Find \(f^{-1}(1)\)

🎥 Video Explanation

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📝 Question

Let \(f\) be an injective map from \(\{x,y,z\}\) to \(\{1,2,3\}\).

Exactly one of the following is true:

• \(f(x)=1\)
• \(f(y)\ne 1\)
• \(f(z)\ne 2\)

Find \(f^{-1}(1)\).

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✅ Solution

🔹 Step 1: Injective ⇒ Bijective

Since domain and codomain have same number of elements,

\(f\) is bijective.

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🔹 Step 2: Try Cases (Only one true)

Case 1: \(f(x)=1\) is true

Then:

• \(f(y)\ne1\) ⇒ true
• \(f(z)\ne2\) ⇒ may be true

More than one statement becomes true ⇒ ❌ invalid

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Case 2: \(f(y)\ne1\) is true

Then:

• \(f(x)=1\) must be false ⇒ \(f(x)\ne1\)
• \(f(z)\ne2\) must be false ⇒ \(f(z)=2\)

Now remaining value 1 must go to \(x\) or \(y\).

But \(f(y)\ne1\), so:

\[ f(x)=1 \]

Contradiction ⇒ ❌ invalid

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Case 3: \(f(z)\ne2\) is true

Then:

• \(f(x)=1\) is false ⇒ \(f(x)\ne1\)
• \(f(y)\ne1\) is false ⇒ \(f(y)=1\)

Now remaining values are 2 and 3.

Since \(f(z)\ne2\), so:

\[ f(z)=3,\quad f(x)=2 \]

✔️ Only one statement true ⇒ valid

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🔹 Final Answer

\[ f^{-1}(1)=y \]

\[ \boxed{\text{Option B}} \]