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    Question

    If |x + 1| + |x βˆ’ 2| β‰₯ 4, then x lies in:

    A x ≀ βˆ’1 or x β‰₯ 3 Correct Answer Incorrect Answer
    B x ≀ 0 or x β‰₯ 4 Correct Answer Incorrect Answer
    C x ≀ βˆ’2 or x β‰₯ 2 Correct Answer Incorrect Answer
    D x ≀ βˆ’1.5 or x β‰₯ 2.5 Correct Answer Incorrect Answer

    Solution

    Given Inequality: |x + 1| + |x βˆ’ 2| β‰₯ 4 Identify the critical points where the expressions inside absolute values change sign. The critical points are:

    • x + 1 = 0 β†’ x = -1
    • x - 2 = 0 β†’ x = 2
    These points divide the real line into three intervals: x < -1, -1 ≀ x < 2, and x β‰₯ 2. Analyze the inequality in each interval. Case 1: x < -1 When x < -1:
    • x + 1 < 0, so |x + 1| = -(x + 1) = -x - 1
    • x - 2 < 0, so |x - 2| = -(x - 2) = -x + 2
    The inequality becomes: (-x - 1) + (-x + 2) β‰₯ 4 -2x + 1 β‰₯ 4 -2x β‰₯ 3 x ≀ -3/2 Since we're in the case x < -1, the solution in this interval is: x ≀ -3/2 Case 2: -1 ≀ x < 2 When -1 ≀ x < 2:
    • x + 1 β‰₯ 0, so |x + 1| = x + 1
    • x - 2 < 0, so |x - 2| = -(x - 2) = -x + 2
    The inequality becomes: (x + 1) + (-x + 2) β‰₯ 4 3 β‰₯ 4 This is never true, so there are no solutions in this interval. Case 3: x β‰₯ 2 When x β‰₯ 2:
    • x + 1 > 0, so |x + 1| = x + 1
    • x - 2 β‰₯ 0, so |x - 2| = x - 2
    The inequality becomes: (x + 1) + (x - 2) β‰₯ 4 2x - 1 β‰₯ 4 2x β‰₯ 5 x β‰₯ 5/2 Since we're in the case x β‰₯ 2, and 5/2 = 2.5 > 2, the solution in this interval is: x β‰₯ 5/2 Step 3: Combine the solutions. From Case 1: x ≀ -3/2 From Case 2: No solutions From Case 3: x β‰₯ 5/2 Therefore, the solution is: x ∈ (-∞, -3/2] βˆͺ [5/2, +∞) So the answer is x ≀ βˆ’1.5 or x β‰₯ 2.5.

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