If $\alpha ,\beta$  be the roots of the equation $4x^2-16x+\lambda =0,\lambda \in R$  such that $1<\alpha <2$  and $2<\beta <3$ , then the number of integral solutions of $\text{'}\lambda \text{'}$  is

$4{\alpha }^2-16\alpha +\lambda =0$          $1<\alpha <2$

$\alpha =1$                       $\alpha =2$

$4-16+\lambda =0$          $16-32+\lambda =0$

$\lambda =12$                     $\lambda =16$

$1<\alpha <2$

$12<\lambda <16$

$4{\beta }^2-16\beta +\lambda =0$     $\beta =3$

$\beta =2$

$16-32+\lambda =0$         $36-48+\lambda =0$

$\lambda =16$                      $\lambda =12$

                                $12<\lambda <16$

$\lambda =13,14,15\to 3$  solutions.