Question

Let $f\left(\frac{x+y}{2}\right)=\frac{1}{2}\left(f\right(x)+f(y\left)\right)$ for real x and y. If f' (0) = – 1 and f(0) = 1 then f(2) is

• A. 1
• B. -1
• C. $\frac{1}{2}$
• D. 2

Answer 1

The correct option is B

-1

$f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}={lim}_{h\to 0}\frac{\frac{f\left(2x\right)+f\left(2h\right)}{2}}{h}{f}^{\prime }\left(x\right)=1;f\left(2x\right)=2\left(x\right)-1\Rightarrow f\left(x\right)=1-x$