Question

If Rolle's theorem holds for the function $f\left(x\right)=2x^3+b{x}^2+cx,x\in [-1,1],$ at the point $x=\frac{1}{2},$ then $2b+c$ equals:

• A. $-1$
• B. $1$
• C. $-3$
• D. $2$

Answer 1

The correct option is A $-1$

$f\left(x\right)=2x^3+b{x}^2+cx$ and $x\in [-1,1]$
For rolles theorem to be applicable $f\left(1\right)=f(-1)$ and $f^{\prime }\left(\frac{1}{2}\right)=0$
$f\left(1\right)=f(-1)$
$\Rightarrow 2+b+c=-2+b-c\Rightarrow c=-2$
and $f^{\prime }\left(\frac{1}{2}\right)=0$
$\Rightarrow 6{\left(\frac{1}{2}\right)}^2+2b\left(\frac{1}{2}\right)+c=0\Rightarrow \frac{3}{2}+b-2=0\Rightarrow 2b=1$
Hence $2b+c=-1$