Question
The solution set of $$\frac{{{x^2} - 3x + 4}}{{x + 1}} > 1,x \in R,$$ is
A.
$$\left( {3, + \infty } \right)$$
B.
$$\left( { - 1,1} \right) \cup \left( {3, + \infty } \right)$$
C.
$$\left[ { - 1,1} \right] \cup \left[ {3, + \infty } \right)$$
D.
None of these
Answer :
$$\left( { - 1,1} \right) \cup \left( {3, + \infty } \right)$$
Solution :
$$\eqalign{
& \frac{{{x^2} - 3x + 4}}{{x + 1}} > 1 \cr
& \Rightarrow \frac{{{x^2} - 3x + 4}}{{x + 1}} - 1 > 0 \cr
& \Rightarrow \frac{{{x^2} - 4x + 3}}{{x + 1}} > 0 \cr
& \Rightarrow \frac{{\left( {x + 1} \right)\left( {x - 1} \right)\left( {x - 3} \right)}}{{{{\left( {x + 1} \right)}^2}}} > 0 \cr
& \Rightarrow \left( {x + 1} \right)\left( {x - 1} \right)\left( {x - 3} \right) > 0\,\,{\text{and }}x \ne - 1 \cr} $$
Using method of interval, we get,
$$x \in \left( { - 1,1} \right) \cup \left( {3,\infty } \right)$$