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If F : [ 1 , ∞ ) → [ 2 , ∞ ) is Given by F ( X ) = X + 1 X , T H E N F − 1 ( X ) (A) X + √ X 2 − 4 2 (B) X 1 + X 2 (C) X − √ X 2 − 4 2 (D) 1 + √ X 2 − 4

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Question

If  \[F : [1, \infty ) \to [2, \infty )\] is given by

\[f\left( x \right) = x + \frac{1}{x}, then f^{- 1} \left( x \right)\]

 

Options

  • \[\frac{x + \sqrt{x^2 - 4}}{2}\]

  • \[\frac{x}{1 + x^2}\]

  • \[\frac{x - \sqrt{x^2 - 4}}{2}\]

  • \[1 + \sqrt{x^2 - 4}\]

MCQ
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Solution

\[\text{Let } f^{- 1} \left( x \right) = y\] 
\[ \Rightarrow f\left( y \right) = x\] 
\[ \Rightarrow y + \frac{1}{y} = x\] 
\[ \Rightarrow y^2 + 1 = xy\] 
\[ \Rightarrow y^2 - xy + 1 = 0\] 
\[ \Rightarrow y^2 - 2 \times y \times \frac{x}{2} + \left( \frac{x}{2} \right)^2 - \left( \frac{x}{2} \right)^2 + 1 = 0\] 
\[ \Rightarrow y^2 - 2 \times y \times \frac{x}{2} + \left( \frac{x}{2} \right)^2 = \frac{x^2 - 1}{4}\] 
\[ \Rightarrow \left( y - \frac{x}{2} \right)^2 = \frac{x^2 - 1}{4}\] 
\[ \Rightarrow y - \frac{x}{2} = \frac{\sqrt{x^2 - 4}}{2}\] 
\[ \Rightarrow y = \frac{x}{2} + \frac{\sqrt{x^2 - 4}}{2}\] 
\[ \Rightarrow y = \frac{x + \sqrt{x^2 - 4}}{2}\] 
\[ \Rightarrow f^{- 1} \left( x \right) = \frac{x + \sqrt{x^2 - 4}}{2}\]

So, the answer is (a) .

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Chapter 2: Functions - Exercise 2.6 [Page 78]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 2 Functions
Exercise 2.6 | Q 38 | Page 78

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