Question

lf $2a+3b+6c=0$, then the equation $a{x}^2+bx+c=0$ has atleast one root in

• A. $(0,1)$
• B. $(1,1)$
• C. $[-1,1]$
• D. $[1,2]$

Answer 1

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

Given:

$2a+3b+6c=0$

Using Rolle's theorem,

Let $f^{\prime }\left(x\right)=a{x}^2+bx+c$

Integrating both sides w.r.t. $x$
$f\left(x\right)=\frac{a{x}^3}{3}+\frac{b{x}^2}{2}+cx+d$
$f\left(0\right)=d$

$f\left(1\right)=\frac{a}{3}+\frac{b}{2}+c+d=\frac{(2a+3b+6c)}{6}+d=d$
$\Rightarrow f\left(0\right)=f\left(1\right)$
So, according to the Rolle's theorem, we can say that $f\left(x\right)$ is a continuous function in $[0,1]$, differentiable in $(0,1)$

$\therefore$ We have a $c$ in $(0,1)$ such that $f^{\prime }\left(c\right)=0$.

Hence, option $A.$