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Berikut ini adalah pertanyaan dari digaaurya123 pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Pertama

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

$\large\text{$\begin{aligned}&f(x)=x\,,\ \ \forall\,x\in\mathbb{R}^{+}\\&\ \sf\ or\\&f(x) =\frac{1}{x}\,,\ \ \forall\,x\in\mathbb{R}^{+}\end{aligned}$}$
 

Problem

Find all functions $f: (0, \infty) \to (0, \infty)$ (so, $f$ is a function from the positive real numbers to the positive real numbers) such that
$\displaystyle\frac {\left( f(w) \right)^2 + \left( f(x) \right)^2}{f(y^2) + f(z^2) } = \frac {w^2 + x^2}{y^2 + z^2}$
for all positive real numbers $w,x,y,z$ satisfying $wx = yz$ .

Solution

First, by letting $(w,x,y,z) = (t,t,t,t)$ which satisfies $wx=yz=t^2$ , we get:

$\begin{aligned}\frac{\left(f(t)\right)^2+\left(f(t)\right)^2}{f\left(t^2\right)+f\left(t^2\right)}&=\frac{t^2+t^2}{t^2+t^2}\\\Rightarrow \frac{\cancel{2}\left(f(t)\right)^2}{\cancel{2}f\left(t^2\right)}&=1\\\therefore\ \left(f(t)\right)^2&=f\left(t^2\right)\end{aligned}$

It can also be easily concluded that $f(1) = 1$ .

Next, let $(w,x,y,z) = \left(t,1,\sqrt{t},\sqrt{t}\right)$ so that $wx=yz=t$ .

$\begin{aligned}&\frac{\left(f(t)\right)^2+\left(f(1)\right)^2}{f\left(\left(\sqrt{t}\right)^2\right)+f\left(\left(\sqrt{t}\right)^2\right)}=\frac{t^2+1^2}{\left(\sqrt{t}\right)^2+\left(\sqrt{t}\right)^2}\\&{\Rightarrow\ }\frac{\left(f(t)\right)^2+1}{\cancel{2}f(t)}=\frac{t^2+1}{\cancel{2}t}\\&{\Rightarrow\ }t\left(f(t)\right)^2+t=\left(t^2+1\right)f(t)\\&{\Rightarrow\ }t\left(f(t)\right)^2-\left(t^2+1\right)f(t)+t=0\end{aligned}$

We get a quadratic equation. Its roots are $f_1(t)$ and $f_2(t)$ which satisfy:

$\begin{aligned}\bullet\ &f_1(t)\cdot f_2(t)=\frac{t}{t}=1\\\bullet\ &f_1(t)+f_2(t)=\frac{t^2+1}{t}=t+\frac{1}{t}\end{aligned}$

Hence,

$\begin{aligned}&f_1(t)=t\,,\ f_2(t)=\frac{1}{t}\\&\therefore\ f(t)\in\left\{t\,,\ \frac{1}{t}\right\}\end{aligned}$

We still need to prove that $f(t)$ holds $\forall\,t\in\mathbb{R}^{+}$ .

Prove

Let $f(x)=x\ \ \forall\,x\in\mathbb{R}^{+}$ .
It’s obvious that $f(x)=x$ is a solution based on the original equation.

Let $f(x) = 1/x\ \ \forall\,x\in\mathbb{R}^{+}$ .

$\begin{aligned}\frac{\left(f(w)\right)^2+\left(f(x)\right)^2}{f\left(y^2\right)+f\left(z^2\right)}&=\frac{\dfrac{1}{w^2}+\dfrac{1}{x^2}}{\dfrac{1}{y^2}+\dfrac{1}{z^2}}\\&=\frac{\ \dfrac{w^2+x^2}{w^2x^2}\ }{\dfrac{y^2+z^2}{y^2z^2}}\\&=\frac{w^2+x^2}{y^2+z^2}\cdot\frac{y^2z^2}{w^2x^2}\\&=\frac{w^2+x^2}{y^2+z^2}\cdot\frac{\cancel{(yz)^2}}{\cancel{(wx)^2}}\\\frac{\left(f(w)\right)^2+\left(f(x)\right)^2}{f\left(y^2\right)+f\left(z^2\right)}&=\frac{w^2+x^2}{y^2+z^2}\end{aligned}$

Therefore, $f(x) = 1/x\ \ \forall\,x\in\mathbb{R}^{+}$ is also a solution.

CONCLUSION

We have 2 solutions:
$\boxed{\:\large\text{$\begin{aligned}&f(x)=x\,,\ \ \forall\,x\in\mathbb{R}^{+}\\&\ \sf\ or\\&f(x) =\frac{1}{x}\,,\ \ \forall\,x\in\mathbb{R}^{+}\end{aligned}$}\:}$