Functions Question 465
Question: Let $ f:[4,\infty )\to [1,\infty ) $ be a function defined by $ f(x)={5^{x(x-4)}}, $ then $ {f^{-1}}(x) $ is
Options:
A) $ 2-\sqrt{4+{\log_5}x} $
B) $ 2+\sqrt{4+{\log_5}x} $
C) $ {{( \frac{1}{5} )}^{x(x-4)}} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Let $ y={5^{x(x-4)}}\Rightarrow x(x-4)=log_5y $
$ \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} $
BETA
