The set of all real values of

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 :

  1. (1) $(0,2)$

  2. (2) $(2,4]$

  3. (3) $(1,3]$

  4. (4) $(-3,-1)$


Correct Option: , 3

Solution:

The given quadratic equation is

$\left(\lambda^{2}+1\right) x^{2}-4 \lambda x+2=0$

$\because$ One root is in the interval $(0,1)$

$\therefore f(0) f(1) \leq 0$

$\Rightarrow 2\left(\lambda^{2}+1-4 \lambda+2\right) \leq 0$

$\Rightarrow 2\left(\lambda^{2}-4 \lambda+3\right) \leq 0$

$(\lambda-1)(\lambda-3) \leq 0 \Rightarrow \lambda \in[1,3]$

But at $\lambda=1$, both roots are 1 so $\lambda \neq 1$

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

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