Question

Let there are $n$ number of polynomial functions $f_i:R\to R,i\in N$ satisfying the equation
$f(a+f(a+b\left)\right)+f\left(ab\right)-f(a+b)-bf\left(a\right)=a$ $\forall a,b\in R$. Then

• A. there exists at least one odd function.
• B. there exists at least one even function.
• C. $f_i\left(1\right)=1\forall i\in N$
• D. $\sum _{i=1}^n{f}_i\left(x\right)=k$, where $k$ is a constant.

Answer 1

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

The correct options are
A there exists at least one odd function.
C $f_i\left(1\right)=1\forall i\in N$
D $\sum _{i=1}^n{f}_i\left(x\right)=k$, where $k$ is a constant.

$f(a+f(a+b\left)\right)+f\left(ab\right)-f(a+b)-bf\left(a\right)=a...\left(1\right)$

Put $a=b=0$, we get
$f(0+f(0\left)\right)+f\left(0\right)-f\left(0\right)=0$
$\Rightarrow f\left(f\right(0\left)\right)=0$
Let $f\left(0\right)=k$, then $f\left(k\right)=0$

Put $a=0,b=k$, we get
$f(0+f(k\left)\right)+f\left(0\right)-f\left(k\right)-kf\left(0\right)=0$
$\because f\left(k\right)=0$ and $f\left(0\right)=k$
$\Rightarrow k+k-0-k^2=0$
$\Rightarrow k(2-k)=0$
$\Rightarrow k=f\left(0\right)=0,2$

Case $1$ : $f\left(0\right)=0$
Put $a=0$ in the eqn $\left(1\right)$, we get
$f\left(f\right(b\left)\right)+f\left(0\right)-f\left(b\right)-bf\left(0\right)=0$
$\Rightarrow f\left(f\right(b\left)\right)=f\left(b\right)$
Let $f\left(b\right)=y$, then we get $f\left(y\right)=y$
$\therefore f\left(x\right)=x$ which is an odd function.

Case $2$ : $f\left(0\right)=2$
Put $a=0$ in the eqn $\left(1\right)$, we get
$f\left(f\right(b\left)\right)+f\left(0\right)-f\left(b\right)-bf\left(0\right)=0$
$\Rightarrow f\left(f\right(b\left)\right)+2-f\left(b\right)-2b=0$
$\Rightarrow f\left(f\right(b\left)\right)=f\left(b\right)+2b-2$
Replace $b$ by $x$,
$\Rightarrow f\left(f\right(x\left)\right)=f\left(x\right)+2x-2$
From above equation, we observe that $f\left(x\right)$ must be a polynomial of degree $1$ as any other power will make the $\text{LHS}$ and $\text{RHS}$ of different powers and will not have any non-trivial solutions.
Now, put $x=0$, we get
$\Rightarrow f\left(f\right(0\left)\right)=f\left(0\right)-2$
$\Rightarrow f\left(2\right)=0(\because f(0)=2)$
Hence, only $f\left(x\right)=2-x$ satisfies the above conditions.

Hence, the functions are $f\left(x\right)=x$ and $f\left(x\right)=2-x$ that satisfying eqn $\left(1\right)$.

$f\left(1\right)=1$ for both the functions.

$\sum _{i=1}^n{f}_i\left(x\right)=x+(2-x)=2\left(\text{constant}\right)$