Question

If $2a + 3b + 6c = 0$ then at least one root of the equation $a{x^2} + bx + c = 0$ lies in the interval
A. (0,1)
B. (1,2)
C. (2,3)
D. (3,4)

Answer 1

Hint: Integrate the equation $a{x^2} + bx + c$ 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 {({y^n} + b{y^m})dy = \dfrac{{{y^{n + 1}}}}{{n + 1}} + \dfrac{{b{y^{m + 1}}}}{{m + 1}} + c} $ where c is called the constant of integration

Complete step-by-step solution:
Let $f'(x) = a{x^2} + bx + c$
On integrating $f'(x)$, we get
$f(x) = a\dfrac{{{x^3}}}{3} + b\dfrac{{{x^2}}}{2} + cx + d$
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 exists a number $c \in (a,b)$, such that $f'(c) = 0$.
$f(1) = \dfrac{a}{3} + \dfrac{b}{2} + c + d$
$f(1) = \dfrac{{2a + 3b + 6c}}{6} + d$
$f(1) = d$
$f(0) = d$
$f(x) = a\dfrac{{{x^3}}}{3} + b\dfrac{{{x^2}}}{2} + cx + d$ is continuous on $[0,1]$ and differentiable on $(0,1)$ because it is a polynomial functional.
Since $f(x) = a\dfrac{{{x^3}}}{3} + b\dfrac{{{x^2}}}{2} + cx + d$ 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 $a{x^2} + bx + c$. The quadratic equation $a{x^2} + bx + c = 0$ therefore, has at least one root in (0,1).

Therefore, the correct answer is A. (0,1).

Note: The Rolle’s theorem can only be applied if all the following three conditions are 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.