If the function `f:[1,oo)->[1,oo)` is defined by `f(x)=2^(x(x-1)),` then `f^-1(x)` is (A) `(1/2)^(x(x-1))` (B) `1/2 sqrt(1+4log_2x)` (C) `1/2(1-sqrt(1+4log_2x))` (D) not definedA. `((1)/(2))^(x(x-1))`B. `(1)/(2)(1+sqrt(1+4log_(2)x))`C. `(1)/(2)(1-sqrt(1+4log_(2)x))`D. not defined

If the function `f:[1,oo)->[1,oo)` is defined by `f(x)=2^(x(x-1)),` th

1.	If the function `f:[1,oo)->[1,oo)` is defined by `f(x)=2^(x(x-1)),` then `f^-1(x)` is (A) `(1/2)^(x(x-1))` (B) `1/2 sqrt(1+4log_2x)` (C) `1/2(1-sqrt(1+4log_2x))` (D) not definedA. `((1)/(2))^(x(x-1))`B. `(1)/(2)(1+sqrt(1+4log_(2)x))`C. `(1)/(2)(1-sqrt(1+4log_(2)x))`D. not defined
Answer» Correct Answer - B Let `y=2^(x(x-1)), " where " y ge 1 " as " x ge 1` Taking `log_(2)` on both sides, we get `log_(2)y=log_(2)2^(x(x-1))` `rArr log_(2)y=x(x-1)` `rArr x^(2) -x-log_(2)y =0` `rArr x=(1pm sqrt(1+4log_(2)y))/(2)` For ` y ge 1, log_(2)y ge 0 rArr 4 log_(2) y ge 0 rArr 1+4 log_(2)y ge 1` `rArr sqrt(1+4log_(2)y) ge 1` `rArr -sqrt(1+4log_(2)y) le -1` `rArr 1-sqrt(1+4log_(2)y) le 0` But `x ge 1` so, `x=1- sqrt(1+4log_(2)y)` is not possible. Therefore, we take `x=(1)/(2)(1+sqrt(1+4log_(2)y))` `rArr f^(-1)(y) =(1)/(2)(sqrt(1+4log_(2)y))` `rArr f^(-1)(x)=(1)/(2)(sqrt(1+4log_(2)x))`