Question

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

• A. $1$
• B. $-1$
• C. $1/2$
• D. $2$

Answer 1

The correct option is B $-1$

Putting y=0 in the given functional equation, We get

$f\left(x/2\right)=\frac{f\left(x\right)+f\left(0\right)}{2}=\frac{1}{2}[1+f\left(x\right)](f\left(0\right)=1)$

$\Rightarrow$ $f\left(x\right)=2f\left(x/2\right)-1$ ---------(i)

Since $f^{\prime }\left(0\right)=-1$ we get $\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}=-1$

$\Rightarrow$ $\underset{h\to 0}{lim}\frac{f\left(h\right)-1}{h}=-1$

Let $xϵR$, then
$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$

$\underset{h\to 0}{lim}\frac{\frac{f\left(2x\right)+f\left(2h\right)}{2}-f\left(x\right)}{h}$

$=\underset{h\to 0}{lim}\frac{1}{h}[\frac{1}{2}\{2f\left(x\right)-1+2f\left(h\right)-1\}-f\left(x\right)]$ -------- using (i)

$=\underset{h\to 0}{lim}\frac{1}{h}[f\left(x\right)-1+f\left(h\right)-f\left(x\right)]$

$=\underset{h\to 0}{lim}\frac{f\left(h\right)-1}{h}=-1$
Thus $f^{\prime }\left(x\right)=-1$, so we get $f\left(x\right)=-x+C.$

But f(0)=1, therefore, $1=f\left(0\right)=-0+C.$
$\Rightarrow C=1$

Thus,$f\left(x\right)=1-x$, in particular
$f\left(2\right)=1-2=-1$