Determine Nature of \(f(x)=x+\sqrt{x^2}\)

📝 Question

Let:

\[ f:\mathbb{R}\to\mathbb{R}, \quad f(x)=x+\sqrt{x^2} \]

Determine whether \(f\) is one-one, many-one, onto or into.

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✅ Solution

🔹 Step 1: Simplify function

We know:

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So,

\[ f(x)=x+|x| \] —

🔹 Step 2: Consider cases

Case 1: \(x \ge 0\)

\[ f(x)=x+x=2x \]

Case 2: \(x < 0\)

\[ f(x)=x-x=0 \] —

🔹 Step 3: Check one-one / many-one

For all \(x<0\), \(f(x)=0\).

Different inputs give same output.

Hence, \(f\) is many-one.

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🔹 Step 4: Find range

From above:

• \(x<0 \Rightarrow f(x)=0\)
• \(x\ge0 \Rightarrow f(x)=2x \ge 0\)

So range is:

\[ [0,\infty) \] —

🔹 Step 5: Check onto / into

Codomain is \(\mathbb{R}\), but range is \([0,\infty)\).

Negative values are not obtained.

Hence, function is into (not onto).

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🎯 Final Answer

\[ \boxed{\text{f is many-one and into}} \]

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🚀 Exam Shortcut

• Replace \(\sqrt{x^2}\) with \(|x|\)
• Split into cases: \(x\ge0\), \(x<0\)
• Same output for many inputs ⇒ many-one
• Range ≠ codomain ⇒ into