Question

If a, b, c are real numbers such that $a + b + c = 0$, then the quadratic equation $3a{x^2} + 2bx + c = 0$ has
A. At least one root in [0,1]
B. At least one root in [1,2]
C. At least one root in [-1,0]
D. None of these

Answer 1

Hint: Integrate the given equation and find two values a and b such that the integrated function is equal at a and b. Check whether Rolle’s theorem is applicable and use it if it is applicable.

Formula used: $\int {({x^n} + b{x^m})dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + \dfrac{{b{x^{m + 1}}}}{{m + 1}} + c} $ where c is called the constant of integration

Complete step-by-step solution:
Let $f'(x) = 3a{x^2} + 2bx + c$
On integrating $f'(x)$, we get
$f(x) = a{x^3} + b{x^2} + cx + d$
Rolle’s theorem states that if a function is continuous on $[a,b]$ and differentiable on $(a,b)$ and if $f(a) = f(b)$, then there exist a number $c \in (a,b)$ such that $f'(c) = 0$.
We know that $f(0) = d$ and $f(1) = a + b + c + d = 0 + d = d$.
$f(x) = a{x^3} + b{x^2} + cx + d$ is continuous on $[0,1]$ because it is a polynomial function. It is also differentiable on $(0,1)$ for the same reason.
Since $f(x) = a{x^3} + b{x^2} + cx + d$ is continuous on $[0,1]$ and differentiable on $(0,1)$ and $f(0) = f(1)$, there exists a number $c \in (0,1)$ such that $f'(c) = 0$.
$f'(c) = 0$ means that c is a root of $3a{x^2} + 2bx + c$. The quadratic equation $3a{x^2} + 2bx + c = 0$ therefore, has at least one root in $[0,1]$
Therefore, the correct answer is A. At least one root in $[0,1]$.

Note: The Rolle’s theorem cannot be applied if any of the following three conditions are not met: 1) The function needs to be continuous on the given interval, 2) The function needs to be differentiable on the given interval (except the end points of the interval), 3) The value of the function needs to be the same at the end points of the interval being considered.