Mathematics - Functions Question with Solution | TestHub

Let $f(0)=k$. Setting $y=0$ in the given equation, we get $f(x^2)-f(0)=x(f(x)-f(0))$, so $f(x^2)=xf(x)-xk+k$. Let $g(x)=f(x)-k$. Then $f(x)=g(x)+k$. Substituting $f(x)=g(x)+k$ into the original equation: $(g(x^2)+k)-(g(y^2)+k)=(x+y)((g(x)+k)-(g(y)+k))$. This simplifies to $g(x^2)-g(y^2)=(x+y)(g(x)-g(y))$. Using $f(x^2)=xf(x)-xk+k$, we get $g(x^2)+k=x(g(x)+k)-xk+k$, which means $g(x^2)=xg(x)$.

Substitute $g(x^2)=xg(x)$ and $g(y^2)=yg(y)$ into the simplified equation: $xg(x)-yg(y)=(x+y)(g(x)-g(y))$. Expanding the right side gives $xg(x)-yg(y)=xg(x)-xg(y)+yg(x)-yg(y)$. This simplifies to $0=-xg(y)+yg(x)$, or $xg(y)=yg(x)$.

For $x,y\ne 0$, we have $\frac{g(x)}{x}=\frac{g(y)}{y}$. This implies $\frac{g(x)}{x}=A$ (a constant) for $x\ne 0$. Since $g(0)=f(0)-k=k-k=0$, we have $g(x)=Ax$ for all $x\in \mathbb{R}$.

Therefore, $f(x)=Ax+k$. This is a polynomial (A is correct) and a continuous function (D is correct). If $f(0)=0$, then $k=0$, so $f(x)=Ax$. This is an odd function ($f(-x)=-Ax=-f(x)$), so B is correct. It is an even function only if $A=0$, i.e., $f(x)=0$, which is not generally true for all $A$. So C is incorrect.