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. $(0,1)$
• B. $(1,2)$
• C. $(2,3)$
• D. $(1,3)$

Answer 1

The correct option is A

$(0,1)$

Explanation for the correct option:

Step 1: Finding the integration

It is given that $f\text{'}\left(x\right)=a{x}^2+bx+c$ (i)

By integrating both sides, we get

$\Rightarrow f\left(x\right)=\frac{a{x}^3}{3}+\frac{b{x}^2}{2}+cx+d$ (ii)

$\Rightarrow f\left(x\right)=\frac{2a{x}^3+3b{x}^2+6cx+6d}{6}$

Step 2: Calculating the value of $f\left(0\right)$

The value of $f\left(0\right)$ can be calculated by substituting the value of $x$ as $0$.

$\Rightarrow f\left(0\right)=\frac{6d}{6}$

$\Rightarrow f\left(0\right)=d$

Step 3: Calculating the value of $f\left(1\right)$

The value of $f\left(1\right)$ can be calculated by substituting the value of $x$ as $1$.

$f\left(1\right)=\frac{2a+3b+6c+6d}{6}$ (iii)

Since,

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

on substituting the value in (iii), we get

$\Rightarrow f\left(1\right)=\frac{6d}{6}$

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

Hence,

$f\left(0\right)=f\left(1\right)$

Therefore one of the roots of the given equation lies between $(0,1)$.

Hence, Option (A) is correct.