Question

Let $f:R\to R$ be a differentiable function satisfying $f\left(\frac{x+y}{3}\right)=\frac{2+f\left(x\right)+f\left(y\right)}{3}\forall x,y\in R$ and $f^{\prime }\left(2\right)=2$, then answer the following questions:

The function $h\left(x\right)=|f\left(\left|x\right|\right)-4|$ is

• A. non-differentiable and discontinuous at $-1,0,1$
• B. differentiable for all real numbers
• C. non-differentiable but continuous at $-1,0,1$
• D. non-differentiable only at $-1$ and $1$

Answer 1

The correct option is C non-differentiable but continuous at $-1,0,1$

Substitute $x=3x,y=0$ in $f\left(\frac{x+y}{2}\right)=\frac{2+f\left(x\right)+f\left(y\right)}{3}$, we get

$f\left(\frac{3x+0}{3}\right)=\frac{2+f\left(3x\right)+f\left(0\right)}{3}\Rightarrow f\left(x\right)=\frac{2+f\left(3x\right)+f\left(0\right)}{3}$ ...(1)

Substitute $x=y=0$

$f\left(0\right)=\frac{2+f\left(0\right)+f\left(0\right)}{3}\Rightarrow 3f\left(0\right)=2+2f\left(0\right)\Rightarrow f\left(0\right)=2$ ....(2)

From (1)

$f\left(x\right)=\frac{2+f\left(3x\right)+2}{3}\Rightarrow 3f\left(x\right)=f\left(3x\right)+4$ ...(3)

Now $f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}=\underset{h\to 0}{lim}\frac{f\left(\frac{3x+3h}{3}\right)-f\left(x\right)}{h}$

$=\underset{h\to 0}{lim}\left\{\frac{\frac{2+f\left(3x\right)+f\left(3h\right)}{3}-f\left(x\right)}{h}\right\}=\underset{h\to 0}{lim}\left[\frac{2+f\left(3x\right)+f\left(3h\right)-3f\left(x\right)}{3h}\right]$

$=\underset{h\to 0}{lim}\left[\frac{2+3f\left(x\right)-4+f\left(3h\right)-3f\left(x\right)}{3h}\right]$

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\left[\frac{f\left(3h\right)-2}{3h}\right]$

Since $f\left(x\right)$ is differentiable $\forall x;{lim}_{h\to 0}f\left(3h\right)=2$

$\Rightarrow f\left(0\right)=2\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{3f^{\prime }\left(3h\right)}{3}=f^{\prime }\left(0\right)=k$ (say)

$\Rightarrow f^{\prime }\left(x\right)=k\Rightarrow f\left(x\right)=kx+c$

$f^{\prime }\left(x\right)=k\Rightarrow f^{\prime }\left(2\right)=k$

But $f^{\prime }\left(2\right)=2\Rightarrow k=2$

$\therefore f\left(x\right)=2x+c$

Also $f\left(0\right)=2\Rightarrow c=2$

$\therefore f\left(x\right)=2x+2$

$h\left(x\right)=|f\left(\left|x\right|\right)-4|=\left|2\right|x|+2-4|=\left|2\right|x|-2|$

$=2|\left|x\right|-1|=\{\begin{array}{c}\begin{array}{cc}2|x-1|& x\ge 0\end{array}\\ \begin{array}{cc}2|-x-1|& x<0\end{array}\end{array}$

$=\{\begin{array}{c}\begin{array}{cc}2(x-1)& x\ge 1\end{array}\\ \begin{array}{cc}2(-x+1)& 0\le x<1\end{array}\\ \begin{array}{cc}2(x+1)& -1\le x<0\end{array}\\ \begin{array}{cc}2(-x-1)& x<-1\end{array}\end{array}$