If $a+b+c=0$ then the equation $3{ ax }^{ 2 }+2bx+c=0$ has

Given $\mathrm{a}+\mathrm{b}+\mathrm{c}=0$ and

Let $\mathrm{f}^{\prime}(\mathrm{x})=3 a \mathrm{x}^{2}+2 \mathrm{bx}+\mathrm{c}$

$\Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{ax}^{3}+\mathrm{bx}^{2}+\mathrm{cx}+\mathrm{d} \text { (on integration) }$

Rolle's theorem states that if $\mathrm{f}(\mathrm{x})$ be continuous on $[\mathrm{a}, \mathrm{b}]$, differentiable on $(\mathrm{a}, \mathrm{b})$ and $f(a)=f(b)$ then there exists some $c$ between a and $\mathrm{b}$ such that $\mathrm{f}^{\prime}(\mathrm{c})=0$

from the options let $(\mathrm{a}, \mathrm{b})=(0,1)$

$f(0)=d$ and $f(1)=a+b+c+d=d$ (since, $\mathrm{a}+\mathrm{b}+\mathrm{c}=0$ )

Therefore, $f(0)=f(1)$. Hence, there exists some c between 0 and 1 such that $\mathrm{f}^{\prime}(\mathrm{c})=0$

Therefore, the equation has atleast one root lying on $(0,1)$