The set of all real values of λ for which the quadratic equation (...

$f\left(0\right) \cdot f\left(1\right) \le 0$

$⇒ 2\left({\lambda}^{2} + 1 - 4\lambda + 2\right) \le 0 ⇒ 2\left({\lambda}^{2} - 4\lambda + 3\right) \le 0$

$⇒ \left(\lambda - 1\right)\left(\lambda - 3\right) \le 0$

$⇒ \lambda \in \left[1, 3\right]$

But at $\lambda = 1$, both roots are $1$.

So, $\lambda \ne 1$