Question

If $27a+9b+3x+d=0$, then the equation $4a{x}^3+3b{x}^2+2cx+d=0$ has atleast one root lying between

• A. $0$ and $1$
• B. $1$ and $3$
• C. $0$ and $3$
• D. $1$ and $2$

Answer 1

Answer: (C)

$0$ and $3$

Consider the polynomial $f\left(x\right)$ given by
$f\left(x\right)=a{x}^4+b{x}^3+c{x}^2+dx$
$f^{\prime }\left(x\right)=4a{x}^3+3b{x}^2+2cx+d$
We have, $f\left(0\right)=0$
and $f\left(3\right)=81a+27b+9c+3d$
$=3(27a+9b+3c+d)=0$ Given
$\therefore$ $0$ and $3$ are the roots of $f\left(x\right)=0$
Consequently, by Rolle's theorem
$f^{\prime }\left(x\right)=0$ ie $4a{x}^3+3b{x}^2+2cx+d$ has atleast one real root between $0$ and $3$.