O‛zb е kiston r е spublikasi oliy va o‛rta maxsus ta’lim vazirligi

1-variant. |y|=Y bitta noldan farqli qiymatga ega va bunda X≠0 . Bu holda Y ga nisbatan kvadrat tenglama bitta ildizga ega bo’lishi kerak. Demak,


.
Demak, a=−1/2 bo’lganda berilgan sistema 4 ta (1.5, ±0.1), (0.5, ±0.1) yechimga ega bo’ladi.
2-variant. |y|=Y noldan farqli qiymatga ega bo’lib, unda X=0 va |y|=Y=0 bo’lib, unda X≠0. Buning uchun Y ga nisbatan kvadrat tenglamaning ozod hadi a=0 bo’lishi kerak. Bu holda 50Y²−10Y=0 => Y=1/5 va Y=0. Bunda Y=1/5 yoki y=±1/5 bo’lganda X=1−5Y=0 => x=1 va bu holda (1, ±1/5) sistemaning ikkita yechimi bo’ladi. Agar Y=0=>y=0 bo’lsa, unda X=1 => x−1=±1 => x=2 yoki x=0 va bu holda sistemaning yana ikkita (2,0), (0, 0) yechimini olamiz.
Demak, a=0 bo’lganda ham sistema 4 ta (1, ±1/5), (0,0), (−2, 0) yechimga ega bo’ladi.
Javob: {−1/2, 0} , ammo testda bunday javob yo’q. A) javobda keltirilgan a=−1 holda sistema umuman yechimga ega bo’lmaydi.
25) Soddalashtiring: .
A) 2 B) 1 C) 1/2 D) 1/4
Yechim: Berilgan kasrning surat va maxraji logarifmlarini topamiz:

.
Javob: B)
26) ABC to’g’ri burchakli uchburcak katetlari AC=15, BC=20. AB gipotenuzada AD=4 kesma ajratigan. CD kesma uzunligini toping.
A) 13 B) 12 C) 10 D) 11

Yechim: ABC uchburchakdan, Pifagor teoremasiga asosan, AB=25 ekanligini topamiz.. Bu holda (chizmaga qarang)
cosA=AC/AB=15/25=3/5.


ACD: CD²=AC²+AD²−2AB∙AD∙cosA=

=225+16−2∙15∙4∙(3/5)=169=> CD=13.

Javob: A)
27) Teng yonli trapetsiya diagonali uning o’tmas burchagini teng ikkiga bo’ladi. Trapetsiyaning kichik asosi 3, perimetri 42. Trapetsiya yuzini toping.
A) 86 B) 96 C) 90 D) 92
Yechim: Masala shartiga ko’ra BC=3 va ABD= DBC=α deb olsak, unda
BDA= DBC=α =>
=> ABD-teng yonli va AB=AD.
Bu holda, masala shartiga asosan,
p=2AB+AD+BC=3AB+3=42=>
=>AB=13=AD.
Bu holda
AH=РD=(AD−HP)/2=
=(AD−BC)/2 (13−3)/2=5,
ABH: BH²=AB²−AH²=
=169−25=144 => BH=12.
Trapetsiya yuzasi formulasiga asosan S=(AD+BC)BH/2=(13+3)∙12/2=96.
Javob: B)

28) ABC uchburchakda AB=BC, BF va AE balandliklar bo’lib, AE:BF=1:2 . Asosdagi burchakning kosinusini toping.

A) 0 B) 1/2 C) 1/4 D) −1/4