Question

The set of all real values of $\lambda$ for which the quadratic equations,$({\lambda }^2+1)x^2-4\lambda x+2=0$ 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

$(1,3)$

Explanation for the correct option:

Apply the condition for the given function $f\left(x\right)=$$({\lambda }^2+1)x^2-4\lambda x+2=0$ to have exactly one root in the interval $(0,1)$

$f\left(0\right)\times f\left(1\right)\le 0$

$\Rightarrow 2[{\lambda }^2-4\lambda +3]\le 0$

$(\lambda -1)(\lambda -3)\le 0$

$\Rightarrow \lambda \in [1,3]$

At $\lambda =1$,

$2x^2-4x+2=0$

$\Rightarrow {(x-1)}^2=0$

$x=1,1$

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

Hence, option C is the correct option.