Question
The number of integral solutions of $$\frac{{x + 2}}{{{x^2} + 1}} > \frac{1}{2}$$ is
A.
4
B.
5
C.
3
D.
None of these
Answer :
3
Solution :
$$\eqalign{
& \frac{{x + 2}}{{{x^2} + 1}} > \frac{1}{2} \cr
& \Rightarrow \,\,\frac{{ - {x^2} + 2x + 3}}{{{x^2} + 1}} > 0 \cr
& \Rightarrow \,\, - {x^2} + 2x + 3 > 0\left\{ {\because \,\,{x^2} + 1\,\,{\text{is always positive}}} \right\}. \cr} $$
By sign scheme, $$ - 1 < x < 3.\,{\text{As }}x\,{\text{is integer,}}\,x = 0,1,2.$$