For ⅓ < x < y < 2, which of the following statements is/are always correct? I. x + 1/x < y + 1/y II. √(1 + y²)/y < √(1 + x²)/x Select the answer using the code given below.

To determine the correct answer, we must analyze the monotonicity (increasing or decreasing nature) of the mathematical functions underlying both statements within the interval t ∈ (⅓, 2).

Why the correct option is correct: Option B is correct because Statement II is universally true for this interval. The expression √(1+t²)/t can be algebraically rewritten as √((1+t²)/t²) = √(1/t² + 1). By the fundamental properties of inequalities, if x 1/y. Squaring both sides yields 1/x² > 1/y², and adding 1 results in 1/x² + 1 > 1/y² + 1. Taking the principal square root proves that √(1/x² + 1) > √(1/y² + 1). Therefore, the expression evaluated for x will always be strictly greater than the expression evaluated for y, confirming Statement II is correct.

Why the wrong options are wrong: Options A and C are incorrect because Statement I assumes uniform behavior for the function f(t) = t + 1/t. Applying the First Derivative Test from calculus, f'(t) = 1 - 1/t². This means the function decreases when t 1. If we take x = 0.5 and y = 0.8 (both valid within the ⅓ f(y), which actively disproves the statement's claim. Option D is incorrect because Statement II is demonstrably true.

Takeaway: When evaluating inequalities over intervals crossing the value of 1, always test fractional values on both sides of 1. Rewriting compound algebraic fractions into segregated forms—like converting √(1+t²)/t to √(1/t² + 1)—greatly simplifies tracing whether a function strictly increases or decreases.