For , which of the following
statements is/are always correct?
I.
II.
Select the answer using the code given
below.
- ((a))
I only
- ((b))
II only
- ((c))
Both I and II
- ((d))
Neither I nor II
Show Answer
II only

Shortcut Trick
For I: The function f(t) = t + 1/t has its minimum at t = 1. Since (1/3, 2) includes 1, the function decreases then increases, so it is not always increasing.
For II: Let g(t) = √(1 + t2) / t = √(1/t2 + 1). As t increases, 1/t2 strictly decreases.
Since x < y, it follows that 1/x2 > 1/y2 ⇒ √(1/x2 + 1) > √(1/y2 + 1).
Thus, g(x) > g(y) is always true for the given range.
∴ The correct answer is II only.

Alternate Method
Given: 1/3 < x < y < 2

For Statement I: x + 1/x < y + 1/y
⇒ Let x = 0.5 and y = 1.0 (both in range 1/3 < x < y < 2)
⇒ x + 1/x = 0.5 + 2 = 2.5
⇒ y + 1/y = 1 + 1 = 2
⇒ 2.5 is not less than 2, so Statement I is False.
For Statement II: √(1 + y2)/y < √(1 + x2)/x
⇒ Let g(t) = √(1 + t2)/t = √((1 + t2)/t2) = √(1/t2 + 1)
⇒ Since x < y ⇒ x2 < y2 ⇒ 1/x2 > 1/y2
⇒ 1/x2 + 1 > 1/y2 + 1
⇒ √(1/x2 + 1) > √(1/y2 + 1)
⇒ Therefore, Statement II is True.
∴ The correct answer is II only.

Additional Information
Monotonic Functions
A function is monotonic if its derivative does not change sign. f(t) = t + 1/t has f'(t) = 1 − 1/t2, which is negative for t < 1 and positive for t > 1.
Inequality Property
For positive values, if a > b, then 1/a < 1/b and √a > √b. These properties are essential for comparing fractional expressions.
AM-GM Inequality
For any positive number x, x + 1/x ≥ 2. The minimum value 2 occurs exactly at x = 1.



































