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Cari semua $f: \mathbb{R} \to \mathbb{R}$ yang memenuhi $$f(x^3+y^3+xy)=x^2f(x)+y^2f(y)+f(xy)$$ untuk semua $x, y \in \mathbb{R}$.


International Zhautykov Olympiad 2015, no. 3

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*Ambil $(x,y)=(1,0)$, maka $f(0)=0$

*$(x,y)=(x,0)$ , $f(x^3)=x^2f(x)$


*$P(x,y)=(x,-x)$  , maka utk $x=/=0, f(x)+f(-x)=0$,. Di kasus pisah y, $x=0$ memenuhi jg sih...


*Liat $(x,y)=(a,b)$ sama $(x,y)=(a,-b)$ di $f(x^3+y^3+xy)=x^2f(x)+y^2f(y)+f(xy)$, terus dijumlahin:







Kliatan utk stp $p,q \in{R}$, ada $a,b\in{R}$ shg $p=a^3$ dan $q= b^3+ab$

[ Set aja $a=p^{\frac{1}{3}}$ sama nilai b shg $b^3+ab-q=0$. Liat aja grafik $Q(x)=b^3+ab-q$ udh kontinu dan lalala..]

Implies $f(p+q)+f(p-q)=2a^2f(a)=2f(p)$ yg ekuiv $f(x)+f(y)=2f(\frac{x+y}{2})$, ...............1

Nah ambil y nya $-y$, dpt $f(x)+f(-y)=2f(\frac{x-y}{2})$ <=> $f(x)-f(y)=2f(\frac{x-y}{2})$, ..............2


Skrg (1)+(2): $f(x)=f(\frac{x+y}{2})+f(\frac{x-y}{2})$, means $f(k+l)=f(k)+f(l)$ utk $k,l$ real


*$f(k+l)=f(k)+f(l)$ kaliin $(k+l)^2$, jd $(k+l)^2[f(k)+f(l)] = (k+l)^2f(k+l) = f((k+l)^3)$

Ambil $l=1$:

$(k^2+2k+1)[f(k)+f(1)] = f(k^3+3k^2+3k+1)$

Mis $f(1)=a$

$(k^2+2k+1)[f(k)+a] = f(k^3)+3f(k^2)+3f(k)+f(1)$

$(k^2+2k+1)f(k)+(k^2+2k+1)a = k^2f(k)+3f(k^2)+3f(k)+a$

$(2k-2)f(k)+(k^2+2k)a = 3f(k^2)$..........(*)


*Ganti k nya sama $k^3$:

$2(k^3-1)f(k^3)+(k^6+2k^3)a = 3 f(k^6) = 3k^4f(k^2)$

$2(k^3-1)k^2f(k)+(k^6+2k^3)a = k^4[(2k-2)f(k)+(k^2+2k)a ]$............. dari (*)

$2(k^4+k^3+k^2)(k-1)f(k)+(k^6+2k^3)a = 2k^4(k-1)f(k) + (k^6+2k^5)a ]$

$2(K^4+k^3+k^2-k^4)(k-1)f(k) = 2k^3(k+1)(k-1)a$

Utk $ k=/=1 , -1$:

$2(k+1)k^2.f(k) = 2k^3(k+1)a$



Utk f(1), jelas $f(1)=a.1$, dan $f(-1)=-f(1)=-a$ dr $f(x)+f(-x)=0.$

jd $f(x)=ax$ utk semua $x$ real, a konstan.

Cek trivial bener kok



Edited by Hisagi
  • Upvote 1

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