|
51.
Parabola uchining koordinatalarini toping:
1)
y
=
x
2
– 4
x –
5;
2)
y
= –
x
2
– 2
x
+ 3;
3)
y
=
õ
2
– 6
x
+ 10;
4)
y
=
x
2
+
x
+
4
5
;
5)
y
= –2
õ
(
x
+ 2);
6)
y
= (
x
– 2)(
x
+ 3).
52.
Funksiyaning grafigini yasang va grafik bo‘yicha uning xossa-
larini aniqlang:
1)
y
=
x
2
– 5
x +
6;
2)
y
=
x
2
+ 10
x
+ 30;
3)
y
= –
õ
2
– 6
x
– 8;
4)
y
= 2
x
2
– 5
x
+ 2;
5)
y
= –3
õ
2
– 3
x
+ 1;
6)
y
= –2
x
2
– 3
x
– 3.
25
O‘ZINGIZNI TEKSHIRIB KO‘RING!
1.
ó = x
2
– 6
õ +
5 funksiyaning grafigini yasang va uning eng
kichik qiymatini toping.
2.
y =
–
x
2
+ 2
x
+ 3 funksiya grafigi yordamida
x
ning qanday
qiymatida funksiyaning qiymati 3 ga teng bo‘lishini toping.
3.
y =
1
– x
2
funksiyaning grafigi bo‘yicha
x
ning funksiya mus-
bat; manfiy qiymatlar qabul qiladigan qiymatlarini toping.
4.
y =
2
x
2
funksiya qanday oraliqlarda o‘sadi? Kamayadi? Shu
funksiyaning grafigini yasang.
5.
ó =
(
õ
– 3)
2
parabola uchining koordinatalarini toping va
uning grafigini yasang.
53
. Funksiyaning grafigini yasamasdan, uning eng katta yoki eng
kichik qiymatini toping:
1)
y
=
x
2
+ 2
x +
3;
2)
y
= –
x
2
+ 2
x
+ 3;
3)
y
= –3
õ
2
+ 7
x
;
4)
y
= 3
x
2
+ 4
x
+ 5.
54.
To‘g‘ri to‘rtburchakning perimetri 600
m. To‘g‘ri to‘rtburchak-
ning yuzi eng katta bo‘lishi uchun uning asosi bilan balandligi
qanday bo‘lishi kerak?
55.
To‘g‘ri to‘rtburchak uning tomonlaridan biriga parallel bo‘lgan
ikkita kesma bilan uch bo‘lakka bo‘lingan. To‘g‘ri to‘rtburchak
perimetri bilan shu kesmalar uzunliklarining yig‘indisi 1600 m
ga teng. Agar to‘g‘ri to‘rtburchakning yuzi eng katta bo‘lsa,
uning tomonlarini toping.
56.
Agar
y
=
x
2
+
px + q
kvadrat funksiya:
1)
x
= 0 bo‘lganda 2 ga teng qiymatni,
x
= 1 bo‘lganda esa 3 ga
teng qiymatni qabul qilsa,
p
va
q
koeffitsiyentlarni toping;
2)
x
= 0 bo‘lganda 0 ga teng qiymatni,
x
= 2 bo‘lganda esa 6 ga
teng qiymatni qabul qilsa,
p
va
q
koeffitsiyentlarni toping.
57
. Agar
y
=
x
2
+ px + q
parabola:
1) abssissalar o‘qini
x
= 2 va
x
= 3 nuqtalarda kessa;
2) abssissalar o‘qini
x
= 1 nuqtada va ordinatalar o‘qini
y
= 3
nuqtada kessa;
3) abssissalar o‘qiga
x
= 2 nuqtada urinsa,
p
va
q
larni toping.
26
58
.
x
ning qanday qiymatlarida funksiyalar teng qiymatlar qabul
qiladi:
1)
y
=
x
2
+ 3
x
+ 2 va
y
= |7 –
x
|;
2)
y
= 3
x
2
– 6
x
+ 3 va
y =
|3
x –
3| ?
59.
Agar:
1) parabolaning (0; 0), (2; 0), (3; 3) koordinatali nuqtalardan o‘tishi;
2) (1; 3) nuqta parabolaning uchi bo‘lishi, (–1; 7) nuqtaning esa
parabolaga tegishli bo‘lishi;
3)
y
=
ax
2
+
bx
+
c
funksiyaning nollari
x
1
= 1 va
x
2
= 3 sonlari
ekani, funksiyaning eng katta qiymati esa 2 ga teng ekani
ma’lum bo‘lsa,
y
=
ax
2
+
bx
+
c
parabolani yasang.
I bobga doir sinov (test) mashqlari
Sinov mashqlarining har biriga 5 ta dan «javob» berilgan. 5 ta
«javob»ning faqat bittasi to‘g‘ri, qolganlari esa noto‘g‘ri. O‘quvchilar-
dan sinov mashqlarini bajarib yoki boshqa mulohazalar yordamida
ana shu to‘g‘ri javobni topish (uni belgilash) talab qilinadi.
1.
a
ning shunday qiymatini topingki,
y
=
ax
2
parabola bilan
y
= 5
x
+ 1 to‘g‘ri chiziqning kesishish nuqtalaridan birining
abssissasi
x
= 1 bo‘lsin.
A)
a
= 6;
B)
a
= –6;
C)
a
= 4;
D)
a
= –4;
E)
a
= 7.
2.
k
ning shunday qiymatini topingki,
y
= –
x
2
parabola bilan
y
=
kx
– 6 to‘g‘ri chiziqning kesishish nuqtalaridan birining
abssissasi
x
= 2 bo‘lsin.
A)
k
= –1;
B)
k
= 1;
C)
k
= 2;
D)
k
= –2;
E)
k
= –6.
3.
b
ning shunday qiymatini topingki,
y
= 3
x
2
parabola bilan
y
= 2
x
+
b
to‘g‘ri chiziqning kesishish nuqtalaridan birining
abssissasi
x
= 1 bo‘lsin.
A)
b
= 2;
B)
b
= –1;
C)
b
= 1;
D)
b
= –2;
E)
b
= 3.
Parabolaning koordinata o‘qlari bilan kesishish nuqtalarining
koordinatalarini toping
(4–7)
:
4.
y
=
x
2
– 2
x
+ 4.
A) (–1; 3);
B) (3; 1);
C) (1; 3);
D) (0; 4);
E) (4; 0).
27
5.
y
=
–x
2
– 4
x
– 5.
A) (–1; 2); B) (2; –1);
C) (5; 0);
D) (–5; 0); E) (0; –5).
6.
y
= 6
x
2
– 5
x
+ 1.
A) (
3
1
; 0), (
2
1
; 0), (0; 1);
B) (–
3
1
; 0), (–
2
1
; 0), (1; 0);
C) (0;
3
1
), (0;
2
1
), (0; 1);
D) (
3
1
; 0), (–
2
1
; 0), (0; –1);
E) to‘g‘ri javob berilmagan.
7.
y
= –
x
2
+ 6
x
+ 7.
A) (–1; 0), (–7; 0), (0; –7);
B) (–1; 0), (7; 0), (0; 7);
C) (1; 0), (7; 0), (0; –7);
D) (–1; 2), (7; –1), (7; 0); E) (3; 16).
Parabola uchining koordinatalarini toping
(8–11)
:
8.
y
=
x
2
– 4
x
.
A) (0; 4);
B) (4; 2);
C) (2; –4);
D) (–4; 2);
E) (0; –4).
9.
y
=
–x
2
+ 2
x
.
A) (–1; –1);
B) (1; –2);
C) (0; 2);
D) (1; 1);
E) (1; –1).
10.
y
=
x
2
+ 6
x
+ 5.
A) (3; –4); B) (–5; –1);
C) (–1; –5); D) (3; 4); E) (–3; –4).
11.
y
= –5
x
2
+ 4
x
+ 1.
A) (
5
2
;
5
9
); B) (–
5
2
;
5
9
);
C) (–
5
9
;
5
2
); D) (2; 9); E) (9; 5).
12.
Abssissalar o‘qini
x
= 1 va
x
= 2 nuqtalarda, ordinatalar o‘qini esa
2
1
=
y
nuqtada kesib o‘tuvchi parabolaning tenglamasini yozing.
A)
2
1
4
3
2
1
2
+
=
-
x
x
y
;
B)
2
1
4
3
4
1
2
+
=
-
x
x
y
;
C)
y
=
x
2
– 3
x
+ 2;
D)
2
1
2
3
2
+
-
=
x
x
y
;
E) to‘g‘ri javob berilmagan.
13.
Abssissalar o‘qini
x
= –1 va
x
= 3 nuqtalarda, ordinatalar o‘qini
esa
y
= 1 nuqtada kesib o‘tuvchi parabolaning tenglamasini yozing.
A)
y
= –
x
2
+ 2
x
+ 3;
B)
1
2
3
2
+
+
-
=
x
y
x
;
28
C)
1
3
2
3
2
+
+
-
=
x
y
x
;
D)
1
3
2
3
2
-
-
=
x
y
x
;
E) to‘g‘ri javob berilmagan.
Parabola qaysi choraklarda joylashgan?
(14–18)
:
14.
y
= 3
x
2
+ 5
x
– 2.
A) I, II, III;
B) II, III, IV;
C) I, III, IV;
D) I, II, III, IV;
E) I, II, IV.
15.
y
=
x
2
– 4
x
+ 6.
A) I, IV;
B) II, III;
C) I, II, III, IV; D) II, III, IV; E) I, II.
16.
y
= –
x
2
– 6
x
– 11.
A) III, IV;
B) I, II, III;
C) II, III, IV;
D) I, III, IV;
E) I, II.
17.
y
= –
x
2
+ 5
x
.
A) I, II, III;
B) I, III, IV;
C) I, II, III, IV;
D) II, III, IV;
E) to‘g‘ri javob berilmagan.
18.
y
=
x
2
– 4
x
.
A) I, II, III;
B) II, III, IV;
C) I, II, IV;
D) III, IV;
E) I, II.
19.
Ikki musbat sonning yig‘indisi 160 ga teng. Agar shu sonlar
kublarining yig‘indisi eng kichik bo‘lsa, shu sonlarni toping.
A) 95; 65;
B) 155; 5;
C) 75; 85;
D) 80; 80; E) 90; 70.
20.
Ikki musbat sonning yig‘indisi
a
ga teng. Agar shu sonlar
kvadratlarining yig‘indisi eng kichik bo‘lsa, shu sonlarni toping.
A)
5
3
5
2
,
a
a
; B)
a
a
a
-
3
3
,
; C)
4
4
3
,
a
a
; D)
a
2
;
a
–
a
2
; E)
2
2
,
a
a
.
29
KVADRAT TENGSIZLIK VA
UNING YECHIMI
1- m a s a l a .
To‘g‘ri to‘rtburchakning tomonlari 2 dm va 3 dm ga
teng. Uning har bir tomoni bir xil sondagi detsimetrlarga shunday
orttirildiki, natijada to‘g‘ri to‘rtburchakning yuzi 12 dm
2
dan ortiq
bo‘ldi. Har bir tomon qanday o‘zgargan?
To‘g‘ri to‘rtburchakning har bir tomoni
x
detsimetrga orttirilgan
bo‘lsin. U holda yangi to‘g‘ri to‘rtburchakning tomonlari (2 +
x
) va
(3 +
x
) detsimetrga, uning yuzi esa (2 +
x
)(3 +
x
) kvadrat detsimetrga teng
bo‘ladi. Masala shartiga ko‘ra (2 +
x
)(3 +
x
) > 12, bundan
x
2
+ 5
x
+ 6 > 12
yoki
x
2
+ 5
x
– 6 > 0.
Bu tengsizlikning chap qismini ko‘paytuvchilarga ajratamiz:
(
x
+ 6)(
x
– 1) > 0.
Masala shartiga ko‘ra,
x
> 0 bo‘lgani uchun
x
+ 6 > 0.
Tengsizlikning ikkala qismini
x
+ 6 musbat songa bo‘lib,
x
– 1 > 0,
ya’ni
x
> 1 ni hosil qilamiz.
J a v o b :
To‘g‘ri to‘rtburchakning har bir tomoni 1 dm dan ko‘p-
roqqa orttirilgan.
x
2
+ 5
x
– 6 > 0 tengsizlikda
x
bilan noma’lum son belgilangan.
Bu – kvadrat tengsizlikka misol.
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