Functions Question 425
Question: The domain of the function $ f(x)=lo{g_{3+x}}(x^{2}-1) $ is
Options:
A) $ (-3,-1)\cup (1,\infty ) $
B) $ [-3,-1)\cup [1,\infty ) $
C) $ (-3,-2)\cup (-2,-1)\cup (1,\infty ) $
D) $ [-3,-2)\cup (-2,-1)\cup [1,\infty ) $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] f(x) is to be defined when $ x^{2}-1>0 $ and $ 3+x>0 $ and $ 3+x\ne 1. $ i.e., $ x^{2}>1 $ and $ x>-3 $ and $ x\ne -2 $ , i.e., $ x<-1orx>1 $ and $ x>-3andx\ne -2 $
$ \therefore D_{f}=(-3,-2)\cup (-2,-1)\cup (1,\infty ) $
BETA
