Question

Number of integral solutions of $\frac{[x+2]}{[x^2+1]}>\frac{1}{2}$ is / are

• A. $0$
• B. $1$
• C. $2$
• D. $3$
• E. $4$

Answer 1

The correct option is D

$3$

Explanation for the correct option:

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

Given: $\frac{[x+2]}{[x^2+1]}>\frac{1}{2}$

Solving $\frac{[x+2]}{[x^2+1]}>\frac{1}{2}$ to find the values of $x$

$$\begin{gathered} \frac{[x+2]}{[x^2+1]}>\frac{1}{2} \\ \Rightarrow 2x+4>x^2+1 \\ \Rightarrow x^2–2x–3<0 \\ \Rightarrow (x–3)(x+1)<0 \\ \Rightarrow –1<x<3 \end{gathered}$$

The integral solution is $3$

$\therefore$The value of $x$ is $0,1,2.$

Hence the correct answer is option (D)