Given quadratic equation is: $x^2-m x+4=0$ Both the roots are real and distinct. $$ \begin{aligned} & \therefore m^2-4 \cdot 1 \cdot 4>0 \\ & \therefore(m-4)(m+4)>0 \\ & \therefore m \in(-\infty,-4) \cup(4, \infty) \end{aligned} $$ ∵ both roots lies in [1, 5] $$ \begin{aligned} & \therefore-\frac{-m}{2} \in(1,5) \\ & \Rightarrow m \in(2,10) \end{aligned} $$ And 1 $(1-m+4)>0 \Rightarrow m<5 m \in(-x, 5)$ And $1 \cdot(25-5 m+4)>0=$ $$ m<\frac{29}{5} $$ $$ \therefore \quad m \in\left(-x, \frac{29}{5}\right) $$