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Question: The number of integral solution of \(\frac{x + 1}{x^{2} + 2} > \frac{1}{4}\) is...

The number of integral solution of x+1x2+2>14\frac{x + 1}{x^{2} + 2} > \frac{1}{4} is

A

1

B

2

C

5

D

None of these

Answer

5

Explanation

Solution

x+1x2+2−14>0\frac{x + 1}{x^{2} + 2} - \frac{1}{4} > 0 ⇒ x2−4x−2x2+2<0\frac{x^{2} - 4x - 2}{x^{2} + 2} < 0

⇒ (x−(2+6))(x−(2−6))<0(x - (2 + \sqrt{6}))(x - (2 - \sqrt{6})) < 0

⇒ 2−6<x<2+62 - \sqrt{6} < x < 2 + \sqrt{6}

Approximately, −0.4<x<4.4- 0.4 < x < 4.4

Hence, integral values of x are 0, 1, 2, 3, 4

Hence, number of integral solution = 5