Question

Let $f: R \rightarrow R$ 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. $ \displaystyle\sum_{ r =0}^{10}( f ( r ))^2=286$
• D. $\tan ^{-1}( f ( x ))$ is derivable $\forall x \in R$.

Answer 1

Answer: (C), (D)

$f^{\prime}(x)=\underset{h \rightarrow 0}{\text{Lim}} \frac{f(x+h)-f(x)}{h}=\underset{h \rightarrow 0}{\text{Lim}} \frac{f\left(\frac{2 x+2 h}{2}\right)-f\left(\frac{2 x+2 \times 0}{2}\right)}{h}$
$ =\underset{h \rightarrow 0}{\text{Lim}}\frac{ f (2 x )+ f (2 h )- f (2 x )- f (0)}{2 h }=\underset{h \rightarrow 0}{\text{Lim}} \frac{ f (2 h )- f (0)}{2 h } $
$\therefore f ^{\prime}( x ) = f ^{\prime}(0)=-1 \text { (Given) }$
Now on integrating both sides, we get
$ f(x)=-x+C$
$\Rightarrow f(x)=-x+1(\operatorname{As~} f(0)=1)$
(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) $( f (0))^2+( f (1))^2+( f (2))^2+\ldots \ldots .+( f (10))^2$
$=1+0+1^2+2^2+3^2+\ldots \ldots .+9^2=1+\frac{10 \times 9 \times 19}{6}=1+(15 \times 19)=1+285=286$
(D) $\tan ^{-1}( f ( x ))=\tan ^{-1}(1- x )$, which is derivable $\forall x \in R$.