If 2a+3b+6c=0, the at least one root of the equation ax2+bx+c=0 lies in the interval

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

1. A

$\left(0,1\right)$

2. B

3. C

4. D

none of these

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### Solution:

Consider the function

$f\left(x\right)=\frac{a{x}^{3}}{3}+\frac{b{x}^{2}}{2}+cx$

We find that $f\left(0\right)=0$ and,

$f\left(1\right)=\frac{a}{3}+\frac{b}{2}+c=\frac{2a+3b+6c}{6}=\frac{0}{6}=0\left[\because 2a+3b+6c=0\right]$

Therefore, 0 and 1 are roots of the polynomial

Consequently, there exists at least one root of the polynomial

${f}^{\mathrm{\prime }}\left(x\right)=a{x}^{2}+bx+c$ lying between 0 and 1.

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