Question

If $2a+3b+6c=0$, then at least one root of the equation $a{x}^2+bx+c=0$ lies in the interval

• A. (2, 3)
• B. (1, 2)
• C. (0, 1)
• D. (1, 3)

Answer 1

Answer: (C)

(0, 1)

Let us consider $f\left(x\right)=\frac{a{x}^3}{3}+\frac{b{x}^2}{2}+cx$
$\therefore f\left(0\right)=0$ and $f\left(1\right)=\frac{a}{3}+\frac{b}{2}+c$
$=\frac{2a+3b+6c}{6}$
$=0$ given.
As$f\left(0\right)=f\left(1\right)=0$ and $f\left(x\right)$ is continuous and also differentiable in [0, 1].
$\therefore$ By Rolle's theorem $f^{\prime }\left(x\right)=0$ in (0,1)
$\Rightarrow$ $a{x}^2+bx+c=0$ has at least one root in the interval (0, 1)