If  both the roots of the quadratic equation  $x^2-mx+4=0$  are real and distinct and they lie in the interval  $[1,5]$ , then m lies in the interval : 

Given quadratic equation is  $x^2-mx+4=0$
Both the roots are real and distinct.
So, discriminant  $b^2-4ac>0$ .
 $\therefore m^2-\mathrm{4.1.4}>0$
$\therefore (m-4)(m+4)>0;\therefore m\in (-\infty ,-4)\cup (4,\infty )$                -----------------(i)
Since, both roots lies in  $[1,5]$
 $1<\frac{-b}{2a}<5;\therefore -\frac{-m}{2}\in (1,5)$
$\Rightarrow m\in (2,10)$                                                                                    ----------------(ii)
And  $f\left(1\right)>0\Rightarrow (1-m+4)>0\Rightarrow m<5$
$\therefore m\in (-\infty ,5)$                                                                   ----------------(iii)
And  $f\left(5\right)>0\Rightarrow (25-5m+4)>0\Rightarrow m<\frac{29}{5}$
$\therefore m\in (-\infty ,\frac{29}{5})$                                                                 ----------------(iv)
From (i), (ii), (iii) and (iv), m  $\in$  (4,5)