Official solution
x2 -mx + 4 = 0
Case-I :
D > 0
m2 - 16 > 0
\[ \Rightarrow \] m \[ \in \] (- \[\infty \], - 4) \[ \cup \] (4, \[\infty \])
Case-II :
\[ \Rightarrow \,\,1 < {{ - b} \over {2a}} < 5\]
\[ \Rightarrow \,\,1 < {m \over 2} < 5 \Rightarrow \,m \in \left( {2,10} \right)\]
Case-III :
f(1) > 0 and f(5) > 0
1 - m + 4 > 0 and 25 - 5m + 4 > 0
m < 5 and m < \[{{29} \over 5}\]
Case-IV :
Let one root is x = 1
1 - m + 4 = 0
m = 5
Now equation x2 - 5x + 4 = 0
(x - 1) (x - 4) = 0
x = 1 i.e. m = 5 is also included
hence m \[ \in \] (4, 5]
So given option is (4, 5)