Question

If $2a + 3b + 6c = 0$ then at least one root of the equation $ax^{2} + bx + c = 0$ lies in the interval

• A. (0, 1)
• B. (1, 2)
• C. (2, 3)
• D. (3, 4)

Answer 1

Answer: (A)

(0, 1)

Answer 2

Let $f^{'}(x) = ax^{2} + bx + c$

∴ $f(x) = \int_{}^{}{f^{'}(x)dx} = \frac{ax^{3}}{3} + \frac{bx^{2}}{2} + cx$

Clearly $f(0) = 0$, $f(1) = \frac{a}{3} + \frac{b}{2} + c = \frac{2a + 3b + 6c}{6} = \frac{0}{6} = 0$

Since, $f(0) = f(1) = 0$. Hence, there exists at least one point c in between 0 and 1, such that $f^{'}(x) = 0$, by Rolle’s theorem.

Trick: Put the value of $a = - 3,b = 2,c = 0$ in given equation

$- 3x^{2} + 2x = 0$ ⇒ $3x^{2} - 2x = 0$ ⇒ $x(3x - 2) = 0$

$x = 0,x = 2/3$, which lie in the interval (0, 1)