Mathematics - Quadratic Equation Question with Solution | TestHub

Let $g(x)=\frac{a{x}^3}{3}+b\cdot \frac{x^2}{2}+cx$ $${\mathrm{g}}^{\prime }(x)=a{x}^2+bx+c$$ Given: $a{x}^2+bx+c=0$ and $2a+3b+6c=0$ Statement-2: (i) $g(0)=0$ and $g(1)$ $$\begin{array}{rl}& =\frac{a}{3}+\frac{b}{2}+c=\frac{2a+3b+6c}{6}\\ & =\frac{0}{6}=0\\ & \Rightarrow g(0)=g(1)\end{array}$$ (ii) $g$ is continuous on $[0,1]$ and differentiable on $(0,1)$ $\therefore$ By Rolle's theorem $\exists k\in (0,1)$ such that $g^{\prime }(k)=0$ This holds the statement 2. Also, from statement-2, we can say $a{x}^2+bx+c=0$ has at least one root in $(0,1)$. Thus statement-1 and 2 both are true and statement-2 is a correct explanation for statement-1.