$\text{ The set of all real values of }\lambda \text{ for which the quadratic equations. }$$({\lambda }^2+1)x^2-4\lambda x+2=0$always have exactly one root in the interval (0,1) is :

$\text{ Let }f\left(x\right)=\left({\lambda }^2+1\right)x^2-4\lambda x+2$

If f(x) = 0 has exactly one root in (0,1) then 

$\begin{array}{c}f\left(0\right)\cdot f\left(1\right)\le 0\\ 2({\lambda }^2+1-4\lambda +2)\le 0\\ (\lambda -1)(\lambda -3)\le 0\end{array}$

$\text{ But for }\lambda =1\Rightarrow n=1\text{ is a repeat root, }$

$\text{ Therefore, }\lambda \in (1,3]$