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    Question

    f(x)=x+ ∣ x ∣ , then the function is:

    A Injective Correct Answer Incorrect Answer
    B Bijective Correct Answer Incorrect Answer
    C Onto but not one-one Correct Answer Incorrect Answer
    D Not injective, not surjective Correct Answer Incorrect Answer

    Solution

    We are given the function: f(x) = x + |x| , defined from ℝ → ℝ Let’s analyze this function. Understand the function piecewise Recall:

    • |x| = x if x ≥ 0
    • |x| = −x if x < 0
    So: f(x) = x + |x| =
    • 2x if x ≥ 0
    • 0 if x < 0
    Thus:
    • For x ≥ 0, f(x) = 2x
    • For x < 0, f(x) = 0
    Check for Injectivity (One-one) A function is injective if f(x₁) = f(x₂) ⇒ x₁ = x₂ Now:
    • f(x) = 0 for all x < 0
    • f(0) = 2×0 = 0
    So, all negative x-values and x = 0 are mapped to the same output (0) ⇒ Multiple inputs give the same output ⇒ Not one-one Check for Surjectivity (Onto) Codomain is ℝ. But what is the range ? From piecewise definition:
    • For x < 0 → f(x) = 0
    • For x ≥ 0 → f(x) = 2x ⇒ outputs all real numbers ≥ 0
    So, range = [0, ∞) Since ℝ ≠ [0, ∞) , it does not cover entire ℝ ⇒ Not onto Therefore, the final answer is:
    • Not injective
    • Not surjective

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