Question

The set of all real values of $\lambda$ for which the quadratic equation, $({\lambda }^2+1)x^2-4\lambda x+2=0$ always have exactly one root in the interval $(0,1)$ is:

• A. $(-3,-1)$
• B. $(2,4]$
• C. $(1,3]$
• D. $(0,2)$

Answer 1

The correct option is C $(1,3]$

$f\left(0\right)f\left(1\right)\le 0$
$\Rightarrow \left(2\right)({\lambda }^2-4\lambda +3)\le 0$
$\Rightarrow (\lambda -1)(\lambda -3)\le 0$
$\Rightarrow \lambda \in [1,3]$
But at $\lambda =1$, both roots are $1$ so $\lambda \ne 1$
$\therefore \lambda \in (1,3]$