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Q. The number of integral solutions of $\frac{x + 2}{x^2 + 1} > \frac{1}{2}$ is

Linear Inequalities

Solution:

$\frac{x + 2}{x^2 + 1} > \frac{1}{2} \:\:\: \Rightarrow \frac{-x^2 + 2x + 3}{ x^2 + 1} > 0$
$\Rightarrow \:\: - x^2 +2x + 3 > 0 [ \because \, x^2 + 1 > 0, \forall \, x \in R] $
$\Rightarrow \:\: x^2 - 2x - 3 < 0 \Rightarrow - 1 < x < 3$
$\because$ x is integer $\therefore $ x = 0,1, 2, hence three values.