If $f:[1,\; + \infty ) \to [2,\; + \infty )$ is given by $f(x) = x + \frac{1}{x}$ then ${f^{ - 1}}$ equals
If $f:[1,\; + \infty ) \to [2,\; + \infty )$ is given by $f(x) = x + \frac{1}{x}$ then ${f^{ - 1}}$ equals
- [IIT 2001]
- A
$\frac{{x + \sqrt {{x^2} - 4} }}{2}$
- B
$\frac{x}{{1 + {x^2}}}$
- C
$\frac{{x - \sqrt {{x^2} - 4} }}{2}$
- D
$1 + \sqrt {{x^2} - 4} $
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- [IIT 2005]
It is easy to see that $f$ is one-one and onto, so that $f$ is invertible with the inverse $f^{-1}$ of $f$ given by $f^{-1}=\{(1,2),(2,1),(3,1)\}=f$
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The inverse of the function $\frac{{{{10}^x} - {{10}^{ - x}}}}{{{{10}^x} + {{10}^{ - x}}}}$ is
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