Question

The complete set of values of $x$ satisfying $x^2+\frac{x^2}{(x+1{)}^2}<\frac{5}{4}$ is

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

Answer 1

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

$\text{Given}:x^2+\frac{x^2}{(x+1{)}^2}<\frac{5}{4}$
$\Rightarrow {(x-\frac{x}{x+1})}^2+\frac{2x^2}{x+1}<\frac{5}{4}$
$\Rightarrow {\left(\frac{x^2}{x+1}\right)}^2+\left(\frac{2x^2}{x+1}\right)<\frac{5}{4}$
$\text{Let}\frac{x^2}{x+1}=t$
$\therefore t^2+2t-\frac{5}{4}<0$
$\Rightarrow 4t^2+8t-5<0$
$\Rightarrow 4t^2+10t-2t-5<0$
$\Rightarrow 2t(2t+5)-1(2t+5)<0$
$\Rightarrow (2t-1)(2t+5)<0$
$\Rightarrow \frac{-5}{2}<t<\frac{1}{2}$
Case $I:t<\frac{1}{2}$
$\Rightarrow \frac{x^2}{x+1}<\frac{1}{2}$

$\Rightarrow \frac{2x^2-x-1}{(x+1)}<0$

$\Rightarrow \frac{(2x+1)(x-1)}{x+1}<0$


$\Rightarrow x\in (-\infty ,-1)\cup (-\frac{1}{2},1)\cdots \left(i\right)$

Case $II.t>\frac{5}{2}$
$\frac{x^2}{x+1}>\frac{-5}{2}$

$\Rightarrow \frac{2x^2+5x+5}{2(x+1)}>0$
$\Rightarrow \frac{2x^2+5x+5}{(x+1)}>0\cdots \left(A\right)$

Now, $2x^2+5x+5>0\forall x\in R$
$\Rightarrow \frac{2x^2+5x+5}{(x+1)}>0$
Thus, $\left(A\right)$ becomes:
$\Rightarrow \frac{1}{(x+1)}>0$
$\Rightarrow x\in (-1,\infty )\cdots \left(ii\right)$
Thus, from both the solution set $\left(i\right)\&\left(ii\right),$ we get:
$x\in (-\frac{1}{2},1)$