Question

The solution set of $ \frac{x^{2} - 3x + 4}{ x + 1} > 1 , x \, \in \, R$, is

• A. $ (3 , + \infty )$
• B. $ ( -1, 1) \cup (3, + \infty ) $
• C. $ [ -1, 1 ] \cup [3, + \infty ) $
• D. none of theses

Answer 1

Answer: (B)

$\frac{x^2-3 x+4}{x+1}>1 \Rightarrow \frac{x^2-3 x+4}{x+1}-1>0 $
$\Rightarrow \frac{x^2-4 x+3}{x+1}>0$
$\Rightarrow \frac{(x+1)(x-1)(x-3)}{(x+1)^2}>0 $
$\Rightarrow(x+1)(x-1)(x-3)>0 \text { and } x \neq-1$
Using method of interval, we get,
$x \in(-1,1) \cup(3, \infty)$