if `f:[1,oo)->[2,oo)` is given by `f(x)=x+1/x` then `f^-1(x)` equals to : a) `(x+sqrt(x^2-4))/2` b) `x/(1+x^2)` c) `(x-sqrt(x^2-4))/2` d) `1+sqrt(x^2-4)`A. `(x+sqrt(x^(2)-4))/(2)`B. `(x)/(1+x^(2))`C. `(x-sqrt(x^(2)-4))/(2)`D. `1+sqrt(x^(2)-4)`

if `f:[1,oo)->[2,oo)` is given by `f(x)=x+1/x` then `f^-1(x)` equals t

1.	if `f:[1,oo)->[2,oo)` is given by `f(x)=x+1/x` then `f^-1(x)` equals to : a) `(x+sqrt(x^2-4))/2` b) `x/(1+x^2)` c) `(x-sqrt(x^2-4))/2` d) `1+sqrt(x^2-4)`A. `(x+sqrt(x^(2)-4))/(2)`B. `(x)/(1+x^(2))`C. `(x-sqrt(x^(2)-4))/(2)`D. `1+sqrt(x^(2)-4)`
Answer» Correct Answer - A Let `y=x+(1)/(x) rArr y=(x^(2)+1)/(x)` `rArr xy=x^(2)+1` `rArr x^(2)-xy+1=0 rArr x=(y pm sqrt(y^(2)-4))/(2)` `rArr f^(-1)(y)=(y pm sqrt(y^(2)-4))/(2)` `rArr f^(-1)(x)=(x pm sqrt(x^(2)-4))/(2)` Since, the range of the inverse function is `[1, oo),` then we take `f^(-1)(x)=(x+sqrt(x^(2)-4))/(2)` If we consider `f^(-1)(x)=(x-sqrt(x^(2)-4))/(2), " then " f^(-1)(x) gt 1` This possible only if `(x-2)^(2) gt x^(2)-4` `rArr x^(2)+4-4x gt x^(2)-4` `rArr 8 gt 4x` `rArr x lt 2, " where " x gt 2` Therefore, (a) is the answer.