If the Rolle’s theorem holds for the function $f\left( x \right)=2{{x}^{3}}+a{{x}^{2}}+bx$ in interval [-1,1] for the point $c=\dfrac{1}{2}$ , then find the value of 2a + b?
(a) 1
(b) -1
(c) 2
(d) -2

Hint: Use the conditions of Rolle’s Theorem to get two equations in terms of a and b. Once we get the two equations, solve them to get a and b and hence, the value of the expression given in the question.

Complete step-by-step answer:
Before starting with the solution, let us discuss Rolle’s Theorem. The theorem states that if a function is f is continuous in the interval [a,b], is differentiable in the interval (a,b), and f(a)=f(b) then there exists at least one c lying in the interval [a,b] such that f’(c)=0.
Now starting with the solution. It is given in the question that Rolle’s theorem holds true for the function $f\left( x \right)=2{{x}^{3}}+a{{x}^{2}}+bx$ in the interval [-1,1]. Therefore, we can say that:
$f(-1)=f(1)$
$\Rightarrow -2+a-b=2+a+b$
$\Rightarrow 2b=-4$
$\Rightarrow b=-2$
Now we know $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$ , so, we can say:
$f'\left( x \right)=6{{x}^{2}}+2ax+b$
Now we will substitute the value of b from the above result, and use the other condition of the Rolle’s theorem that derivative is zero for c and it is given that c is equal to $\dfrac{1}{2}$ . So, we get
$f'\left( c \right)=6{{c}^{2}}+2ac-2$
$\Rightarrow f'\left( \dfrac{1}{2} \right)=6{{\left( \dfrac{1}{2} \right)}^{2}}+2\times \dfrac{1}{2}\times a-2$
$\Rightarrow 0=6\times \dfrac{1}{4}+a-2$
$\Rightarrow a=\dfrac{1}{2}$
Now using a and b to find the value of the expression 2a+b, which is equal to $2\times \dfrac{1}{2}-2=1-2=-1$ .
Therefore, the answer to the above question is option (b).

Note: While using Rolle’s Theorem, don’t forget to ensure that the function is differentiable and continuous in the given interval, as it is a necessary condition for Rolle’s Theorem to hold true.