1. Muammo: Parabolani $b(-3;4)$ vektor bo'yicha parallel ko'chirish natijasida hosil bo'lgan parabolaning tenglamasini toping. Formulasi: Agar $y=f(x)$ parabola $b(h;k)$ vektor bo'yicha ko'chirilsa, yangi parabola tenglamasi $y=f(x-h)+k$ bo'ladi. Berilgan: $y=3x^2+2x-7$, $b(-3;4)$ Yangi tenglama: $$y=3(x+3)^2+2(x+3)-7+4=3(x^2+6x+9)+2x+6-3=3x^2+18x+27+2x+6-3=3x^2+20x+30$$ Javob: D) $y=3x^2+20x+30$ 2. Muammo: $y=-3x^2+4x-5$ funksiyaning grafigi qaysi choraklardan o'tadi? Parabola ochilishi pastga, $a=-3<0$. Y kesishish nuqtasi: $y(-0)= -5$ (III yoki IV chorak) Nol nuqtalar: $-3x^2+4x-5=0$ diskriminant $D=16-4ullet(-3)ullet(-5)=16-60=-44<0$ nol nuqtalar yo'q. Demak, parabola OX o'qini kesmaydi, faqat II va IV choraklardan o'tadi. Javob: D) II, III, IV 3. Muammo: $f(x)=x^2-2x+m^2+5m+3$ parabola OX o'qini $-1$ nuqtada kesadi. OY o'qini qaysi $y$ qiymatda kesadi? OX kesishish: $f(-1)=0$ Hisoblaymiz: $(-1)^2 - 2(-1) + m^2 + 5m + 3=0 o 1+2 + m^2 + 5m + 3=0 o m^2 + 5m + 6=0$ $D=25-24=1$, ildizlar $m=-2$ yoki $m=-3$ OY kesishish: $f(0)=m^2 + 5m + 3$ $m=-2 o 4 -10 +3 = -3$ $m=-3 o 9 -15 +3 = -3$ Javob: B) -3 4. Muammo: $x=-1$ to'g'ri chiziq parabola simmetriya o'qi bo'lishi uchun $a$ ni toping. Parabola: $y=(a+1)x^2 - 2(a+3)x + a + 1$ Simmetriya o'qi: $x = -rac{b}{2a}$ Bu $x=-1$ bo'lishi kerak. $a = a+1$, $b = -2(a+3)$ $$-1 = -\frac{-2(a+3)}{2(a+1)} = -\frac{-2a -6}{2a + 2} = -\frac{-2a -6}{2a + 2}$$ $$-1 = -\frac{-2a -6}{2a + 2} o -1 = \frac{2a + 6}{2a + 2} o -(2a + 2) = 2a + 6 o -2a - 2 = 2a + 6$$ $$-4a = 8 o a = -2$$ Javob: D) -2 5. Muammo: $y=2x^2 - 5x + 3$ parabolani $b(5; -4)$ vektor bo'yicha ko'chirish. Yangi tenglama: $y=2(x-5)^2 - 5(x-5) + 3 - 4$ Hisoblaymiz: $$2(x^2 - 10x + 25) - 5x + 25 + 3 - 4 = 2x^2 - 20x + 50 - 5x + 25 - 1 = 2x^2 - 25x + 74$$ Javob: B) $y=2x^2 - 25x + 74$ 6. Muammo: $y=(a+1)x^2 - 2(a+3)x + a + 1$ parabola $x$ o'qiga urinsa, $a$ ni toping. Parabola $x$ o'qiga urishi uchun $y=0$ da bitta ildiz bo'lishi kerak, ya'ni diskriminant $D=0$. $a = a+1$, $b = -2(a+3)$, $c = a+1$ $$D = b^2 - 4ac = [-2(a+3)]^2 - 4(a+1)(a+1) = 4(a+3)^2 - 4(a+1)^2 = 4[(a+3)^2 - (a+1)^2]$$ $$=4[(a^2 + 6a + 9) - (a^2 + 2a + 1)] = 4(4a + 8) = 16(a + 2)$$ $D=0 o 16(a+2)=0 o a=-2$ Javob: D) -2 7. Muammo: $y = kx + 2k$ funksiya $(2;12)$ nuqtadan o'tadi. Substitutsiya: $$12 = k imes 2 + 2k = 2k + 2k = 4k o k = 3$$ Javob: A) 3; -2 (tekshirish uchun $k=-2$ ni ham tekshiramiz: $y=-2x -4$, $y(2)=-4-4=-8 eq 12$ faqat $k=3$ to'g'ri, shuning uchun faqat 3) 8. Muammo: $y=4x^2 - 2kx + 3$ va $y=kx + 2$ kesishmaydi, nechta butun $k$ uchun? Kesishmaslik uchun tenglama: $$4x^2 - 2kx + 3 = kx + 2 o 4x^2 - 3kx + 1 = 0$$ Diskriminant: $$D = (-3k)^2 - 4 imes 4 imes 1 = 9k^2 - 16$$ Kesishmaslik uchun $D < 0 o 9k^2 < 16 o |k| < \frac{4}{3}$$ Butun $k$ qiymatlari: $k = -1, 0, 1$ Javob: B) 3 9. Muammo: Parabola $x=3$ nuqtada OX o'qiga uradi, OY o'qini $y=4.5$ nuqtada kesadi. Parabola tenglamasi: $y = a(x - h)^2 + k$ OX o'qiga urishi: $k=0$ va $h=3$ OY kesishish: $y(0) = a(0-3)^2 + 0 = 9a = 4.5 o a=0.5$ Javob: B) $y=0.5(x-3)^2$ 10. Muammo: $y=6 + 5x - x^2$ parabolaning OX o'qiga nisbatan simmetrigini toping. Simmetriya o'qi: $x = -\frac{b}{2a} = -\frac{5}{2 imes (-1)} = \frac{5}{2} = 2.5$ Javob: $x=2.5$ 11. Muammo: $y = x^2 - 14x + 34$ parabola uchining koordinatalari ko'paytmasi. Uchi koordinatalari: $$x_0 = -\frac{b}{2a} = \frac{14}{2} = 7$$ $$y_0 = f(7) = 49 - 98 + 34 = -15$$ Ko'paytma: $7 imes (-15) = -105$ Javob: B) -105 12. Muammo: $y = -x^2 + 6x - 2$ funksiyaning qiymatlar sohasini toping. Parabola ochilishi pastga ($a=-1<0$), uchi: $$x_0 = -\frac{b}{2a} = -\frac{6}{-2} = 3$$ $$y_0 = f(3) = -9 + 18 - 2 = 7$$ Qiymatlar sohas: $(-1 ext{infty}; 7]$ Javob: C) $(-1 ext{infty}; 7]$ 13. Muammo: $y = ax^2 + bx + c$; $a>0$, $b>0$, $c<0$ grafigi qaysi choraklardan o'tadi? $a>0$ ochilishi yuqoriga, $c<0$ OY kesishish manfiy, $b>0$ grafigi I va IV choraklarga ta'sir qiladi. Javob: D) I, II va IV 14. Muammo: $y = x^2 - 4x + n + 2$ funksiyaning eng kichik qiymati 3 ga teng bo'lishi uchun $n$ ni toping. Eng kichik qiymat: $$x_0 = \frac{4}{2} = 2$$ $$y_{min} = f(2) = 4 - 8 + n + 2 = n - 2$$ $$n - 2 = 3 o n = 5$$ Javob: D) 5 15. Muammo: $y = -x^2 + 10x + 16$ funksiyaning qiymatlar sohasini toping. Uchi: $$x_0 = -\frac{b}{2a} = -\frac{10}{-2} = 5$$ $$y_0 = f(5) = -25 + 50 + 16 = 41$$ Qiymatlar sohas: $(-1 ext{infty}; 41]$ Javob: A) $(-1 ext{infty}; 41]$ 16. Muammo: To'g'ri fikrni toping. A) noto'g'ri, OX o'qini har doim 2 nuqtada kesmaydi. B) to'g'ri, to'g'ri chiziq parabola bilan bitta nuqtada kesishi mumkin (teginish). C) to'g'ri, ikki parabola eng ko'pi bilan 2 nuqtada kesishadi. D) noto'g'ri, agar shox tepaga qaragan bo'lsa, bosh koeffitsiyent musbat bo'ladi. Javob: C) Ikki parabola eng ko'pi bilan 2 nuqtada kesishadi. 17. Muammo: $y = x^2 - (m-5)x + 4$ parabola OX o'qining manfiy tarafiga urishi uchun $m$ ni toping. Diskriminant: $$D = (m-5)^2 - 16 > 0$$ Ildizlar: $$x_{1,2} = \frac{m-5 \\pm \sqrt{D}}{2}$$ Ildizlar manfiy bo'lishi kerak, shuning uchun $m=9$ to'g'ri javob. Javob: B) 9 18. Muammo: $f(x) = x^2 - ax + 2 > -7$ barcha $x$ uchun, $a$ ning eng kichik butun qiymati. $$f(x) > -7 o x^2 - ax + 9 > 0$$ Minimum qiymat: $$x_0 = \frac{a}{2}$$ $$f(x_0) = \left(\frac{a}{2}\right)^2 - a \times \frac{a}{2} + 9 = -\frac{a^2}{4} + 9 > 0$$ $$-\frac{a^2}{4} + 9 > 0 o a^2 < 36 o |a| < 6$$ Eng kichik butun $a$: $-5$ Javob: A) -5 19. Muammo: $y = x^2 + 5x - 7$ grafigi qaysi choraklardan o'tadi? Diskriminant: $$D = 25 + 28 = 53 > 0$$ Ildizlar mavjud, $a>0$ ochilishi yuqoriga. OY kesishish: $y(0) = -7 < 0$ Demak, grafigi II, III va IV choraklardan o'tadi. Javob: B) II, III va IV 20. Muammo: $f'(x) = 9x^2 - 4x + 3$, $f(1) = 7$, $f(2)$ ni toping. Integrallash: $$f(x) = \int (9x^2 - 4x + 3) dx = 3x^3 - 2x^2 + 3x + C$$ $$f(1) = 3 - 2 + 3 + C = 4 + C = 7 o C = 3$$ $$f(2) = 24 - 8 + 6 + 3 = 25$$ Javob: A) 25 21. Muammo: $f'(x) = 3x^2 - 4x + 5$, $f(2) = 15$, $f(3)$ ni toping. Integrallash: $$f(x) = x^3 - 2x^2 + 5x + C$$ $$f(2) = 8 - 8 + 10 + C = 10 + C = 15 o C = 5$$ $$f(3) = 27 - 18 + 15 + 5 = 29$$ Javob: C) 29 22. Muammo: $y = 2x^2 + mx + 1$ parabola uchi $y = -x$ to'g'ri chizig'ida yotsa, $m$ qiymatlari yig'indisi. Uchi: $$x_0 = -\frac{m}{4}, y_0 = 2x_0^2 + m x_0 + 1$$ $y_0 = -x_0$ bo'lishi kerak: $$2x_0^2 + m x_0 + 1 = -x_0 o 2x_0^2 + (m + 1) x_0 + 1 = 0$$ Substitutsiya $x_0 = -\frac{m}{4}$: $$2 \left(-\frac{m}{4}\right)^2 + (m+1) \left(-\frac{m}{4}\right) + 1 = 0$$ $$2 \frac{m^2}{16} - \frac{m(m+1)}{4} + 1 = 0 o \frac{m^2}{8} - \frac{m^2 + m}{4} + 1 = 0$$ $$\frac{m^2}{8} - \frac{m^2}{4} - \frac{m}{4} + 1 = 0 o -\frac{m^2}{8} - \frac{m}{4} + 1 = 0$$ $$-m^2 - 2m + 8 = 0 o m^2 + 2m - 8 = 0$$ $$D = 4 + 32 = 36$$ $$m = \frac{-2 \pm 6}{2} o m = 2 \text{ yoki } -4$$ Yig'indisi: $2 + (-4) = -2$ Javob: B) -2 23. Muammo: $y = ax^2 + bx + c$ parabola OX o'qini $(5;0)$ va $(1;0)$ nuqtalarda kesadi, OY o'qini $-3$ nuqtada kesadi. Uchining ordinatasini toping. $y = a(x-5)(x-1) = a(x^2 - 6x + 5)$ OY kesishish: $y(0) = 5a = -3 o a = -\frac{3}{5}$ Uch koordinatalari: $$x_0 = \frac{5+1}{2} = 3$$ $$y_0 = a(3-5)(3-1) = -\frac{3}{5}(-2)(2) = \frac{12}{5} = 2.4$$ Javob: C) 2.4 24. Muammo: $y=4x^2 - 5x + 1$ funksiya haqida to'g'ri fikrlarni toping. 1) OY o'qini kesishish: $y(0) = 1 eq 0$, noto'g'ri. 2) Uchi: $$x_0 = \frac{5}{8} > 0$$ $$y_0 = 4 \left(\frac{5}{8}\right)^2 - 5 \times \frac{5}{8} + 1 = 4 \times \frac{25}{64} - \frac{25}{8} + 1 = \frac{100}{64} - \frac{200}{64} + \frac{64}{64} = -\frac{36}{64} < 0$$ Uchi III chorakda emas, noto'g'ri. 3) Funksiya kamayish oralig'i $(-\infty, \frac{5}{8})$, noto'g'ri. 4) Aniqlanish sohasi $(-\infty, \infty)$, to'g'ri. 5) OX o'qini kesishish: $$4x^2 - 5x + 1 = 0$$ $$D = 25 - 16 = 9 > 0$$ Ildizlar musbat emas, noto'g'ri. To'g'ri javob: B) 1; 4; 5 25. Muammo: Grafikdagi parabola funksiyasini toping. Vertex $(3.5, ext{max})$, parabola pastga ochilgan. Variantlar orasida to'g'ri: B) $y = -\frac{1}{2}(x - 3.5)^2 + 5$ 26. Muammo: $y = -3(x - 2)^2 + 5$ funksiyaning qiymatlar sohasini toping. Parabola ochilishi pastga, maksimum $y=5$. Qiymatlar sohas: $(-\infty, 5]$ Javob: B) $(-\infty; 5]$ 27. Muammo: $y = x^2 - 1$ va $y = -x + 5$ kesishishidan hosil bo'lgan soha yuzini toping. Kesishish nuqtalari: $$x^2 - 1 = -x + 5 o x^2 + x - 6 = 0$$ $$D=1 + 24=25$$ $$x = \frac{-1 \pm 5}{2} o x=2, x=-3$$ Yuzani hisoblash: $$S = \int_{-3}^2 [(-x + 5) - (x^2 - 1)] dx = \int_{-3}^2 (-x + 5 - x^2 + 1) dx = \int_{-3}^2 (-x^2 - x + 6) dx$$ $$= \left[-\frac{x^3}{3} - \frac{x^2}{2} + 6x \right]_{-3}^2 = \left(-\frac{8}{3} - 2 + 12\right) - \left(-\frac{-27}{3} - \frac{9}{2} - 18\right)$$ $$= \left(\frac{22}{3}\right) - \left(9 - 4.5 - 18\right) = \frac{22}{3} - (-13.5) = \frac{22}{3} + 13.5 = \frac{22}{3} + \frac{27}{2} = \frac{44}{6} + \frac{81}{6} = \frac{125}{6}$$ Javob: B) 125/6 28. Muammo: $a<0$, $c<0$ bo'lsa, $y = ax^2 + bx + c$ grafigi qaysi choraklardan o'tadi? $a<0$ ochilishi pastga, $c<0$ OY kesishish manfiy. Demak, grafigi III va IV choraklardan o'tadi. Javob: A) III, IV