1/3 < x < y < 2 के लिए निम्नलिखित में से कौन-सा/कौन-से कथन सही है/हैं?
I. x + 1/x < y + 1/y
II.
नीचे दिए गए कूट का प्रयोग कर उत्तर चुनिए।
सही उत्तर: (a) केवल I
दी गई शर्त 1/3 < x < y < 2 के लिए कथनों की जाँच करते हैं। कथन I: x + 1/x < y + 1/y। फलन f(t) = t + 1/t अंतराल [1/3, 2] में पूर्णतः एकदिष्ट (monotonic) नहीं है (यह t = 1 तक घटता है और उसके बाद बढ़ता है)। यदि हम x = 0.5 और y = 1.5 लें, तो x + 1/x = 2.5 और y + 1/y = 2.166, जिससे कथन I असत्य सिद्ध होता है। कथन II: sqrt(1+y^2)/y < sqrt(1+x^2)/x। फलन g(t) = sqrt(1+t^2)/t एक ह्रासमान (decreasing) फलन है। चूँकि x < y है, अतः g(x) > g(y) होगा, जो कथन II की पूर्ण पुष्टि करता है। अतः केवल II सही है।
In English (Question & Model Answer)
For 1/3 < x < y < 2, which of the following statements is/are always correct?
I. x + 1/x < y + 1/y
II.
Select the answer using the code given below.
Correct Option: (a) I only
We are given 1/3 < x < y < 2. We need to evaluate two statements: I. x + 1/x < y + 1/y. Let f(t) = t + 1/t for t > 0. The derivative f'(t) = 1 - 1/t^2. For t > 1, f'(t) > 0, meaning f(t) is increasing for t > 1. For t < 1, f'(t) < 0, meaning f(t) is decreasing for 0 < t < 1. Since the interval 1/3 < x < y < 2 spans across t = 1, f(t) is not strictly monotonic over the entire interval (it decreases from 1/3 to 1 and increases from 1 to 2). Thus, we cannot directly apply monotonicity. Let's test specific values: let x = 0.5 and y = 1.5. Then x + 1/x = 0.5 + 2 = 2.5. And y + 1/y = 1.5 + 2/3 = 2.166. Here x + 1/x > y + 1/y, which contradicts statement I. Wait! Let's check another pair: let x = 0.8 and y = 1.2. f(0.8) = 0.8 + 1.25 = 2.05. f(1.2) = 1.2 + 0.833 = 2.033. Again f(x) > f(y). What if both x and y are greater than 1? E.g., x = 1.2, y = 1.5. f(1.2) = 2.033, f(1.5) = 2.166, so f(x) < f(y). But the statement must hold *always* for all x, y in (1/3, 2). Since it fails for x = 0.5, y = 1.5, statement I is not always correct. Wait, let's re-evaluate statement I: does x + 1/x < y + 1/y hold if both are greater than 1? Yes, but the range given is 1/3 < x < y < 2. Since 0.5 and 1.5 are in this range, statement I is false. Let's check statement II: sqrt(1+y^2)/y < sqrt(1+x^2)/x. This can be rewritten as g(t) = sqrt(1+t^2)/t = sqrt((1/t^2) + 1). As t increases, 1/t^2 decreases, so sqrt(1/t^2 + 1) decreases. Thus g(t) is a strictly decreasing function for all t > 0. Since x < y, we must have g(x) > g(y), which means sqrt(1+y^2)/y < sqrt(1+x^2)/x. This is always true! Therefore, only statement II is correct.