Check Function \(f(x)=x^3+1\) on \( \mathbb{Q} \)

📺 Video Explanation

📝 Question

Check whether the function

\[ f:\mathbb{Q}\to\mathbb{Q},\quad f(x)=x^3+1 \]

is:

• injection (one-one)
• surjection (onto)
• bijection

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✅ Solution

🔹 Step 1: Check Injection (One-One)

Assume:

\[ f(x_1)=f(x_2) \]

Then:

\[ x_1^3+1=x_2^3+1 \]

So:

\[ x_1^3=x_2^3 \]

Thus:

\[ x_1=x_2 \]

✔ Function is one-one.

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🔹 Step 2: Check Surjection (Onto)

Let:

\[ y\in\mathbb{Q} \]

Need:

\[ x^3+1=y \]

So:

\[ x^3=y-1 \]

\[ x=\sqrt[3]{y-1} \]

But cube root of a rational number need not be rational.

Example:

Take:

\[ y=3 \]

Then:

\[ x^3=2 \]

No rational \(x\) satisfies this.

So:

\[ 3 \] is not in range.

❌ Not onto.

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🎯 Final Answer

\[ \boxed{\text{f is one-one but not onto}} \]

So:

✔ Injection
❌ Surjection
❌ Bijection

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🚀 Exam Shortcut

• Cubic function is injective
• Over rationals, cube roots may be irrational
• Find one missing rational value to disprove onto