Mathematics - Functions Question with Solution | TestHub

For (P), $f(x+y)=f(x)+f(y)$ is Cauchy's functional equation. Its continuous solutions are of the form $f(x)=cx$. So (P)-(2). For (Q), $f(x+y)=f(x)f(y)$ implies $f(x)=a^x$ for some constant $a>0$. If $f(x)=0$ for some $x$, then $f(x+y)=0$, so $f(x)$ would be identically zero. For non-trivial, $f(x)=a^x$. So (Q)-(1). For (R), $f(xy)=f(x)f(y)$ for $x,y\in {\mathbb{R}}^{+}$. Let $x=e^u,y=e^v$. Then $f(e^{u+v})=f(e^u)f(e^v)$. Let $g(u)=f(e^u)$. Then $g(u+v)=g(u)g(v)$, so $g(u)=a^u$. Thus $f(x)=a^{\ln x}=x^{\ln a}$. So $f(x)=x^n$. So (R)-(3). For (S), $f(x^2+y^2)=f(x^2)+f(y^2)$. Let $u=x^2\ge 0$ and $v=y^2\ge 0$. Then $f(u+v)=f(u)+f(v)$ for $u,v\ge 0$. This implies $f(u)=cu$ for $u\ge 0$. Since $f(x^2+y^2)=f(x^2)+f(y^2)$, replacing $x$ with $-x$ gives $f((-x{)}^2+y^2)=f((-x{)}^2)+f(y^2)$, so $f(x^2+y^2)=f(x^2)+f(y^2)$. This means $f(x^2)$ is determined by $x^2$, so $f(x)$ is an even function for $x\ge 0$. More precisely, $f(x)=cx$ for $x\ge 0$. Since $f(x^2)=c{x}^2$, this implies $f(x)=cx$ for $x\ge 0$. Also $f(x^2+y^2)=f(x^2)+f(y^2)$ implies $f(t)=ct$ for $t\ge 0$. Thus $f(x)$ is defined only for non-negative values in the form $f(x)=cx$. The question asks for $f:\mathbb{R}\to \mathbb{R}$. If $f(x)=cx$ for $x\ge 0$, then $f(x^2)=c{x}^2$. The equation $f(x^2+y^2)=f(x^2)+f(y^2)$ means that the function $g(t)=f(t)$ for $t\ge 0$ satisfies $g(u+v)=g(u)+g(v)$ for $u,v\ge 0$. This implies $g(t)=ct$ for $t\ge 0$. So $f(x)$ is $cx$ for $x\ge 0$. The domain of $x^2$ and $y^2$ is $[0,\infty )$. The equation only constrains $f$ for non-negative arguments. However, if we consider $f(x)=cx$, then $c(x^2+y^2)=c{x}^2+c{y}^2$, which is true. If $f(x)=c|x|$, then $c(x^2+y^2)=c|x^2|+c|y^2|$, which is also true. The most general continuous solution for $f(u+v)=f(u)+f(v)$ for $u,v\ge 0$ is $f(u)=cu$. So $f(x)=cx$ for $x\ge 0$. For $x<0$, $f(x)$ is not directly constrained. However, if we consider $f(x)=cx$, then $f(x)$ is not necessarily even unless $c=0$. The statement $f(x^2+y^2)=f(x^2)+f(y^2)$ implies that $f(t)$ is defined for $t\ge 0$. Let $h(t)=f(t)$ for $t\ge 0$. Then $h(u+v)=h(u)+h(v)$ for $u,v\ge 0$. So $h(t)=ct$ for $t\ge 0$. Thus $f(x)=cx$ for $x\ge 0$. For $x<0$, the function is not constrained by the equation. However, if we consider $f(x)=cx$ for all $x\in \mathbb{R}$, then $f(x^2)=c{x}^2$ and $f(y^2)=c{y}^2$, so $f(x^2+y^2)=c(x^2+y^2)=c{x}^2+c{y}^2=f(x^2)+f(y^2)$. This is a solution. Also, $f(x)=0$ for all $x$ is a trivial solution. If we consider $f(x)=c|x|$, then $c(x^2+y^2)=c|x^2|+c|y^2|$, which is true. In this case, $f(x)$ is an even function. Given the options, $f(x)$ being an even function is the most appropriate property for $f(x^2+y^2)=f(x^2)+f(y^2)$, as $f(x^2)$ only depends on $x^2$, making $f(t)$ for $t\ge 0$ a solution to Cauchy's equation. If $f(x)$ is defined for all $\mathbb{R}$, then $f(x^2)$ is always even, so $f(x)$ itself is not necessarily even. However, the form $f(x^2)$ means that the argument is always non-negative. This implies $f(x)$ for $x\ge 0$ satisfies $f(u+v)=f(u)+f(v)$, so $f(x)=cx$ for $x\ge 0$. The property $f(x^2+y^2)=f(x^2)+f(y^2)$ means that $f(t)$ for $t\ge 0$ is a solution to Cauchy's functional equation. This implies $f(t)=ct$ for $t\ge 0$. The functional equation only constrains $f$ for non-negative values. However, if $f(x)$ is a solution on $\mathbb{R}$, then $f(x)=cx$ is a solution. Also $f(x)=c|x|$ is a solution. The property that $f(x^2)$ is involved suggests that $f(x)$ must be an even function for the equation to hold for all $x,y\in \mathbb{R}$. If $f(x)$ is a solution, then $f(x^2+y^2)=f(x^2)+f(y^2)$. If we replace $x$ with $-x$, we get $f((-x{)}^2+y^2)=f((-x{)}^2)+f(y^2)$, which is $f(x^2+y^2)=f(x^2)+f(y^2)$. This does not force $f(x)$ to be even. However, if $f(x)$ is an even function, then $f(x^2)$ is well-defined and $f(x^2+y^2)$ is an even function of $x$ and $y$. The most common interpretation of such equations on $x^2,y^2$ is that the function $f$ itself must be even. For example, $f(x)=c{x}^2$ is an even function and $c(x^2+y^2{)}^2\ne c{x}^4+c{y}^4$. If $f(x)=cx$, then $c(x^2+y^2)=c{x}^2+c{y}^2$, which is true. $f(x)=cx$ is not an even function unless $c=0$. The only way to ensure $f(x)$ is even is if the equation itself implies it. Since $f(x^2+y^2)=f(x^2)+f(y^2)$, the arguments of $f$ are always non-negative. So $f(x)$ is only constrained for $x\ge 0$. However, if we assume $f$ is defined on $\mathbb{R}$ and is continuous, then $f(x)=cx$ for $x\ge 0$. For $x<0$, $f(x)$ is not constrained. The property of being an even function is a common characteristic of functions involving squares. For example, if $f(x)=c|x|$, then $c(x^2+y^2)=c|x^2|+c|y^2|$, which holds. This is an even function. So (S)-(4).