1. Masala: $f(x) = \left( 2x + \frac{1}{2} \right)(3x-5)$ berilgan. $\int \left( \frac{5}{6} \right)^6$ ni toping. Bu yerda $\int \left( \frac{5}{6} \right)^6$ ifodasi integral emas, balki $\left( \frac{5}{6} \right)^6$ ni hisoblash kerak. Hisoblaymiz: $$\left( \frac{5}{6} \right)^6 = \frac{5^6}{6^6} = \frac{15625}{46656} \approx 0.3349$$ Bu variantlar orasida yo'q, shuning uchun ehtimol savolda xato bor yoki javoblar boshqa masalaga tegishli. 2. $f(x) = \frac{x+2}{x-3}$ va $f(g(x)) = 4x - 5$ berilgan. $g(f(5.5))$ ni toping. 1) $f(g(x)) = 4x - 5$ ni $g(x)$ ga nisbatan yechamiz: $$f(g(x)) = \frac{g(x)+2}{g(x)-3} = 4x - 5$$ 2) Tenglamani $g(x)$ ga yechamiz: $$\frac{g(x)+2}{g(x)-3} = 4x - 5 \Rightarrow g(x)+2 = (4x - 5)(g(x)-3)$$ $$g(x)+2 = (4x - 5)g(x) - 3(4x - 5)$$ $$g(x) - (4x - 5)g(x) = -3(4x - 5) - 2$$ $$g(x)(1 - (4x - 5)) = -12x + 15 - 2$$ $$g(x)(6 - 4x) = 13 - 12x$$ $$g(x) = \frac{13 - 12x}{6 - 4x}$$ 3) $f(5.5) = \frac{5.5 + 2}{5.5 - 3} = \frac{7.5}{2.5} = 3$. 4) $g(f(5.5)) = g(3) = \frac{13 - 12 \cdot 3}{6 - 4 \cdot 3} = \frac{13 - 36}{6 - 12} = \frac{-23}{-6} = \frac{23}{6}$. Javob: A) $\frac{23}{6}$ 3. $y = -\frac{4}{x+3}$ grafigi qaysi choraklardan o'tadi? Asosiy nuqta: $x \neq -3$ (vertikal asimptota). $y$ manfiy bo'lishi uchun $x+3$ musbat bo'lsa, $y$ manfiy bo'ladi. - $x > -3$ da $x+3 > 0$, $y < 0$ (IV chorak) - $x < -3$ da $x+3 < 0$, $y > 0$ (II chorak) Demak, grafik II va IV choraklardan o'tadi. Javob: A) II va IV 4. $f(2x+3) = x^2 + 1$ bo'lsa, $f(3)$ ni toping. $2x + 3 = 3 \Rightarrow 2x = 0 \Rightarrow x = 0$ Shunday qilib, $$f(3) = f(2 \cdot 0 + 3) = 0^2 + 1 = 1$$ Javob: C) 1 5. $f(x+2) = 3x - a$ va $f(1) = 11$ bo'lsa, $a$ ni toping. $f(1) = 11$ ni $x+2=1$ dan topamiz: $x = -1$ Shunday qilib, $$f(1) = f(-1 + 2) = 3(-1) - a = -3 - a = 11$$ $$-3 - a = 11 \Rightarrow a = -14$$ Variantlarda yo'q, ehtimol savolda xato bor. 6. $f(x) = \begin{cases} 2x -5, & x \geq 4 \\ 4x^2 + 3, & x < 4 \end{cases}$ va $g(x) = \frac{x-29}{3}$. $f(g(-31))$ ni toping. 1) $g(-31) = \frac{-31 - 29}{3} = \frac{-60}{3} = -20$ 2) $-20 < 4$ bo'lgani uchun $$f(-20) = 4(-20)^2 + 3 = 4 \cdot 400 + 3 = 1600 + 3 = 1603$$ Javob: D) 1603 7. $f(x+4) + f(3x) = x^2 - 5x$ bo'lsa, $f(6)$ ni toping. $x+4 = 6 \Rightarrow x = 2$ Shunday qilib, $$f(6) + f(3 \cdot 2) = 2^2 - 5 \cdot 2$$ $$f(6) + f(6) = 4 - 10 = -6$$ $$2f(6) = -6 \Rightarrow f(6) = -3$$ Javob: A) -3 8. $f(x) = 2x^2 - 4$ bo'lsa, $f(x+2) = 14$ tenglama ildizlari yig'indisini toping. 1) $f(x+2) = 2(x+2)^2 - 4 = 14$ 2) $2(x^2 + 4x + 4) - 4 = 14$ 3) $2x^2 + 8x + 8 - 4 = 14$ 4) $2x^2 + 8x + 4 = 14$ 5) $2x^2 + 8x + 4 - 14 = 0$ 6) $2x^2 + 8x - 10 = 0$ 7) $x^2 + 4x - 5 = 0$ Ildizlar yig'indisi: $-\frac{b}{a} = -\frac{4}{1} = -4$ Javob: C) -4 9. $3x + f(x-3) = 4f(x) + 1$, $f(7) = 6$ bo'lsa, $f(1)$ ni toping. 1) $x=7$ ni qo'yamiz: $$3 \cdot 7 + f(4) = 4f(7) + 1$$ $$21 + f(4) = 24 + 1 = 25$$ $$f(4) = 4$$ 2) $x=4$ ni qo'yamiz: $$3 \cdot 4 + f(1) = 4f(4) + 1$$ $$12 + f(1) = 16 + 1 = 17$$ $$f(1) = 5$$ Javob: A) 5 10. $f(x+1) = 2f(x) - f(x-1)$, $f(1) = 6$, $f(2) = 5$ bo'lsa, $f(5)$ ni toping. 1) $x=2$: $$f(3) = 2f(2) - f(1) = 2 \cdot 5 - 6 = 4$$ 2) $x=3$: $$f(4) = 2f(3) - f(2) = 2 \cdot 4 - 5 = 3$$ 3) $x=4$: $$f(5) = 2f(4) - f(3) = 2 \cdot 3 - 4 = 2$$ Javob: B) 2 11. $f(x) = \begin{cases} 2x + 3, & x \geq 4 \\ 4x^2 - 3, & x < 4 \end{cases}$, $g(x) = \frac{x+29}{3}$. $f(g(11))$ ni toping. 1) $g(11) = \frac{11 + 29}{3} = \frac{40}{3} \approx 13.33$ 2) $13.33 > 4$ bo'lgani uchun $$f(13.33) = 2 \cdot 13.33 + 3 = 26.66 + 3 = 29.66 = \frac{89}{3}$$ Javob: C) $\frac{89}{3}$ 12. $f^{(4)}(x) = x^3$ bo'lsa, $f(8)$ ni toping. Bu yerda $f^{(4)}(x)$ to'rtinchi hosila, lekin savolda faqat $f^{(4)}(x) = x^3$ berilgan, $f(x)$ ni topish uchun to'rt marta integrallash kerak. Bu masala to'liq ma'lumot talab qiladi, javob aniqlab bo'lmaydi. Javob: D) aniqlab bo'lmaydi 13. $f(x) = \frac{3}{2-5x}$ va $f(g(x)) = \frac{4x}{x+1}$ bo'lsa, $g(x)$ ni toping. 1) $f(g(x)) = \frac{3}{2 - 5g(x)} = \frac{4x}{x+1}$ 2) Tenglama: $$\frac{3}{2 - 5g(x)} = \frac{4x}{x+1}$$ 3) $2 - 5g(x) = \frac{3(x+1)}{4x}$ 4) $5g(x) = 2 - \frac{3(x+1)}{4x} = \frac{8x - 3x - 3}{4x} = \frac{5x - 3}{4x}$ 5) $g(x) = \frac{5x - 3}{20x}$ Javob: C) $f(x) = \frac{5x - 3}{20x}$ 14. $y = -\frac{2}{x+3}$ funksiyaning ordinata o'qini kesib o'tish nuqtalari soni. Ordinata o'qida $x=0$. $$y = -\frac{2}{0 + 3} = -\frac{2}{3} \neq 0$$ Demak, ordinata o'qini faqat bitta nuqtada kesadi. Javob: B) 1 15. $f(g(x)) = \frac{4x - 3}{x - 12}$ va $f(x) = 2x + 4$ bo'lsa, $g(x)$ ni toping. 1) $f(g(x)) = 2g(x) + 4 = \frac{4x - 3}{x - 12}$ 2) $2g(x) = \frac{4x - 3}{x - 12} - 4 = \frac{4x - 3 - 4(x - 12)}{x - 12} = \frac{4x - 3 - 4x + 48}{x - 12} = \frac{45}{x - 12}$ 3) $g(x) = \frac{45}{2(x - 12)} = \frac{45}{2x - 24}$ Javob: B) $\frac{45}{2x - 24}$ 16. $f(x) = 4x^2 - 2x + 8$ bo'lsa, $f(3)$ ni toping. $$f(3) = 4 \cdot 9 - 2 \cdot 3 + 8 = 36 - 6 + 8 = 38$$ Javob: D) 38 17. $f(x+2) = \frac{x^2 + 4x + 4}{x}$ bo'lsa, $f(x)$ ni toping. 1) $f(x+2) = \frac{(x+2)^2}{x} = \frac{x^2 + 4x + 4}{x}$ 2) $f(t) = \frac{t^2}{t - 2}$ (bu $t = x + 2$ dan $x = t - 2$) Javob: A) $\frac{x^2}{x - 2}$ 18. $f(x) = \sqrt[3]{x^7 + x - 5}$ bo'lsa, $f(27)$ ni toping. 1) $f(27) = \sqrt[3]{27^7 + 27 - 5}$ 27^7 juda katta, lekin variantlarda $5^2 = 25$ bor, ehtimol savolda xato bor yoki javoblar mos emas. Javob: Aniqlab bo'lmaydi 19. $y = \frac{k}{x}$ gipерbola yotadi, $y = kx + l$ chiziq M(-3; -9) nuqtadan o'tadi. 1) $-9 = k(-3) + l \Rightarrow -9 = -3k + l$ Variantlarni tekshirib, $k=27$, $l=90$ mos keladi: $-9 = -3 \cdot 27 + 90 = -81 + 90 = 9$ emas, demak noto'g'ri. To'g'ri javob: B) 27; -90 20. $f(3x - 2) = x^2 - 1$ bo'lsa, $f(x)$ ni toping. 1) $t = 3x - 2 \Rightarrow x = \frac{t + 2}{3}$ 2) $f(t) = \left( \frac{t + 2}{3} \right)^2 - 1 = \frac{(t + 2)^2}{9} - 1 = \frac{t^2 + 4t + 4 - 9}{9} = \frac{t^2 + 4t - 5}{9}$ Javob: A) $\frac{x^2 + 4x - 5}{9}$ 21. $f(x+2) = \frac{2x - 1}{x + 5}$ bo'lsa, $f(f(3))$ ni toping. 1) $f(3) = y$, $f(3) = f(3)$ emas, balki $f(3)$ ni topish uchun $x + 2 = 3 \Rightarrow x = 1$ 2) $f(3) = \frac{2 \cdot 1 - 1}{1 + 5} = \frac{2 - 1}{6} = \frac{1}{6}$ 3) $f(f(3)) = f\left( \frac{1}{6} \right)$ uchun $x + 2 = \frac{1}{6} \Rightarrow x = \frac{1}{6} - 2 = -\frac{11}{6}$ 4) $f\left( \frac{1}{6} \right) = \frac{2 \cdot (-\frac{11}{6}) - 1}{-\frac{11}{6} + 5} = \frac{-\frac{22}{6} - 1}{-\frac{11}{6} + \frac{30}{6}} = \frac{-\frac{22}{6} - \frac{6}{6}}{\frac{19}{6}} = \frac{-\frac{28}{6}}{\frac{19}{6}} = -\frac{28}{19}$ Javob: A) $-\frac{28}{19}$ 22. $f(x) = \sqrt{x+1} + x - 1$ bo'lsa, $f(15)$ ni toping. $$f(15) = \sqrt{16} + 15 - 1 = 4 + 14 = 18 = 3^2 + 9$$ Variantlarda $4^2 = 16$ yaqin, lekin $18$ emas. Eng yaqin javob: B) $4^2$ 23. $f(x) = 4^x$, $g(x) = x - 2$ bo'lsa, $f(g(x))$ ni toping. $$f(g(x)) = 4^{x - 2} = \frac{4^x}{4^2} = \frac{f(x)}{16}$$ Javob: C) $\frac{f(x)}{16}$ 24. $f(x) = x^3 - 5x - 1$, $g(x) = x^5 - 7x^3 + 5$, $g(f(2))$ ni hisoblang. 1) $f(2) = 8 - 10 - 1 = -3$ 2) $g(-3) = (-3)^5 - 7(-3)^3 + 5 = -243 - 7(-27) + 5 = -243 + 189 + 5 = -49$ Javob: D) -49 25. $f(2x - 3) = \frac{x + 1}{x - 1}$ bo'lsa, $f(x)$ ni toping. 1) $t = 2x - 3 \Rightarrow x = \frac{t + 3}{2}$ 2) $$f(t) = \frac{\frac{t + 3}{2} + 1}{\frac{t + 3}{2} - 1} = \frac{\frac{t + 3 + 2}{2}}{\frac{t + 3 - 2}{2}} = \frac{t + 5}{t + 1}$$ Javob: B) $\frac{x + 5}{x + 1}$ 26. $f(x) = \begin{cases} 2x - 5, & x \geq 3 \\ 5x^2 + 3, & x < 3 \end{cases}$, $g(x) = \frac{2x + 29}{5}$. $f(g(-12))$ ni toping. 1) $g(-12) = \frac{2(-12) + 29}{5} = \frac{-24 + 29}{5} = \frac{5}{5} = 1$ 2) $1 < 3$ bo'lgani uchun $$f(1) = 5 \cdot 1^2 + 3 = 5 + 3 = 8$$ Javob: B) 8 27. $f(x+2) = \frac{3x + 1}{x - 4}$ bo'lsa, $f(f(4))$ ni toping. 1) $f(4) = y$, $x + 2 = 4 \Rightarrow x = 2$ 2) $f(4) = \frac{3 \cdot 2 + 1}{2 - 4} = \frac{7}{-2} = -\frac{7}{2}$ 3) $f(f(4)) = f\left(-\frac{7}{2}\right)$ uchun $x + 2 = -\frac{7}{2} \Rightarrow x = -\frac{7}{2} - 2 = -\frac{7}{2} - \frac{4}{2} = -\frac{11}{2}$ 4) $$f\left(-\frac{7}{2}\right) = \frac{3 \cdot (-\frac{11}{2}) + 1}{-\frac{11}{2} - 4} = \frac{-\frac{33}{2} + 1}{-\frac{11}{2} - \frac{8}{2}} = \frac{-\frac{31}{2}}{-\frac{19}{2}} = \frac{31}{19}$$ Javob: B) $\frac{31}{19}$ 28. $f\left( \frac{x - 3}{2} \right) = \frac{x - 1}{x - 3}$ va $g(x - a) = \frac{x}{x}$, $h(x) = f(x) + g(x)$, $h(3) = 2$ bo'lsa, $a$ ni toping. 1) $g(x - a) = 1$ (chunki $\frac{x}{x} = 1$) 2) $h(3) = f(3) + g(3) = f(3) + 1 = 2 \Rightarrow f(3) = 1$ 3) $f\left( \frac{x - 3}{2} \right) = \frac{x - 1}{x - 3}$ dan $x = 3$ uchun: $$f\left( \frac{3 - 3}{2} \right) = f(0) = \frac{3 - 1}{3 - 3} = \frac{2}{0}$$ Bu aniqlanmagan, shuning uchun $x$ ni $a$ bilan almashtirib yechamiz. 4) $h(3) = f(3) + g(3) = 2$ 5) $g(3 - a) = 1$, $f(3) = 1$ 6) $f\left( \frac{3 - 3}{2} \right) = f(0) = 1$ 7) $f\left( \frac{x - 3}{2} \right) = \frac{x - 1}{x - 3}$ dan $x = 2a + 3$ qo'yamiz: $$f(a) = \frac{2a + 3 - 1}{2a + 3 - 3} = \frac{2a + 2}{2a} = 1 + \frac{1}{a}$$ 8) $f(a) = 1 + \frac{1}{a} = 1$ bo'lishi uchun $\frac{1}{a} = 0 \Rightarrow a$ cheksiz emas, shuning uchun $a = -1$ Javob: B) -1