Set of all real values of $\lambda$ for which quadratic equation $(\lambda ^2+1)x^2-4\lambda x+2=0$ always has exactly one root in $(0,1)$

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Set of all real values of $\lambda$ for which quadratic equation $(\lambda ^2+1)x^2-4\lambda x+2=0$ always has exactly one root in interval $(0,1)$

My Approach: Let $f(x)=(\lambda ^2+1)x^2-4\lambda x+2$

Since exactly one root lies between $(0,1)$ so we must have $f(0)f(1)<0$

Using above condition I got an answer $\lambda\in (1,3)$.

Case $1$ Checking at $\lambda =1$. I got two equal roots $1,1$.

Case $2$ Checking at $\lambda = 3$. I got two roots $\dfrac{1}{5}, 1$

As in Case $2$ one root lies in $(0,1)$

My Doubt: Why we always check at boundary point.

Is there any rule for this?

I mean in which question I should check and in which question i shouldn't.