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Q.
The set of all real values of $\lambda$ for which the
quadratic equations, $\left(\lambda^{2}+1\right) x^{2}-4 \lambda x+2=0$ always have exactly one root in the interval (0,1) is:
JEE MainJEE Main 2020Complex Numbers and Quadratic Equations
Solution:
If exactly one root in $(0, 1)$ then
$\Rightarrow f (0) \cdot f (1)<\,0$
$ \Rightarrow 2\left(\lambda^{2}-4 \lambda+3\right)<\,0$
$ \Rightarrow 1< \,\lambda < \,3$
Now for $ \lambda=1, 2 x ^{2}-4 x +2=0$
$( x -1)^{2}=0, x =1, 1$
So both roots doesn't lie between $(0, 1)$
$\therefore \lambda \neq 1$
Again for $\lambda=3$
$10 x^{2}-12 x+2=0$
$\Rightarrow x=1, \frac{1}{5}$
so if one root is 1 then second root lie between $(0, 1)$
so $\lambda=3$ is correct
$\therefore \lambda \in (1,3]$
