Complex Numbers And Quadratic Equations question 685
Question: The solution set of $ \frac{x^{2}-3x+4}{x+1}>1, $ $ x\in R, $ is
Options:
A) $ (3,+\infty ) $
B) $ (-1,1)\cup (3,+\infty ) $
C) $ [-1,1]\cup (3,+\infty ) $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \frac{x^{2}-3x+4}{x+1}>1\Rightarrow \frac{x^{2}-3x+4}{x+1}-1>0 $
$ \Rightarrow \frac{x^{2}-4x+3}{x+1}>0\Rightarrow \frac{(x+1)(x-1)(x-3)}{{{( x+1 )}^{2}}}>0 $
$ \Rightarrow (x+1)(x-1)(x-3)>0andx\ne -1 $ Using method of interval, we get, $ x\in ( -1,1 )\cup ( 3,\infty )) $
BETA
