Question

Find the number of integral values of $k$ for which $e ^{\lambda^2-2 \lambda+1+\ln 3}$ and $e ^{-\left(\lambda^2-2 \lambda+1\right)+\ln 2}$, where $\lambda \in R -\{1\}$ are the roots of the equation $x ^2-(3 k +1) x +3 k ^2- k +2=0$.

Answer 1

$\text { Product of roots }=3 k ^2- k +2= e ^{\ln 3+\ln 2}=6 $
$\Rightarrow 3 k ^2- k -4=0 $
$\Rightarrow 3 k ^2-4 k +3 k -4=0 $
$( k +1)(3 k -4)=0 $
$k =-1, k =\frac{4}{3}$
$\text { sum of roots }=3 k +1=3 \cdot e ^{\lambda^2-2 \lambda+1}+2 \cdot e ^{-\left(\lambda^2-2 \lambda+1\right)}$
$\because k =-1$ not satisfy given relation
So number of integral value of $k$ is zero.