Question

If $4a+2b+c=0$ then the equation $3a{x}^2+2bx+c=0$ has at least one real root lying between

• A. $0$ and $1$
• B. $1$ and $2$
• C. $0$ and $2$
• D. none of these

Answer 1

Answer: (C)

$0$ and $2$

Consider
$f\left(x\right)=a{x}^3+b{x}^2+cx$
$f\left(0\right)=0$
And
$f\left(2\right)=8a+4b+2c$
$=2[4a+2b+c]$
Now $4a+2b+c=0$
Or
$f\left(2\right)=0$
Hence $x=2$ and $x=0$ are roots of $f\left(x\right)$.
Now
$f^{\prime }\left(x\right)=3a{x}^2+2bx+c$.
Hence there lies atleast one root of $f^{\prime }\left(x\right)$ between $[0,2]$ ...(Rolle's theorem).