Question

If $(\lambda -2)x^2+(\lambda -1)x+3<0,\forall x\in R,$ then the range of $\lambda$ is

• A. $(\frac{14-\sqrt{96}}{2},2)$
• B. $(\frac{14-\sqrt{96}}{2},\frac{14+\sqrt{96}}{2})$
• C. $(-\infty ,2)$
• D. $\varphi$

Answer 1

The correct option is D $\varphi$

$(\lambda -2)x^2+(\lambda -1)x+3<0,\forall x\in R$
So, $a<0,D<0$
$\Rightarrow \lambda -2<0\Rightarrow \lambda <2\cdots \left(1\right)$

$D<0\Rightarrow (\lambda -1{)}^2-12(\lambda -2)<0\Rightarrow {\lambda }^2-2\lambda +1-12\lambda +24<0\Rightarrow {\lambda }^2-14\lambda +25<0$

Now, ${\lambda }^2-14\lambda +25=0$
$\Rightarrow \lambda =\frac{14\pm \sqrt{196-100}}{2}\Rightarrow \lambda =\frac{14\pm \sqrt{96}}{2}=7\pm 2\sqrt{6}$
So, ${\lambda }^2-14\lambda +25<0$
$\Rightarrow 7-2\sqrt{6}<\lambda <7+2\sqrt{6}\cdots \left(2\right)$

From equations $\left(1\right)$ and $\left(2\right)$, we get
$\lambda \in \varphi$