The set of all real values of for which (^2+1)x^2-4 x+2=0 always has exactly one root in t…

Question

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

Accepted Answer

$f(0)f(1)\le0$: $2(\lambda^2+1-4\lambda+2)\le0 \implies (\lambda-1)(\lambda-3)\le0 \implies \lambda\in[1,3]$.

At $\lambda=1$ both roots equal 1 (not interior), so excluded: $\lambda\in(1,3].$

Solutions · 1