Question

If $3(a+2c)=4(b+3d),$ then the equation $a{x}^3+b{x}^2+cx+d=0$ will have atleast one real root in

• A. $(0,1)$
• B. $(-1,0)$
• C. $(3,4)$
• D. $(1,2)$

Answer 1

The correct option is B $(-1,0)$

Let $f\left(x\right)=a{x}^3+b{x}^2+cx+d$
$g\left(x\right)=\int f\left(x\right)dx=\frac{a{x}^4}{4}+\frac{b{x}^3}{3}+\frac{c{x}^2}{2}+dx+e$
$g\left(0\right)=e$
$g(-1)=\frac{a}{4}-\frac{b}{3}+\frac{c}{2}-d+e$
$=\frac{a+2c}{4}-\frac{b+3d}{3}+e=e$
$\{\because 3(a+2c)=4(b+3d\left)\right\}$
$g\left(x\right)$ is continuous and differentiable everywhere.
and $g\left(0\right)=g(-1)$
$\therefore$ From RMVT, $g^{\prime }\left(x\right)=0$ have at least one real root in $(-1,0)$
$\Rightarrow f\left(x\right)=0$ have at least one real root in $(-1,0)$