f(x)=x+ ∣ x ∣ , then the function is:

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