Let $f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2}$ for all real values of x and y. If f'(0) exists and equals -1 and f(0) = 1, then f'(2) is equal to

-1

We have,

$f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2}$ for all $x, y \in R$              .......(i)

$\Rightarrow f\left(\frac{x+0}{2}\right)=\frac{f(x)+f(0)}{2}$ for all $x \in R$          [Putting y = 0]

$\Rightarrow f\left(\frac{x}{2}\right)=\frac{f(x)+1}{2}$ for all $x \in R$

$\Rightarrow f(x)=2 f\left(\frac{x}{2}\right)-1$ for all $x \in R$              .......(ii)

Now,

$f'(2) =\lim\limits_{h \rightarrow 0} \frac{f(2+h)-f(2)}{h}$

$\Rightarrow f'(2) =\lim\limits_{h \rightarrow 0} \frac{f\left(\frac{4+2 h}{2}\right)-f(2)}{h}$

$\Rightarrow f'(2) =\lim\limits_{h \rightarrow 0} \frac{\frac{f(4)+f(2 h)}{2}-f(2)}{h}$               [Using (i)]

$\Rightarrow f'(2)=\lim\limits_{h \rightarrow 0} \frac{f(4)+f(2 h)-2 f(2)}{2 h}$

$\Rightarrow f'(2) =\lim\limits_{h \rightarrow 0} \frac{2 f(2)-1+2 f(h)-1-2 f(2)}{2 h}$            [Putting x = 4 and 2h respectively in (ii)]

$\Rightarrow f'(2)=\lim\limits_{h \rightarrow 0} \frac{f(h)-1}{h}$

$\Rightarrow f'(2)=\lim\limits_{h \rightarrow 0} \frac{f(h)-f(0)}{h}=f'(0)=-1$