If $a + b + c = 0$ then the equation $3a{x^2} + 2bx + c = 0$ has, in the interval $(0,1)$.
A) At least one root
B) At most one root
C) No root
D) Exactly one root exists

Hint: To find the answer to this question, first we have to know about Rolle’s theorem then consider our given equation is in one function. After that, put values of $x$ according to our problem then apply Rolle’s theorem to our question, then we will find that the answer after applying Rolle’s theorem is given an interval or not.

Complete step by step answer:
First of all write our equation once again,
$ \Rightarrow 3a{x^2} + 2bx + c = 0$
Now, let’s consider given equation is a function,
$ \Rightarrow f'(x) = 3a{x^2} + 2bx + c$
So that’s why we get,
$ \Rightarrow f(x) = a{x^3} + b{x^2} + cx + d$
We can see that $f(x)$ is polynomial, it is continuous and differentiable for all $x \in [0,1]$ .
Now, put value of $x = 0$ in $f(x)$ ,
$ \Rightarrow f(0) = a(0) + b(0) + c(0) + d$
$ \Rightarrow f(0) = d$
Now, put value of $x = 1$ in $f(x)$ ,
$ \Rightarrow f(1) = a(1) + b(1) + c(1) + d$
$ \Rightarrow f(1) = a + b + c + d$
But it’s given that $a + b + c = 0$ so we can say that,
$ \Rightarrow f(1) = d$
We can see that $f(0) = f(1)$
Now, we can apply Rolle’s theorem for above equation in the interval of $(0,1)$
Hence,
$ \Rightarrow f'(c) = 0$
Above equation is correct for $0 < c < 1$
Hence, there exists at least one root of the equation $3a{x^2} + 2bx + c = 0$ in the interval $(0,1)$. So, the correct option is option (A).

Note:
For better understanding of Rolle's theorem see the following statement. Rolle’s theorem states that if a function f is continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$ such that $f(a) = f(b)$ , then $f'(x) = 0$ for some $x$ with $a \leqslant x \leqslant b$ .