x < y$ , then $g (x) > g (y)$ .
This matches the statement: $g (y) < g (x)$ .
Therefore, Statement II is always correct.

Why the other options are incorrect

Option (a) - I only: This is incorrect because Statement I fails when x and y are chosen such that x is significantly less than 1 (where the function is decreasing) and y is closer to 1.
Option (c) - Both I and II: This is incorrect because, as demonstrated above, Statement I is not universally true for the given interval.
Option (d) - Neither I nor II: This is incorrect because Statement II is mathematically certain due to the strictly decreasing nature of the function $\sqrt{1 / t^{2} + 1}$ for positive values.

Key Concept

Monotonicity of functions: A function $f (x)$ is increasing if its derivative is positive, and decreasing if its derivative is negative, determining the direction of inequalities.