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Q.
If $2a + 3b + 6c = 0,$ then at least one root of the equation $ax^{2} + bx + c = 0$ lies in the interval
Application of Derivatives
Solution:
Let $f^{'}\left(x\right)=ax^{2}+bx +c$
$\Rightarrow f\left(x\right)=\frac{ax^{3}}{3}+\frac{bx^{2}}{2}+cx+ d$
$\Rightarrow f\left(x\right)=\frac{2ax^{3}+3bx^{2}+6cx+6d}{6}$
$\therefore f\left(1\right)=\frac{2a+3b+6c+6d}{6}=\frac{6d}{6}=d$
$\left(\because2a+3b+6c=0\right)$ and $f\left(0\right)=d$
So from Rolle’s theorem there exists at least one$\alpha$ in $\left(0, 1\right)$for which $f'\left(x\right) = 0.$
Or there is at least one root of $ax^{2} + bx + c = 0$ in $\left(0,1\right).$