If the Rolle’s theorem holds for the function  $f\left(x\right)=x^4+a{x}^3+bx$, in  $-1\le x\le 1$, and $f^{\text{'}}\left(\frac{1}{2}\right)=0$ , then  $ab=$

We have,
  $f\left(x\right)=x^4+a{x}^3+bx,[-1,1]$
Since, RMVT is applicable on  $f\left(x\right)$
 $$\begin{gathered} \therefore f(-1)=f\left(1\right) \\ \Rightarrow 1-a-b=1+a+b \\ \Rightarrow a+b=0 \end{gathered}$$
Also, we have
  $$\begin{gathered} f^{\text{'}}\left(\frac{1}{2}\right)=0 \\ \therefore 4{\left(\frac{1}{2}\right)}^3+3a{\left(\frac{1}{2}\right)}^2+b=0 \\ \Rightarrow \frac{1}{2}+\frac{3}{4}a+b=0 \\ \Rightarrow 2+3a+4b=0 \\ \Rightarrow 3a+4b=-2 \end{gathered}$$
On solving equation (i) and (ii), we get
 $$\begin{gathered} a=2,b=-2 \\ \therefore ab=\left(2\right)(-2)=-4 \end{gathered}$$