Question

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

• A. $f\left(|x|\right)$ is not differentiable at only one point
• B. Number of solutions of the equation $f\left(x\right)=f^{-1}\left(x\right)$ is exactly one
• C. $\sum _{r=0}^{10}\left(f\right(r){)}^2=286$
• D. Area bounded between the curves $y=f\left(x\right)$ and $y=g\left(x\right)$ is $\frac{1}{24}\text{sq. units}$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $f\left(|x|\right)$ is not differentiable at only one point
C $\sum _{r=0}^{10}\left(f\right(r){)}^2=286$
D Area bounded between the curves $y=f\left(x\right)$ and $y=g\left(x\right)$ is $\frac{1}{24}\text{sq. units}$

Given, $f$ is differentiable and
$f\left(\frac{x+y}{2}\right)=\frac{f\left(x\right)+f\left(y\right)}{2}$
$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$
$=\underset{h\to 0}{lim}\frac{f\left(\frac{2x+2h}{2}\right)-f\left(\frac{2x+2\times 0}{2}\right)}{h}$
$=\underset{h\to 0}{lim}\frac{f\left(2x\right)+f\left(2h\right)-f\left(2x\right)-f\left(0\right)}{2h}$
$=\underset{h\to 0}{lim}\frac{f\left(2h\right)-f\left(0\right)}{2h}\left[\frac{0}{0}\right]\text{form}$
Applying L'Hopitals' rule, we get
$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{2f^{\prime }\left(2h\right)}{2}$
$\Rightarrow f^{\prime }\left(x\right)=f^{\prime }\left(0\right)=-1$
So, $f\left(x\right)=-x+C$
But $f\left(0\right)=1\Rightarrow C=1$
$\therefore f\left(x\right)=-x+1$

$f\left(|x|\right)=1-|x|$ which is not differentiable at $x=0$

$f\left(x\right)=1-x$
$\therefore f^{-1}\left(x\right)=1-x$
$f\left(x\right)=f^{-1}\left(x\right)$ for all $x\in R$
$\Rightarrow f\left(x\right)=f^{-1}\left(x\right)$ will have infinitely many solutions.

$\sum _{r=0}^{10}\left(f\right(r){)}^2$
$=\left(f\right(0){)}^2+(f\left(1\right){)}^2+\left(f\right(2){)}^2+\cdots +(f\left(10\right){)}^2$
$=1+0+(-1{)}^2+(-2{)}^2+\cdots +(-9{)}^2$
$=1+\frac{9\times 10\times 19}{6}$
$=1+285=286$

Area bounded between $y=f\left(x\right)$ and $y=g\left(x\right)$
Finding intersection point :
$1-x=2x^2-2x+1\Rightarrow 2x^2-x=0\Rightarrow x=0,\frac{1}{2}$

Area $=\underset{0}{\overset{1/2}{\int }}\left(f\right(x)-g(x\left)\right)dx$
$=\underset{0}{\overset{1/2}{\int }}(-2x^2+x)dx={[-\frac{2}{3}{x}^3+\frac{x^2}{2}]}_0^{1/2}$
$=\frac{1}{24}$