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If (f(x)=\frac{x-1}{x+1}), then for (x\ne -1), what is (f^{-1}(x))?यदि (f(x)=\frac{x-1}{x+1}), तो (x\ne -1) के लिए (f^{-1}(x)) क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: (\frac{1+x}{1-x})(\frac{1+x}{1-x})
Step 1: Let (y=\frac{x-1}{x+1}). Step 2: From (y(x+1)=x-1), we get (x(y-1)=-(1+y)). Step 3: Hence (x=\frac{1+y}{1-y}), so (f^{-1}(x)=\frac{1+x}{1-x}).चरण 1: (y=\frac{x-1}{x+1}) मानें। चरण 2: (y(x+1)=x-1) से (x(y-1)=-(1+y)) मिलता है। चरण 3: इसलिए (x=\frac{1+y}{1-y}), अतः (f^{-1}(x)=\frac{1+x}{1-x})।
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