Question

Complete set of values of $x$, satisfying the in equality $x^2+\frac{x^2}{(x+1{)}^2}<\frac{5}{4},$ is

• A. $(\frac{-1}{2},1)$
• B. $(\frac{-5}{2},\frac{1}{2})$
• C. $(1,\frac{5}{2})$
• D. $(\frac{-5}{2},\frac{-1}{2})$

Answer 1

The correct option is A $(\frac{-1}{2},1)$

$x^2+\frac{x^2}{(x+1{)}^2}<\frac{5}{4}$
$\Rightarrow {(x-\frac{x}{x+1})}^2+\frac{2x^2}{x+1}<\frac{5}{4}$
$\Rightarrow {\left(\frac{x^2}{x+1}\right)}^2+\left(\frac{2x^2}{x+1}\right)<\frac{5}{4}$
$\text{Let}\frac{x^2}{x+1}=t$
$\therefore t^2+2t-\frac{5}{4}<0$
$\Rightarrow 4t^2+8t-5<0$
$\Rightarrow 4t^2+10t-2t-5<0$
$\Rightarrow 2t(2t+5)-1(2t+5)<0$
$\Rightarrow (2t-1)(2t+5)<0$
$\Rightarrow \frac{-5}{2}<t<\frac{1}{2}$
So $\frac{x^2}{x+1}<\frac{1}{2}\text{and}\frac{x^2}{x+1}>\frac{-5}{2}$
$\Rightarrow \frac{2x^2-x-1}{(x+1)}<0,\frac{2x^2+5x+5}{2(x+1)}>0$
$\Rightarrow \frac{(2x+1)(x-1)}{x+1}<0,\frac{2x^2+5x+5}{2(x+1)}>0$
So $x\in (\frac{-1}{2},1)$