Answer: (2)

Solution

Given quadratic equation is: $x^2 - mx + 4 = 0$

Both the roots are real and distinct.

So, discriminant $B^2 - 4AC > 0$.

$\therefore m^2 - 4 \cdot 1 \cdot 4 > 0$

$\therefore (m-4)(m+4) > 0$

$\therefore m \in (-\infty, -4) \cup (4, \infty)$ …(i)

Since, both roots lies in $[1, 5]$

$\therefore \frac{-m}{-2} \in (1, 5)$

$\Rightarrow m \in (2, 10)$ …(ii)

And $1 \cdot (1 - m + 4) > 0 \Rightarrow m < 5$

$\therefore m \in (-\infty, 5)$ …(iii)

And $1 \cdot (25 - 5m + 4) > 0 \Rightarrow m < \frac{29}{5}$

$\therefore m \in \left(-\infty, \frac{29}{5}\right)$ …(iv)

From (i), (ii), (iii) and (iv), $m \in (4, 5)$