Question

If $f(a-b)+f(b-c)+f(c-a)=2f(a+b+c)$ where $a,b,c\in R$ and $ab+bc+ca=0$, then which of the following is/are possible ?

• A. $f\left(x\right)=c_1{x}^2+c_2{x}^4,c_i\ne 0$
• B. $f\left(x\right)=c_0+c_1{x}^2+c_2{x}^4,c_i\ne 0$
• C. $f\left(x\right)=c_1{x}^2+c_2{x}^4+c_3{x}^6,c_i\ne 0$
• D. $f\left(x\right)$ is always an even function

Answer 1

Answer: (A), (D)

$f\left(x\right)$ is always an even function

$f(a-b)+f(b-c)+f(c-a)=2f(a+b+c)$
Put $a=b=c=0$
$\Rightarrow f\left(0\right)=0$

Put $b=c=0$, we have $f\left(a\right)+f(-a)=2f\left(a\right)$
$\Rightarrow f\left(a\right)=f(-a)$
So, $f$ is an even function.

Let $f\left(x\right)=c_0+c_1{x}^2+c_2{x}^4+...+c_m{x}^{2m}$
We know that
$ab+bc+ca=0$
$\Rightarrow a=\frac{-bc}{b+c}$
Put $b=(1+\sqrt{3})x$ and $c=(1-\sqrt{3})x$
$\Rightarrow a=x$

$f(-\sqrt{3}x)+f\left(2\sqrt{3}x\right)+f(-\sqrt{3}x)=2f\left(3x\right)$
Therefore, comparing the coefficient of $x^{2m}$, we get
$2(-\sqrt{3}{)}^{2m}+(2\sqrt{3}{)}^{2m}=2\cdot 3^{2m}$
$\Rightarrow 2\cdot 3^m+{12}^m=2\cdot 3^{2m}$
$\Rightarrow 2+4^m=2\cdot 3^m$
This holds for $m=1$ and $m=2$,
When $m=3$
$4^3=642\cdot 3^3=54$
So, $2+4^m>2\cdot 3^m$
Therefore,
$2+4^m>4^m>2\cdot 3^m$ for $m\ge 3$
Therefore, $f\left(x\right)$ must be of the form
$c_1{x}^2+c_2{x}^4$