JEE Main · 2005 · 08 May 2005 · Offline · Quadratic Equations and Inequalities

For both roots of $x^2-2kx+k^2+k-5=0$ to be less than $5$, three conditions must hold together: the roots must be real, the axis of symmetry must lie below $5$, and the quadratic must be positive at $x=5$. The axis is at $x=k$, so $k<5$. The discriminant condition $4k^2-4(k^2+k-5)\ge 0$ reduces to $k\le 5$. Finally $f(5)=25-10k+k^2+k-5=k^2-9k+20=(k-4)(k-5)>0$ requires $k<4$ or $k>5$. Taking all three together leaves $k<4$, so $k\in (-\infty ,4)$.