Question

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

• A. $\left(0,1\right)$
• B. $\left(1,2\right)$
• C. $\left(2,3\right)$
• D. $\left(1,3\right)$

Answer 1

Answer: (A)

Let $f^{{}^{\prime }}\left(x\right)=a{x}^2+bx+c$
$\Rightarrow f\left(x\right)=\frac{a{x}^3}{3}+\frac{b{x}^2}{2}+cx+d$
$\Rightarrow f\left(x\right)=\frac{2a{x}^3+3b{x}^2+6cx+6d}{6}$
$\therefore f\left(1\right)=\frac{2a+3b+6c+6d}{6}=\frac{6d}{6}=d$
$\left(\because 2a+3b+6c=0\right)$ and $f\left(0\right)=d$
So from Rolle’s theorem there exists at least one$\alpha$ in $\left(0,1\right)$for which $f^{\prime }\left(x\right)=0.$
Or there is at least one root of $a{x}^2+bx+c=0$ in $\left(0,1\right).$