JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Quadratic Equations and Inequalities

Write $f(x)=({\lambda }^2+1)x^2-4\lambda x+2$. At the ends of the interval, $f(0)=2>0$ and $f(1)={\lambda }^2-4\lambda +3=(\lambda -1)(\lambda -3)$. For a single root strictly inside $(0,1)$ the function must change sign across it, $f(0)f(1)<0$, which needs $(\lambda -1)(\lambda -3)<0$, i.e. $1<\lambda <3$. At $\lambda =3$ one root lands exactly at $x=1$ while the other, $x=\frac{1}{5}$, lies inside, so $\lambda =3$ still leaves exactly one root in the open interval and is kept; at $\lambda =1$ there is a double root at $x=1$, outside the interval, so $\lambda =1$ is dropped. The set is $\lambda \in (1,3]$.