Question

If $-3<\frac{x^2-\lambda x-2}{x^2+x+1}<2$ for all $x\in R,$ then the value of $\lambda$ belongs to

• A. $\left(-1,7\right)$
• B. $\left(-6,2\right)$
• C. $\left(-1,2\right)$
• D. $\left(-6,7\right)$

Answer 1

Answer: (C)

$-3<\frac{x^2-\lambda x-2}{x^2+x+1}<2$
$\Rightarrow -3x^2-3x-3<x^2-\lambda x-2<2x^2+2x+2$ $\left(since{x}^2+x+1>0,\forall x\in R\right)$
$\Rightarrow 4x^2+x\left(3-\lambda \right)+1>0,x^2+x\left(2+\lambda \right)+4>0$
$\left(i\right)4x^2-x\left(\lambda -3\right)+1>0$
$\Rightarrow D<0\Rightarrow {\left(\lambda -3\right)}^2-4\times 4\times 1<0$
$\Rightarrow \left(\lambda -3+4\right)\left(\lambda -3-4\right)<0$
$\Rightarrow \left(\lambda +1\right)\left(\lambda -7\right)<0\Rightarrow \lambda \in \left(-1,7\right)$
$\left(ii\right)x^2+x\left(\lambda +2\right)+4>0$
$\Rightarrow D<0\Rightarrow {\left(\lambda +2\right)}^2-4\times 4<0$
$$\begin{gathered} \left(\lambda +2-4\right)\left(\lambda +2+4\right)<0 \\ \left(\lambda -2\right)\left(\lambda +6\right)<0\Rightarrow \lambda \in \left(-6,2\right) \end{gathered}$$
Taking the intersection of the solutions of $(i)$ and $(ii)$ , we get,
$\lambda \in \left(-1,2\right)$