$f\left( x \right) = 2{x^3} + a{x^2} + bx$

let, $a=-1,b=1$

Given that $f\left( x \right)$ satisfy Roll theorem in interval $[-1,1]$

$f\left( x \right)$ must satify two conditions.

$(1)f(a)=f(b)$

$(2)f'(c)=0$

($c$ should be between $a$ and $b$ )

$f\left( a \right) = f\left( 1 \right) – 2{\left( 1 \right)^3} + a{\left( 1 \right)^2} + b\left( 1 \right) = 2 + a + b$

$f\left( b \right) = f\left( { – 1} \right) – 2{\left( { – 1} \right)^3} + a{\left( { – 1} \right)^2} + b\left( { – 1} \right) =  – 2 + a – b$

$f\left( a \right) = f\left( b \right)$

$2 + a + b =  – 2 + a – b$

$2b =  – 4$

$b =  – 2$

( given that $c = \frac{1}{2}$)

$f'\left( x \right) = 6{x^2} + 2ax + b$ 

at $x = \frac{1}{2},f'\left( x \right) = 0$

$0 = 6{\left( {\frac{1}{2}} \right)^2} + 2a\left( {\frac{1}{2}} \right) + b$

$\frac{3}{2} + a + b = 0$

$\frac{3}{2} + a – 2 = 0$

$a = 2 – \frac{3}{2} = \frac{1}{2}$

$2a + b = 2 \times \frac{1}{2} – 2 = 1 –  – 1$