Question:
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 :
Correct Option: , 2
Solution:
If exactly one root in $(0,1)$ then
$\begin{array}{ll}\Rightarrow & \mathrm{f}(0) \cdot \mathrm{f}(1)<0 \\ \Rightarrow & 2\left(\lambda^{2}-4 \lambda+3\right)<0 \\ \Rightarrow & 1<\lambda<3\end{array}$
Now for
$\lambda=1,2 \mathrm{x}^{2}-4 \mathrm{x}+2=0$
$(x-1)^{2}=0, x=1,1$
So both roots doesn't lie between $(0,1)$
$\therefore \lambda \neq 1$
Again for $\lambda=3$
$10 x^{2}-12 x+2=0$
$\Rightarrow \quad x=1, \frac{1}{5}$
so if one root is 1 then second root lie between $(0,1)$ so $\lambda=3$ is correct.
$\therefore \quad \lambda \in(1,3] .$