JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Quadratic Equations and Inequalities

For both roots of $x^2-mx+4=0$ to be real and distinct the discriminant must be positive, $m^2-16>0$; since the product of the roots is $4>0$ and they are to lie in $[1,5]$ the sum $m$ is positive, so $m>4$. The parabola opens upward, so it must be non-negative at the ends of the interval: $f(1)=1-m+4\ge 0$ gives $m\le 5$ and $f(5)=25-5m+4\ge 0$ gives $m\le 5.8$. The vertex abscissa $m/2$ already lies in $[1,5]$. Combining these conditions, $m$ lies in $(4,5)$.