Mathematics - Functions Question with Solution | TestHub

Setting $x=y$ in the given equation, we get $f(0)=(x-x)(f(x)+f(x))=0$.

Setting $y=0$, we get $f(x^2)=xf(x)$.

Replacing $x$ with $-x$,

we have $f((-x{)}^2)=-xf(-x)$, so $f(x^2)=-xf(-x)$.

Thus, $xf(x)=-xf(-x)$ for all $x$. For $x\ne 0$,

this implies $f(x)=-f(-x)$. Since $f(0)=0$, $f$ is an odd function (Option A is correct).

Now, replace $y$ with $-y$ in the original equation: $f(x^2-(-y{)}^2)=(x-(-y))(f(x)+f(-y))$. Since $f$ is odd, $f(-y)=-f(y)$, so $f(x^2-y^2)=(x+y)(f(x)-f(y))$.

Equating this with the original equation, $(x-y)(f(x)+f(y))=(x+y)(f(x)-f(y))$.

Expanding both sides: $xf(x)+xf(y)-yf(x)-yf(y)=xf(x)-xf(y)+yf(x)-yf(y)$.

This simplifies to $2xf(y)=2yf(x)$. For $x\ne 0$ and $y\ne 0$,

we have $\frac{f(x)}{x}=\frac{f(y)}{y}$.

This means $\frac{f(x)}{x}$ is a constant, say $c$, for all $x\ne 0$. So $f(x)=cx$ for $x\ne 0$. Since $f(0)=0$, $f(x)=cx$ for all $x\in \mathbb{R}$.

Hence, $f(x)$ is a linear function (Option B is correct).

If $f(1)=1$, then $c(1)=1⟹c=1$, so $f(x)=x$ is the unique function satisfying this condition (Option C is correct). A linear function $f(x)=cx$ is continuous and differentiable everywhere. Therefore, option D is incorrect.