Question

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

• A. $1$
• B. $-1$
• C. $0$
• D. $8$

Answer 1

The correct option is B $-1$

Since $f\left(\frac{x+y}{2}\right)=\frac{f\left(x\right)+f\left(y\right)}{2}$
Replacing $x$ by $2x$ and $y$ by $0$, then $f\left(x\right)=\frac{f\left(2x\right)+f\left(0\right)}{2}$
$\Rightarrow f\left(2x\right)+f\left(0\right)=2f\left(x\right)$$\Rightarrow f\left(2x\right)-2f\left(x\right)=-f\left(0\right)$
Now, $f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$
$=\underset{h\to 0}{lim}\left\{\frac{f\left(\frac{2x+2h}{2}\right)-f\left(x\right)}{h}\right\}$
$=\underset{h\to 0}{lim}\left\{\frac{\frac{f\left(2x\right)+f\left(2h\right)}{2}-f\left(x\right)}{h}\right\}$
$=\underset{h\to 0}{lim}\left\{\frac{\frac{f\left(2x\right)+f\left(2h\right)-2f\left(x\right)}{2}}{h}\right\}$
$=\underset{h\to 0}{lim}\left\{\frac{f\left(2h\right)-f\left(0\right)}{2h}\right\}$
$=f^{\prime }\left(0\right)$
$=-1\forall x\in R$