For ${\displaystyle \frac{1}{3}}<x<y<2$, which of the following statements is/are always correct?

The correct option is B - II only.

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 $f^{\prime }(t)=1-1/t^2$.
- If $t>1$, $f^{\prime }(t)>0$ (Function is increasing).
- If $t<1$, $f^{\prime }(t)<0$ (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.