For 1/3 < 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.
Answer & explanation
Answer: (b) II only
t + 1/t falls until t = 1 and rises after it, so Statement I breaks when x and y are on opposite sides of 1. Statement II simplifies to √(1/y² + 1) < √(1/x² + 1), which always holds because y > x > 0.
- Statement I: try x = 1/2, y = 1. Left = 0·5 + 2 = 2·5; right = 1 + 1 = 2. 2·5 < 2 is false, so I is not always correct.
- Statement II: divide inside the root by the square of the denominator: √(1 + y²)/y = √(1/y² + 1) and √(1 + x²)/x = √(1/x² + 1).
- Since 0 < x < y, 1/y² < 1/x², so √(1/y² + 1) < √(1/x² + 1). II always holds.
- Check II with numbers: x = 1, y = 1·5 gives 1·202 < 1·414.
- ✗ I Fails for x = 1/2, y = 1: 2·5 is not less than 2. The function t + 1/t decreases on (1/3, 1).
- ✓ II Equivalent to 1/y² < 1/x², true whenever 0 < x < y.
Remember · For 'always correct', hunt for one counter-example; for the other statement, simplify until the inequality is obvious.
Question and answer: UPSC's provisional GS Paper II (2026, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·