Question

If $2a+3b+6c=0$, then atleast 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. $(1,3)$

Answer 1

The correct option is A $(0,1)$

Given, $2a+3b+6c=0$
Let $f^{\prime }\left(x\right)=a{x}^2+bx+c$
$f\left(x\right)=\frac{a{x}^3}{3}+\frac{b{x}^2}{2}+cx+d$
$\Rightarrow f\left(x\right)=\frac{2a{x}^3+3b{x}^2+6cx+6d}{6}$
Now, $f\left(1\right)=\frac{2a+3b+6c+6d}{6}=\frac{6d}{6}=d$
and $f\left(0\right)=\frac{6d}{6}=d$
$\therefore f\left(0\right)=f\left(1\right)$
$\Rightarrow f^{\prime }\left(x\right)$ will vanish atleast once betweeen $0$ and 1 (by Rolle's theorem)
$\therefore$ atleast one of the roots of the equation $a{x}^2+bx+c=0$ lies between $0$ and $1$