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प्रश्न
If \[F : [1, \infty ) \to [2, \infty )\] is given by
\[f\left( x \right) = x + \frac{1}{x}, then f^{- 1} \left( x \right)\]
विकल्प
\[\frac{x + \sqrt{x^2 - 4}}{2}\]
\[\frac{x}{1 + x^2}\]
\[\frac{x - \sqrt{x^2 - 4}}{2}\]
\[1 + \sqrt{x^2 - 4}\]
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उत्तर
\[\text{Let } f^{- 1} \left( x \right) = y\]
\[ \Rightarrow f\left( y \right) = x\]
\[ \Rightarrow y + \frac{1}{y} = x\]
\[ \Rightarrow y^2 + 1 = xy\]
\[ \Rightarrow y^2 - xy + 1 = 0\]
\[ \Rightarrow y^2 - 2 \times y \times \frac{x}{2} + \left( \frac{x}{2} \right)^2 - \left( \frac{x}{2} \right)^2 + 1 = 0\]
\[ \Rightarrow y^2 - 2 \times y \times \frac{x}{2} + \left( \frac{x}{2} \right)^2 = \frac{x^2 - 1}{4}\]
\[ \Rightarrow \left( y - \frac{x}{2} \right)^2 = \frac{x^2 - 1}{4}\]
\[ \Rightarrow y - \frac{x}{2} = \frac{\sqrt{x^2 - 4}}{2}\]
\[ \Rightarrow y = \frac{x}{2} + \frac{\sqrt{x^2 - 4}}{2}\]
\[ \Rightarrow y = \frac{x + \sqrt{x^2 - 4}}{2}\]
\[ \Rightarrow f^{- 1} \left( x \right) = \frac{x + \sqrt{x^2 - 4}}{2}\]
So, the answer is (a) .
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