Write the number of integral solutions of (x+2)/(x^2+1)>1/2

Question

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

(x+2)/(x²+1)>1/2

Answer 1

(x+2)/(x²+1)>1/2

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

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

⇒ \(\frac{-x^2+2x+3}{2(x^2+1)}\) > 0

Here,

Denominator i.e., x² + 1 is always positive and not equal to zero. 

So, neglect it. 

⇒ - x² + 2x + 3 > 0 

⇒ x² – 2x – 3 < 0 

⇒ (x – 3)(x + 1) < 0 

Case I : (x – 3) < 0 and (x + 1) > 0 

⇒ x < 3 and x > -1 

By takin intersection x ∈ (-1, 3) 

Case II : (x – 3) > 0 and (x + 1) < 0 

⇒ x > 3 and x < -1

By taking intersection x ∈ ∅. 

So, case II is irrelevant. 

So, the complete solution is x ∈ (-1, 3) 

The integral solution is 0, 1 and 2. 

So, number of integral solution is 3.