The set of all real values of

The given quadratic equation is

$\left(\lambda^{2}+1\right) x^{2}-4 \lambda x+2=0$

$\because$ One root is in the interval $(0,1)$

$\therefore f(0) f(1) \leq 0$

$\Rightarrow 2\left(\lambda^{2}+1-4 \lambda+2\right) \leq 0$

$\Rightarrow 2\left(\lambda^{2}-4 \lambda+3\right) \leq 0$

$(\lambda-1)(\lambda-3) \leq 0 \Rightarrow \lambda \in[1,3]$

But at $\lambda=1$, both roots are 1 so $\lambda \neq 1$

$\therefore \lambda \in(1,3]$