Functions Question 24
Question: If the function $ f:
[1,\ \infty )\to [1,\ \infty ) $ is defined by $ f(x)={2^{x(x-1)}}, $ then $ {f^{-1}} $ (x) is [IIT 1999]
Options:
A) $ {{( \frac{1}{2} )}^{x(x-1)}} $
B) $ \frac{1}{2}(1+\sqrt{1+4{\log_2}x}) $
C) $ \frac{1}{2}(1-\sqrt{1+4{\log_2}x}) $
D) Not defined
Show Answer
Answer:
Correct Answer: B
Solution:
Given $ f(x)={2^{x(x-1)}},\Rightarrow x,(x-1)={\log_2}f(x) $
$ \Rightarrow x^{2}-x-{\log_2}f(x)=0\Rightarrow x=\frac{1\pm \sqrt{1+4{\log_2}f(x)}}{2} $ Only $ x=\frac{1+\sqrt{1+4{\log_2}f(x)}}{2} $ lies in the domain
$ \therefore {f^{-1}}(x)=\frac{1}{2},[1+\sqrt{1+4,{\log_2}x}] $ .
BETA
