Question

Which of the following is/are true?

• A. If $f:R\to R$ is a function satisfying $f(x-f(y\left)\right)=f\left(f\right(y\left)\right)+xf\left(y\right)+f\left(x\right)-1\forall x,y\in R$ then $f\left(x\right)$ can be $1-\frac{x^2}{2}.$
• B. If $f\left(x\right)=\frac{ax+b}{cx-a},x\ne \frac{a}{c},a,b,c\in R-\left\{0\right\},$ then $f\left(f\right(x\left)\right)=x$
• C. If $f$ be a real valued integrable function satisfying $f\left(x\right)+f(x+4)=f(x+2)+f(x+6)$ and $g\left(x\right)=\underset{x}{\overset{x+8}{\int }}f\left(t\right)dt,\forall x\in R,$ then $g\left(x\right)$ can be a constant function.
• D. Domain of the function $\sqrt{1-f\left(x\right)}$ is $[0,\infty ),$ where $f^3\left(x\right)=1-x^3-3xf\left(x\right),x\in R$ and $f(-1)\ne -1$

Answer 1

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

The correct options are
A If $f:R\to R$ is a function satisfying $f(x-f(y\left)\right)=f\left(f\right(y\left)\right)+xf\left(y\right)+f\left(x\right)-1\forall x,y\in R$ then $f\left(x\right)$ can be $1-\frac{x^2}{2}.$
B If $f\left(x\right)=\frac{ax+b}{cx-a},x\ne \frac{a}{c},a,b,c\in R-\left\{0\right\},$ then $f\left(f\right(x\left)\right)=x$
C If $f$ be a real valued integrable function satisfying $f\left(x\right)+f(x+4)=f(x+2)+f(x+6)$ and $g\left(x\right)=\underset{x}{\overset{x+8}{\int }}f\left(t\right)dt,\forall x\in R,$ then $g\left(x\right)$ can be a constant function.
D Domain of the function $\sqrt{1-f\left(x\right)}$ is $[0,\infty ),$ where $f^3\left(x\right)=1-x^3-3xf\left(x\right),x\in R$ and $f(-1)\ne -1$

$\to$
For $f(x-f(y\left)\right)=f\left(f\right(y\left)\right)+xf\left(y\right)+f\left(x\right)-1\forall x,y\in R$
Put $x=0,f\left(y\right)=0\Rightarrow f\left(0\right)=1$
Put $x=a,f\left(y\right)=a\Rightarrow f\left(a\right)=1-\frac{a^2}{2}$

$\to$
For $f\left(x\right)=\frac{ax+b}{cx-a},f\left(f\right(x\left)\right)=x$

$\to$
For $f\left(x\right)+f(x+4)=f(x+2)+f(x+6)$

$x\to x+2$
$f\left(x\right)=f(x+8)$
$\Rightarrow f\left(x\right)$ is periodic with period $8$
$\underset{x}{\overset{x+8}{\int }}f\left(t\right)dt=\underset{0}{\overset{8}{\int }}f\left(t\right)dt=g\left(x\right)$
$\therefore g\left(x\right)$ is a constant function.

$\to$
Let $a=f\left(x\right),b=x,c=-1$
then $a^3+b^3+c^3=3abc$
$\Rightarrow (a+b+c)\times \frac{1}{2}\left[\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2]=0$
$f(-1)\ne -1\Rightarrow a+b+c=0\Rightarrow f\left(x\right)=1-x$