Question

If one root of $g\left(x\right)=0$ is less than $p$ and other root is greater than $q$, then $\lambda$ lies in the interval

• A. $(-\infty ,\frac{2}{3})$
• B. $(\frac{2}{3},3)$
• C. $(3,\infty )$
• D. $(-\frac{2}{3},\infty )$

Answer 1

The correct option is A $(-\infty ,\frac{2}{3})$

Let $y=\frac{x^2+x-1}{x^2+x+1}$
$\Rightarrow (y-1)x^2+(y-1)x+(y+1)=0$
For $x$ to be real, $D\ge 0$ for the above equation.
$\Rightarrow (y-1{)}^2-4(y-1\left)\right(y+1)\ge 0$
$\Rightarrow (y-1)(-3y-5)\ge 0$
$\Rightarrow (y-1)(3y+5)\le 0$
$\Rightarrow y\in [-\frac{5}{3},1)$ (since $y\ne 1$)
Hence integral values are 'y' are $-1$ and $0$.
Since $p<q$, we get $p=-1$ and $q=0$.

$g(-1)<0\Rightarrow 1+2\lambda +\lambda -3<0\Rightarrow \lambda <\frac{2}{3}$
$g\left(0\right)<0\Rightarrow \lambda <3\therefore \lambda \in (-\infty ,\frac{2}{3})$