`f: R->R` given by `f(x)=x+sqrt(x^2)` is (a) injective (b) surjective (c) bijective (d) none of these

Explanation

To determine whether the function \( f: \mathbb{R} \to \mathbb{R} \) given by \( f(x) = x + \sqrt{x^2} \) is injective, surjective, bijective, or none of these, we can analyze the function step by step.

### Step 1: Simplify the function
The expression \( \sqrt{x^2} \) can be simplified based on the value of \( x \):
- If \( x \geq 0 \), then \( \sqrt{x^2} = x \).
- If \( x < 0 \), then \( \sqrt{x^2} = -x \).

Thus, we can rewrite the function \( f(x) \) as:
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