A) \[2-\sqrt{4+{{\log }_{5}}x}\]
B) \[2+\sqrt{4+{{\log }_{5}}x}\]
C) \[{{\left( \frac{1}{5} \right)}^{x(x-4)}}\]
D) None of these
Correct Answer: B
Solution :
[b] Let \[y={{5}^{x(x-4)}}\Rightarrow x(x-4)=lo{{g}_{5}}y\] \[\Rightarrow {{x}^{2}}-4x-{{\log }_{5}}y=0\] \[\Rightarrow x=\frac{4\pm \sqrt{16+4{{\log }_{5}}y}}{2}=(2\pm \sqrt{4+{{\log }_{5}}y})\] But \[x\ge 4,\] so \[x=(2+\sqrt{4+{{\log }_{5}}y})\] \[\therefore \,\,\,\,{{f}^{-1}}(x)=2+\sqrt{4+{{\log }_{5}}x}\]
You need to login to perform this action.
You will be redirected in
3 sec
