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The Inverse of the Function F : R → { X ∈ R : X < 1 } Given by F ( X ) = E X − E − X E X + E − X is (A) 1 2 Log 1 + X 1 − X (B) 1 2 Log 2 + X 2 − X (C) 1 2 Log 1 − X 1 + X (D) None of These - Mathematics

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Question

The inverse of the function

\[f : R \to \left\{ x \in R : x < 1 \right\}\] given by

\[f\left( x \right) = \frac{e^x - e^{- x}}{e^x + e^{- x}}\] is 

 

Options

  • \[\frac{1}{2} \log \frac{1 + x}{1 - x}\]

  •  \[\frac{1}{2} \log \frac{2 + x}{2 - x}\]

  • \[\frac{1}{2} \log \frac{1 - x}{1 + x}\]

  • none of these

MCQ
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Solution

\[\text{Let} f^{- 1} \left( x \right) = y . . . \left( 1 \right)\] 
\[ \Rightarrow f\left( y \right) = x\] 
\[ \Rightarrow \frac{e^y - e^{- y}}{e^y + e^{- y}} = x\] 
\[ \Rightarrow \frac{e^{- y} \left( e^{2y} - 1 \right)}{e^{- y} \left( e^{2y} + 1 \right)} = x\] 
\[ \Rightarrow \left( e^{2y} - 1 \right) = x\left( e^{2y} + 1 \right)\] 
\[ \Rightarrow e^{2y} - 1 = x e^{2y} + x\] 
\[ \Rightarrow e^{2y} \left( 1 - x \right) = x + 1\] 
\[ \Rightarrow e^{2y} = \frac{1 + x}{1 - x}\] 
\[ \Rightarrow 2y = \log_e \left( \frac{1 + x}{1 - x} \right)\] 
\[ \Rightarrow y = \frac{1}{2}l {og}_e \left( \frac{1 + x}{1 - x} \right)\] 
\[ \Rightarrow f^{- 1} \left( x \right) = \frac{1}{2}l {og}_e \left( \frac{1 + x}{1 - x} \right) [\text{from}\left( 1 \right)]\] 

So, the answer is (a).

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Chapter 2: Functions - Exercise 2.6 [Page 78]

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RD Sharma Mathematics [English] Class 12
Chapter 2 Functions
Exercise 2.6 | Q 33 | Page 78

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