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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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