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)

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