Question

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

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

Answer 1

Answer: (D)

$-1$

$f\left(\frac{x+y}{2}\right)=\frac{f\left(x\right)+f\left(y\right)}{2}$ is only possible if the function is affine, ie both concave $x$ convex hence it is of the form

$f\left(x\right)=ax+b$

Given $f\left(0\right)=1\Rightarrow b=1$

also $f^1\left(0\right)=1$

$\Rightarrow a=-1$

$\therefore$ $f\left(x\right)=-x+1$

$\therefore$ $Ff\left(2\right)=-2+1=-1$ [D]