Question

The set of possible values of $\lambda$ for which $x^2-\left({\lambda }^2-5\lambda +5\right)x+\left(2{\lambda }^2-3\lambda -4\right)=0$ has roots , whose sum and product are both less than $1$ , is

• A. $\left(-1,\frac{5}{2}\right)$
• B. $\left(1,4\right)$
• C. $\left(1,\frac{5}{2}\right)$
• D. $\left[1,\frac{5}{2}\right]$

Answer 1

The correct option is C $\left(1,\frac{5}{2}\right)$

$\begin{array}{c}Given,\\ sumofroots={\lambda }^2-5\lambda +5<1\\ {\lambda }^2-5\lambda +4<0\\ (\lambda -1)(\lambda -4)<0\\ \therefore \lambda \in (1,4)\\ Productofroots=2{\lambda }^2-3\lambda -4<1\\ =2{\lambda }^2-3\lambda -5<0\\ =2{\lambda }^2-5\lambda +2\lambda -5<0\\ =(\lambda +1)(2\lambda -5)<0\\ \therefore \lambda \in (-1,\frac{5}{2})\\ So,thebothconditionswillbetrue:\\ then,\\ \lambda =(1,\frac{5}{2})\\ So,thatthecorrectoptionisC.\end{array}$