Functions Question 382
Question: The domain of the function $ f(x)=log_2( -{\log_{1/2}}( 1+\frac{1}{{x^{1/4}}} )-1 ) $ is
Options:
A) $ ( 0,1 ) $
B) $ ( 0,1 ] $
C) $ [1,\infty ) $
D) $ (1,\infty ) $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ f(x) $ is defined if $ -{\log_{1/2}}( 1+\frac{1}{{x^{1/4}}} )-1>0 $
$ \Rightarrow {\log_{1/2}}( 1+\frac{1}{{x^{1/4}}} )<-1 $
$ \Rightarrow 1+\frac{1}{{x^{1/4}}}>{{( \frac{1}{2} )}^{-1}} $
$ \Rightarrow \frac{1}{{x^{1/4}}}>1\Rightarrow 0<x<1 $
BETA
