# Write the number of integral solutions of

Question:

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

Solution:

We have:

$\frac{x+2}{x^{2}+1}>\frac{1}{2}$

$\Rightarrow \frac{x+2}{x^{2}+1}-\frac{1}{2}>0$

$\Rightarrow \frac{2(x+2)-\left(x^{2}+1\right)}{2\left(x^{2}+1\right)}>0$

$\Rightarrow \frac{2 x+4-x^{2}-1}{2\left(x^{2}+1\right)}>0$

$\Rightarrow \frac{-x^{2}+2 x+3}{2\left(x^{2}+1\right)}>0$

To make the fraction of the left side positive, either the numerator or the denominator should be positive or both should be negative.

Since, it is clear that the denominator is positive, the numerator must be positive.

$-x^{2}+2 x+3>0$

$\Rightarrow x^{2}-2 x-3<0$

$\Rightarrow(x-3)(x+1)<0$

Now, to make the left side negative, one of these $[i . e .(x-3)$ or $(x+1)]$ should be positive and the other should be negative.

Also, $x+1>x-3$

$\therefore x+1>0$ and $x-3<0$

$\Rightarrow x>-1$ and $x<3$

$\Rightarrow x \in(-1,3)$

The integral solution of $x$ is $\{0,1,2\}$.

Hence, there are 3 integral solutions of the given inequation.

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