Question

The quadratic equation $3a{x}^2+2bx+c=0$ has at least one root between $0$ and $1$ if

• A. $a+b+c=0$
• B. $c=0$
• C. $3a+2b+c=0$
• D. $a+b=c$

Answer 1

The correct option is A $a+b+c=0$

$f\left(x\right)=3a{x}^2+2bx+c$
$g\left(x\right)=\int f\left(x\right)=a{x}^3+b{x}^2+cx+d$
$g\left(x\right)$ is continous & differentiable in [0, 1]
$f\left(c\right)=\frac{g\left(1\right)-g\left(0\right)}{1-0}$ where $cϵ[0,1]$ (Lagrange's theorem)
$f\left(c\right)=a+b+c$
$0=a+b+c$