Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# Write the number of integral solutions of x + 2 x 2 + 1 > 1 2 - Mathematics

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

#### Solution

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

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 15 Linear Inequations
Q 7 | Page 31