Functions Question 424
Question: If $ {\log_{1/2}}( x^{2}-5x+7 )>0, $ then exhaustive range of values of x is
Options:
A) $ (-\infty ,2)\cup (3,\infty ) $
B) $ (2,3) $
C) $ (-\infty ,1)\cup (1,2)\cup (2,\infty ) $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ f(x)=lo{g_{1/2}}(x^{2}-5x+7)>0\Rightarrow x^{2}-5x+7>0 $ $ x^{2}-5x+7<1,x\in R\Rightarrow x^{2}-5x+6<0\Rightarrow x\in (2,3) $
BETA
