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 $x^2-(3k+1)x+3k^2-k+2=0$.

Answer 1

Answer: 0

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