Multiple choice

In each of the following questions, two equations are given. You have to solve them and $I. (x^2 - 1 = 0) II. (y^3 - 1 = 0)$ Give answer-

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

  5. If x = y or relationship can not be determined
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

From I: x² - 1 = 0 gives x² = 1, so x = ±1. From II: y³ - 1 = 0 gives y³ = 1, so y = 1 (real cube root). Now compare: if x = -1, then x < y (since -1 < 1). If x = 1, then x = y. Combining both cases, we have x ≤ y (x is always less than or equal to y). Therefore 'x ≤ y' is correct.

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