Check Injective / Surjective
🎥 Video Explanation
📝 Question
Let \( f:\left[-\tfrac12,\tfrac12\right] \to \left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right] \) be defined by
\[ f(x)=\sin^{-1}(3x-4x^3) \]
- A. bijection
- B. injection but not a surjection
- C. surjection but not an injection
- D. neither
✅ Solution
🔹 Step 1: Key Identity
\[ 3x – 4x^3 = \sin(3\theta) \quad \text{if } x=\sin\theta \]
Let \(x=\sin\theta\), where \(\theta \in \left[-\tfrac{\pi}{6},\tfrac{\pi}{6}\right]\)
—🔹 Step 2: Transform Function
\[ f(x)=\sin^{-1}(\sin 3\theta) \]
Since \(3\theta \in \left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]\),
\[ \sin^{-1}(\sin 3\theta)=3\theta \]
—🔹 Step 3: Final Form
\[ f(x)=3\theta = 3\sin^{-1}(x) \]
—🔹 Step 4: Check Injective
\(\sin^{-1}(x)\) is strictly increasing ⇒ so is \(3\sin^{-1}(x)\)
✔️ Injective
—🔹 Step 5: Check Surjective
When \(x \in \left[-\tfrac12,\tfrac12\right]\),
\[ \sin^{-1}(x) \in \left[-\tfrac{\pi}{6},\tfrac{\pi}{6}\right] \]
\[ f(x)=3\sin^{-1}(x) \in \left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right] \]
Matches codomain ⇒ ✔️ Onto
—🔹 Final Answer
\[ \boxed{\text{Option A: bijection}} \]