If $\mathrm{\alpha }, \mathrm{\beta }$ be the roots of the equation 4x² - 16x + $\mathrm{\lambda }$ = 0, $\mathrm{\lambda }$ $\in$ R such that $1<\mathrm{\alpha }<2 \mathrm{and} 2<\mathrm{\beta }<3\mathrm{.}$, then the number of integral solutions of $\mathrm{\lambda }$ is ____

Clearly , roots are real and distinct.
Therefore, $\mathrm{D}>0\Rightarrow 256-16\mathrm{\lambda }>0\Rightarrow \mathrm{\lambda }<16$
Now, one root αlies between 1 and 2.

$\begin{array}{l}\therefore \mathrm{f}\left(1\right)\mathrm{f}\left(2\right)<0\Rightarrow (4-16+\mathrm{\lambda })(16-32+\mathrm{\lambda })<0\\ \Rightarrow (\mathrm{\lambda }-12)(\mathrm{\lambda }-16)<0\Rightarrow 12<\mathrm{\lambda }<16\end{array}$

or 

$$\begin{gathered} f\left(x\right)=4x^2-16x+\lambda . \\ \text{ Clearly, }1\text{ and }3\text{ lie outside the roots and }2\text{ lies between the roots. } \\ \therefore f\left(1\right)>0,f\left(2\right)<0\text{ and }f\left(3\right)>0 \\ \Rightarrow \lambda -12>0\text{ and λ}-16<0\Rightarrow 12<\lambda <16\Rightarrow \\ Integer values of \lambda =13,14,15 \end{gathered}$$