Question

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

Answer 1

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