Mathematics - Functions Question with Solution | TestHub

Let $g(x)=f(x)-x^2$. Substituting this into the functional equation:$g(x+y)+(x+y{)}^2=g(x)+x^2+g(y)+y^2+2xy$ $g(x+y)+x^2+y^2+2xy=g(x)+x^2+g(y)+y^2+2xy$ This simplifies to $g(x+y)=g(x)+g(y)$, which is Cauchy's functional equation.

Since $f(x)$ is a polynomial (as suggested by the $2xy$ term and standard solutions to such equations), $g(x)$ must be continuous, hence $g(x)=cx$ for some constant $c$. Thus, $f(x)=x^2+cx$. Given $f(1)=3$, we have $1^2+c(1)=3⟹1+c=3⟹c=2$. Therefore, $f(x)=x^2+2x$.

A) $f(x)=x^2+2x$ is a polynomial, so it is differentiable for all $x\in \mathbb{R}$. $f^{\prime }(x)=2x+2$. This statement is correct.

B) For $f(x)=0$, we have $x^2+2x=0⟹x(x+2)=0$. The roots are $x=0$ and $x=-2$. These are two distinct real roots. This statement is correct.

C) To find the range, we can rewrite $f(x)$ by completing the square: $f(x)=(x^2+2x+1)-1=(x+1{)}^2-1$. Since $(x+1{)}^2\ge 0$, the minimum value of $f(x)$ is $-1$ (at $x=-1$). The function opens upwards, so its range is $[-1,\infty )$. This statement is correct.

D) For $f(x)$ to be an even function, $f(-x)=f(x)$. Here, $f(-x)=(-x{)}^2+2(-x)=x^2-2x$. Since $x^2-2x\ne x^2+2x$ (e.g., for $x=1$, $f(1)=3$ but $f(-1)=-1$), $f(x)$ is not an even function. This statement is incorrect.