Show \(g^{-1}\) Exists but \(f^{-1}\) Does Not Exist and Find \(g^{-1}\)
📝 Question
Let:
\[ A=\{x\in \mathbb{R} \mid -1 \le x \le 1\} \]
Define functions:
\[ f:A\to A,\quad f(x)=x^2 \]
\[ g:A\to A,\quad g(x)=\sin\left(\frac{\pi x}{2}\right) \]
Show that \(g^{-1}\) exists but \(f^{-1}\) does not exist. Also find \(g^{-1}\).
✅ Solution
🔹 Step 1: Check invertibility of \(f(x)=x^2\)
Take:
\[ f(1)=1,\quad f(-1)=1 \]
Since different inputs give the same output, \(f\) is not one-one.
Hence, \(f^{-1}\) does not exist.
🔹 Step 2: Check invertibility of \(g(x)=\sin\left(\frac{\pi x}{2}\right)\)
The function \(\sin\left(\frac{\pi x}{2}\right)\) is strictly increasing on \([-1,1]\).
Also,
\[ g(-1)=-1,\quad g(1)=1 \]
So range is \([-1,1]\).
Thus, \(g\) is one-one and onto.
Hence, \(g^{-1}\) exists.
🔹 Step 3: Find \(g^{-1}\)
Let:
\[ y=\sin\left(\frac{\pi x}{2}\right) \]
Take inverse sine:
\[ \frac{\pi x}{2}=\sin^{-1}(y) \]
\[ x=\frac{2}{\pi}\sin^{-1}(y) \]
Interchanging \(x\) and \(y\):
:contentReference[oaicite:0]{index=0} —🎯 Final Answer
\[ \boxed{f^{-1} \text{ does not exist}} \]
\[ \boxed{g^{-1}(x)=\frac{2}{\pi}\sin^{-1}(x)} \]
🚀 Exam Shortcut
- Check injectivity: \(x^2\) fails since \(f(-x)=f(x)\)
- \(\sin x\) is increasing on restricted interval ⇒ invertible
- Apply inverse sine and solve directly