Question

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

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

Answer 1

Answer: (B)

$-1$

By observation , we can get

$f\left(x\right)=y=mx+c$

now $f^{\prime }\left(0\right)=-1=m$

$f\left(0\right)=c=1$

Therefore, $f\left(x\right)=-x+1$
$f(-2)=-2+1=-1$