If Rolle's therorem holds for $f(x)=x\left( { x }^{ 2 }+ax+b \right) +2atx=\frac { 1 }{ 2 }$in the interval (-1 , 1), then which of the following (S) is/are corect? ;

Given: $f(x)=x\left(x^{2}+a x+b\right)+2$

nolds Rolle's theorem at $x=\frac{1}{2}$ in interwal $[-1,1]$

then $f(-1)=f(1)$

$\begin{array}{l} \Rightarrow(-1)\left((-1)^{2}+a(-1)+b\right)+2=1\left(1^{2}+a(1)+b\right)+2 \\ \Rightarrow-(1-a+b)+2=1+a+b+2 \\ \Rightarrow 2-b+a-1=3+a+b \\ \Rightarrow b=-1 \\ f^{\prime}(x)=0 \quad \text { at } x=\frac{1}{2} \\ \Rightarrow\left(x^{2}+a x+b\right)+2 x^{2}+a x=0 \end{array}$

$\begin{array}{l} \Rightarrow \frac{1}{4}+\frac{a}{2}+b+\frac{1}{2}+\frac{a}{2}=0 \\ \Rightarrow \frac{3}{4}+a-1=0 \\ \Rightarrow a=-\frac{1}{4} \end{array}$

How

$\begin{aligned} a+b & =\frac{-1}{4}+(-1) \\ & =\frac{-1}{4}-1 \\ & =\frac{-3}{4} \end{aligned}$

Hence option (A) is correct