Question

If $(2a+3b+6c=0)$, $a,b,c\in R$, then quadratic equation $a{x}^2+bx+c=0$ has

• A. At least one root in $[0,1]$
• B. At least one root in $(0,1)$
• C. At least one root in $[2,3]$
• D. At least one root in $[4,5]$

Answer 1

The correct option is B At least one root in $(0,1)$

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 $(0,1)$