Question

Number of integral solutions of $\frac{x + 2}{x^{2} + 1}$>$\frac{1}{2}$is –

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (D)

3

Answer 2

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

⇒ $\frac{- x^{2} - 1 + 2x + 4}{2(x^{2} + 1)}$> 0 ⇒ $\frac{3 + 2x - x^{2}}{2(x^{2} + 1)}$> 0

Since denominator is +ve

∴ 3 + 2x – x² > 0 ⇒ – 1 < x < 3 ⇒ x = 0, 1, 2