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

Answer 1

The correct option is A $(0,1)$

$\frac{a}{3}+\frac{b}{2}+c=0$

Let $f\left(x\right)=a{x}^2+bx+c$

$F\left(x\right)=\int f\left(x\right)=\frac{a{x}^3}{3}+\frac{b{x}^2}{2}+cx+d$

$F\left(x\right)$ is continous & differentiable in [0, 1]

$F^1\left(c\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}$ where $cϵ[0,1]$ (apply LMVT)

$=\frac{a}{3}+\frac{b}{2}+c$

$f\left(c\right)=0$