Question

Given that $x^{2}f\left(x\right)+f(1-x)=2x-x^4$
Prove that $f\left(x\right)=1-x^2$.

Answer 1

Clearly since RHS is a polynomial of degree 4 LHS must also be a polynomial of degree at most 4.

Putting $x=0$ we get $f\left(1\right)=0$

Putting $x=1$ we get $f\left(1\right)+f\left(0\right)=2-1$

$\Rightarrow f\left(0\right)=1$

Putting $x=-1$ we get

$f(-1)+f\left(2\right)=-2-1=-3$.....(1)

Putting $x=2$ we get

$4f\left(2\right)+f(-1)=4-16=-12$....(2)

$\left(1\right)\times 4-\left(2\right)$ gives

$3f(-1)=0\Rightarrow f(-1)=0$

Hence f(x) has 2 roots 1 and -1.

Since it is a quadratic at most these are the only roots it has

$f\left(x\right)=(x-1)(x-(-1\left)\right)$ or $(1-x)(1+x)$

Putting $f\left(x\right)=x^2-1$ in given equation for $f\left(x\right)$, we find $x^4-2x=-(x^4-2x)$, which is not possible

So, $f\left(x\right)=1-x^2$