Question

Out of the two roots of $x^2+(1-2\lambda )x+({\lambda }^2-\lambda -2)=0$ one root is greater than 3 and the other root is less then 3, then the limits of $\lambda$ are

• A. λ<2
• B. 2<λ<5
• C. λ>5
• D. λ=5/2

Answer 1

The correct option is B

2<λ<5

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

a = 1 of $\alpha ,\beta$ are roots of (1),

if $\alpha <3<\beta \Rightarrow a.f\left(3\right)<0$

$\Rightarrow f\left(3\right)<0$

$\Rightarrow 9+(1-2\lambda )3+{\lambda }^2-\lambda -2<0$

$\Rightarrow \lambda \in (2,5)$