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. None of these

Answer 1

Answer: (A)

$(0,1)$

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

$\int f\left(x\right)dx=\int (a{x}^2+bx+c)dx\Rightarrow g\left(x\right)=\frac{a{x}^3}{3}+\frac{b{x}^2}{2}+cx+d\Rightarrow g\left(x\right)=\frac{2a{x}^3+3b{x}^2+6cx}{6}+d\Rightarrow g\left(0\right)=0+d=dg\left(1\right)=\frac{2a+3b+6c}{6}+d$

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

$\Rightarrow g\left(1\right)=d$

Now $g\left(x\right)$ is a cubic polynomial function so it will be continous and differentiable in $(0,1)$

Also $g\left(0\right)=g\left(1\right)=$

So, by using Rolle's theorem $g^{\prime }\left(x\right)$ has at least one root in $(0,1)$

$g^{\prime }\left(x\right)=f\left(x\right)$

$\Rightarrow$ $f\left(x\right)$ has at least one root in $(0,1)$

Hence, option A is correct.