Question

Let $f:R\to R$ be a differentiable function satisfying $f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2}\forall x,y\in R$. If $f(0)=1$ and $f^{\prime }(0)=-1$, then which of the following is (are) correct?

• A. $f(|x|)$ is discontinuous at one point.
• B. Number of solution of the equation $f(x)=f^{-1}(x)$ is exactly one.
• C. $\sum _{r=0}^{10}(f(r){)}^2=286$
• D. ${\tan }^{-1}(f(x))$ is derivable $\forall x\in R$.

Answer 1

Answer: (C), (D)

$f^{\prime }(x)=\underset{h\to 0}{\mathrm{Lim}}\frac{f(x+h)-f(x)}{h}=\underset{h\to 0}{\mathrm{Lim}}\frac{f\left(\frac{2x+2h}{2}\right)-f\left(\frac{2x+2\times 0}{2}\right)}{h}$
$=\underset{h\to 0}{\mathrm{Lim}}\frac{f(2x)+f(2h)-f(2x)-f(0)}{2h}=\underset{h\to 0}{\mathrm{Lim}}\frac{f(2h)-f(0)}{2h}$
$\therefore f^{\prime }(x)=f^{\prime }(0)=-1\text{ (Given) }$
Now on integrating both sides, we get
$f(x)=-x+C$
$\Rightarrow f(x)=-x+1(\mathrm{As}f(0)=1)$
(A) $f(|x|)=1-|x|$, which is continuous $\forall x\in R$.
(B) $f(x)=1-x$
$\therefore f^{-1}(x)=1-x$
$\Rightarrow f(x)=f^{-1}(x)$ will have infinite solutions.
(C) $(f(0){)}^2+(f(1){)}^2+(f(2){)}^2+\dots \dots .+(f(10){)}^2$
$=1+0+1^2+2^2+3^2+\dots \dots .+9^2=1+\frac{10\times 9\times 19}{6}=1+(15\times 19)=1+285=286$
(D) ${\tan }^{-1}(f(x))={\tan }^{-1}(1-x)$, which is derivable $\forall x\in R$.