Check Function \(f(x)=\dfrac{x}{2}\) on \([-1,1]\)

📺 Video Explanation

📝 Question

Let:

\[ A=[-1,1] \]

Discuss whether the function:

\[ f:A\to A,\quad f(x)=\frac{x}{2} \]

is one-one, onto, or bijective.

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✅ Solution

🔹 Check One-One (Injective)

Assume:

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

Then:

\[ \frac{x_1}{2}=\frac{x_2}{2} \]

So:

\[ x_1=x_2 \]

✔ Function is one-one.

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🔹 Check Onto (Surjective)

Since:

\[ x\in[-1,1] \]

Then:

\[ f(x)=\frac{x}{2}\in\left[-\frac12,\frac12\right] \]

So range is:

\[ \left[-\frac12,\frac12\right] \]

But codomain is:

\[ [-1,1] \]

Values like:

\[ \frac34,\ -1 \]

are not attained.

❌ 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

• Linear scaling keeps injective nature
• Compare image interval with codomain
• Smaller image interval means not onto