JEE Main · 2004 · 24 Apr 2004 · Offline · Quadratic Equations and Inequalities

Consider the antiderivative $g(x)=\frac{a}{3}{x}^3+\frac{b}{2}{x}^2+cx$, whose derivative is $g^{\prime }(x)=a{x}^2+bx+c$. Clearly $g(0)=0$, and $g(1)=\frac{a}{3}+\frac{b}{2}+c=\frac{2a+3b+6c}{6}=0$ by the given relation. Since $g$ takes the same value at $0$ and at $1$ and is differentiable, Rolle's theorem supplies a point in $(0,1)$ where $g^{\prime }$ vanishes, that is where $a{x}^2+bx+c=0$. So at least one root lies in $(0,1)$.