Question

If $a+b+c=0$, then the equation $3a{x}^2+2bx+c=0$ has, in the interval $\left(0,1\right)$

• A. At least one root
• B. At most one root
• C. No root
• D. Exactly one root exists

Answer 1

The correct option is A

At least one root

Explanation for the correct option.

Find the number of roots in the equation.

Let $f\text{'}\left(x\right)=3a{x}^2+2bx+c$. Now upon integrating both sides it is found that

$$\begin{gathered} f\left(x\right)=3a·\frac{x^3}{3}+2b·\frac{x^2}{2}+cx+d \\ \Rightarrow f\left(x\right)=a{x}^3+b{x}^2+cx+d \end{gathered}$$

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

Now, in the interval $(0,1)$ and for the function $f\left(x\right)=a{x}^3+b{x}^2+cx+d$ the value of $f\left(0\right)$ is given as:

$\begin{array}{rcl}f\left(0\right)& =& a·0^3+b·0^2+c·0+d\\ & =& d\end{array}$

And the value of $f\left(1\right)$ is given as:

$\begin{array}{rcl}f\left(1\right)& =& a·1^3+b·1^2+c·1+d\\ & =& a+b+c+d\\ & =& 0+d\left[a+b+c=0\right]\\ & =& d\end{array}$

So the function $f\left(x\right)=a{x}^3+b{x}^2+cx+d$ is continous and differentiable in the interval $\left(0,1\right)$ and $f\left(0\right)=f\left(1\right)$. So there exists some $c$ between $0$ and $1$ such that $f\text{'}\left(c\right)=0$.

So there is at least one root in the interval $\left(0,1\right)$ for the equation $f\text{'}\left(x\right)=0$.

So the equation $3a{x}^2+2bx+c=0$ has at least one root, in $\left(0,1\right)$.

Hence, the correct option is A.