Question
If $$f:\left[ {1,\infty } \right) \to \left[ {2,\infty } \right)$$ is given by $$f\left( x \right) = x + \frac{1}{x}$$ then $${f^{ - 1}}\left( x \right)$$ equals
A.
$$\frac{{\left( {x + \sqrt {{x^2} - 4} } \right)}}{2}$$
B.
$$\frac{x}{{\left( {1 + {x^2}} \right)}}$$
C.
$$\frac{{\left( {x - \sqrt {{x^2} - 4} } \right)}}{2}$$
D.
$$1 + \sqrt {{x^2} - 4} $$
Answer :
$$\frac{{\left( {x + \sqrt {{x^2} - 4} } \right)}}{2}$$
Solution :
$$\eqalign{
& f\left( x \right) = x + \frac{1}{x} = y \Rightarrow {x^2} - yx + 1 = 0 \cr
& \Rightarrow x = \frac{{y \pm \sqrt {{y^2} - 4} }}{2} \cr
& \therefore x = \frac{{y + \sqrt {{y^2} - 4} }}{2}\,\,\left( {\because x \geqslant 1{\text{ and }}y \geqslant 2} \right) \cr
& \therefore {f^{ - 1}}\left( x \right) = \frac{{x + \sqrt {{x^2} - 4} }}{2} \cr} $$
NOTE THIS STEP: