# 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 : Option: 1 Option: 2 Option: 3 Option: 4

If exactly one root in (0, 1) then

$\begin{array}{l} \Rightarrow \mathrm{f}(0) \cdot \mathrm{f}(1)<0 \\ \Rightarrow 2\left(\lambda^{2}-4 \lambda+3\right)<0 \\ \Rightarrow 1<\lambda<3 \end{array}$

Now for $\lambda=1$

$\begin{array}{l} 2 x^{2}-4 x+2=0 \\ (x-1)^{2}=0\\ x=1,1 \end{array}$

So both roots doesn’t lie between (0, 1)

$\therefore \lambda\neq1$

\begin{aligned} &\text { Again for } \lambda=3\\ &\begin{aligned} & 10 x^{2}-12 x+2=0 \\ \Rightarrow & x=1, \frac{1}{5} \end{aligned} \end{aligned}

So if one root is 1 then second root lie between (0, 1) so $\lambda=3$

$\therefore \lambda \in(1,3]$

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