If $f\left(\frac{x+y}{2}\right)=\frac{f\left(x\right)+f\left(y\right)}{2},\mathrm{\forall }x,y\in R,f\left(0\right)=1$ $f^{\mathrm{\prime }}\left(0\right)=-1,\text{ then }f\left(2\right)=$

 

Replacing $x$ by 2$x$ and $y$ by $0,$the given equation becomes 

$f\left(x\right)=\frac{f\left(2x\right)+1}{2}$

Replacing $x$ by 2$x$ and $y$ by 2$h,$we also get 

$\begin{array}{l}f(x+h)=\frac{f\left(2x\right)+f\left(2h\right)}{2}\\ \text{ (ii) }-\text{ (i) }\Rightarrow f(x+h)-f\left(x\right)=\frac{f\left(2h\right)-1}{2}\cdots \cdots \cdots \cdots \cdots \cdots \cdots \left(ii\right)\end{array}$

Now $f^{\mathrm{\prime }}\left(x\right)=\underset{h\to 0}{lim} \frac{f(x+h)-f\left(x\right)}{h}=\underset{h\to 0}{lim} \frac{f\left(2h\right)-f\left(0\right)}{2h}$

or $f^{\mathrm{\prime }}\left(x\right)=f^{\mathrm{\prime }}\left(0\right)=-1\therefore f\left(x\right)=1-x,f\left(2\right)=-1.$