QORI NIYOZIY OLIMPIADASINING II-BOSQICH SAVOLLARI
(1) F1 va F2 nuqtalar 4x2 + 9y 2 = 36 ellipsning fokuslari bo’lsin. Ellipsda M F1 :
M F2 = 2 : 1 nisbatni qanoatlantiradigan M nuqta olingan bo’lsin, u holda M F1 F2
uchburchak yuzini toping.
(2) Bir xil tartibli A, B, C kvadrat (haqiqiy) matritsalar berilgan bo’lsin. Agar A
matritsa teskarilanuvchi bo’lib, (A − B)C = BA−1 tenglik o’rinli bo’lsa, u holda
C(A − B) = A−1 B tenglik ham o’rinli bo’lishini isbotlang.
(3) Integralni hisoblang.
π
Z
0
cos2 ( x2 )
dx.
3π 2 + 4πx − 4x2
(4) Quyidagi differensial tenglamani umumiy yechimini toping.
00
0
xy − y + 4x3 y = 0.
(5) Qiymatlar sohasi sanoqsiz va
|f (a1 ) − f (a2 )| ≤ |a1 − a2 |3 ,
∀a1 , a2 ∈ A
shartni qanoatlantiradigan A ⊂ R to‘plam va f : A → R funksiyaga misol keltiring.
1
2
YECHIMLAR
√
(1) Ellipsning fokuslari orasidagi masofa F1 F2 = 2c = 2 5. Shuningdek, M F1 + M F2 =
2a = 6 va F1 M : F2 M = 2 : 1 ekanidan F M1 = 4 va F M2 = 2. U holda M F1 F2
to‘g‘ri burchakli uchburchak bo‘ladi va uning yuzi 4 ga teng.
(2) (A−B)C = BA−1 ⇔ AC−BC−BA−1 +AA−1 = E ⇔ A(C+A−1 )−B(C+A−1 ) =
E ⇔ (A − B)(C + A−1 ) = E. Demak, A − B teskarilanuvchi ekan. U holda
(C + A−1 )(A − B) = E ⇔ C(A − B) + A−1 A − A−1 B = E ⇔ C(A − B) = A−1 B.
(3)
Z
cos2 ( x2 )
π
1 + cos x
1 π
|
dx = −
3π dx = |x = t +
π
2
2
8 0 (x + 2 )(x − 2 )
2
0 3π + 4πx − 4x
Z π
Z π
Z π
1 2
1 2
1 2
1 − sin t
1
sin t
dt = −
+
dt.
=−
π
π
π
8 − 2 (t + π)(t − π)
8 − 2 (t + π)(t − π) 8 − 2 (t + π)(t − π)
Z π
Z π
2
2
1
1
sin t
t − π π2
ln 3
=
ln
|− π = −
,
dt = 0.
2
π (t + π)(t − π)
π
2π
t+π
π
−2
− 2 (t + π)(t − π)
Z π
cos2 ( x2 )
ln 3
dx =
.
2
2
8π
0 3π + 4πx − 4x
√ 0 00
0
00
0
(4) Yangi o‘zgaruvchi kiritib olaylik, ya’ni t = x2 . U holda yx = 2 tyt , yxx = 4tytt + 2yt .
Demak, tenglamani yangi noma’lum bo‘yicha ko‘rinishi quyidagicha bo‘ladi:
√ 0
√
3
00
0
t(4tytt + 2yt ) − 2 tyt + 4t 2 y = 0.
Z
π
Oxirgi tenglamadan
3
00
00
4t 2 (ytt + y) = 0 ⇒ ytt + y = 0.
Oxirgi differensial tenglamani umumiy yechimi y(t) = C1 sin t + C2 cos t ko‘rinishda
ekanligini ko‘rsatish qiyin emas. Demak, y(x) = C1 sin x2 + C2 cos x2 .
P
αk
(5) A ⊂ [0, 1] to‘plamni Kantor to‘plam sifatida tanlab olaylik, ya’ni ∞
k=1 3k , bu yerda
αk ∈ {0, 2}, k ∈ N. f funksiyani quyidagicha aniqlaylik
∞
∞
X
X
αk
αk
)=
.
f(
k
3k+1
3
3
k=1
k=1
(1)
P
P∞ α(2)
αk
k
va
a
=
a1 = ∞
2
k
k=1 3k . Faraz qilaylik a1 < a2 bo‘lsin, u holda shunday
k=1 3
(1)
(2)
(1)
(2)
(1)
(2)
n ∈ N son topilib α1 = α1 , ..., αn−1 = αn−1 , va αn = 0, αn = 2. U holda
∞
X
2
2
1
=
.
a2 − a1 ≥ n −
k
n
3
3
3
k=n+1
Shuningdek,
0 < f (a2 ) − f (a1 ) ≤
∞
X
k=n
2
33k+1
=
9
1
· 3n < |a2 − a1 |3 .
13 3
3
f inektiv va A sanoqsiz ekanligidan, f ning qiymatlar sohasi ham sanoqsiz bo‘ladi.