Question

If Rolle's theorem holds true 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 $(2b+c)$ is equal to

• A. $1$
• B. $-1$
• C. $2$
• D. $-3$

Answer 1

The correct option is B $-1$

Since, Rolle's theorem holds true for $f\left(x\right)$ in the interval $[-1,1]$
$\therefore f\left(1\right)=f(-1)$
$2+b+c=-2+b-c\Rightarrow 2c=-4$
or, $c=-2$
$f^{\prime }\left(\frac{1}{2}\right)=0$
$\because f^{\prime }\left(x\right)=6x^2+2bx+c$
$\therefore f^{\prime }\left(\frac{1}{2}\right)=\frac{6}{4}+b+c=0\Rightarrow b+c=\frac{-3}{2}$
or, $b=\frac{1}{2}$

$\therefore 2b+c=1-2=-1$