Question

Write the number of integral solutions of \[\frac{x + 2}{x^2 + 1} > \frac{1}{2}\]

Answer 1

\[\text{ We have }, \]
\[ \frac{x + 2}{x^2 + 1} > \frac{1}{2}\]
\[ \Rightarrow \frac{x + 2}{x^2 + 1} - \frac{1}{2} > 0\]
\[ \Rightarrow \frac{2\left( x + 2 \right) - \left( x^2 + 1 \right)}{2\left( x^2 + 1 \right)} > 0\]
\[ \Rightarrow \frac{2x + 4 - x^2 - 1}{2\left( x^2 + 1 \right)} > 0\]
\[ \Rightarrow \frac{- x^2 + 2x + 3}{2\left( x^2 + 1 \right)} > 0\]
\[\text{ To make the fraction of the left side positive, either the numerator or the }\]
\[\text{ denominator should be positive or both should be negative } . \]
\[\text{ Since, it is clear that the denominator is positive, the numerator must be positive }. \]
\[ - x^2 + 2x + 3 > 0\]
\[ \Rightarrow x^2 - 2x - 3 < 0\]
\[ \Rightarrow \left( x - 3 \right)\left( x + 1 \right) < 0\]
\[\text{ Now, to make the left side negative, one of these } \left[ i . e . (x - 3) or (x + 1) \right] s\text{ hould be positive and the other should be negative } . \]
\[\text{ Also }, x + 1 > x - 3\]
\[ \therefore x + 1 > 0 \text{ and } x - 3 < 0\]
\[ \Rightarrow x > - 1 \text{ and } x < 3\]
\[ \Rightarrow x \in \left( - 1, 3 \right)\]
\[\text{ The integral solution of x is } \left\{ 0, 1, 2 \right\} . \]
\[Hence, there are 3 integral solutions of the given inequation .\]