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:

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

Answer 1

Answer: (B)

If exactly one root in $(0,1)$ then

$\Rightarrow f(0)\cdot f(1)<0$
$\Rightarrow 2\left({\lambda }^2-4\lambda +3\right)<0$
$\Rightarrow 1<\lambda <3$
Now for $\lambda =1,2x^2-4x+2=0$
$(x-1{)}^2=0,x=1,1$
So both roots doesn't lie between $(0,1)$
$\therefore \lambda \ne 1$
Again for $\lambda =3$
$10x^2-12x+2=0$
$\Rightarrow x=1,\frac{1}{5}$
so if one root is 1 then second root lie between $(0,1)$
so $\lambda =3$ is correct
$\therefore \lambda \in (1,3]$