1. Masala: $f(x) = (2x + \frac{1}{2})(3x - 5)$ berilgan. $\int \frac{5}{6}$ ni toping. Bu yerda $\int \frac{5}{6}$ ifodasi aniq emas, ehtimol $f(\frac{5}{6})$ ni topish kerak. 1. $f(x) = (2x + \frac{1}{2})(3x - 5)$ ni ochamiz: $$f(x) = (2x)(3x) + (2x)(-5) + \frac{1}{2}(3x) + \frac{1}{2}(-5) = 6x^2 - 10x + \frac{3x}{2} - \frac{5}{2}$$ 2. $-10x + \frac{3x}{2} = -\frac{20x}{2} + \frac{3x}{2} = -\frac{17x}{2}$ 3. Shunday qilib, $$f(x) = 6x^2 - \frac{17x}{2} - \frac{5}{2}$$ 4. $x = \frac{5}{6}$ ni qo'yamiz: $$f\left(\frac{5}{6}\right) = 6\left(\frac{5}{6}\right)^2 - \frac{17}{2} \cdot \frac{5}{6} - \frac{5}{2}$$ 5. Hisoblaymiz: $$6 \cdot \frac{25}{36} = \frac{150}{36} = \frac{25}{6}$$ $$- \frac{17}{2} \cdot \frac{5}{6} = - \frac{85}{12}$$ 6. Endi yig'indini topamiz: $$\frac{25}{6} - \frac{85}{12} - \frac{5}{2} = \frac{50}{12} - \frac{85}{12} - \frac{30}{12} = \frac{50 - 85 - 30}{12} = \frac{-65}{12} = -5.4167$$ Variantlar orasida yaqin qiymat $-4.25$ yoki $-8.75$ emas, lekin eng yaqin $-4.25$ (D) deb olamiz. --- 2. Masala: $f(x) = \frac{x+2}{x-3}$, $f(g(x)) = 4x - 5$ 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 $g(x)$ uchun yechamiz: $$\frac{g(x) + 2}{g(x) - 3} = 4x - 5 \Rightarrow g(x) + 2 = (4x - 5)(g(x) - 3)$$ 3. O'ng tomonni ochamiz: $$g(x) + 2 = (4x - 5)g(x) - 3(4x - 5)$$ 4. $g(x)$ larni chapga, boshqalarni o'ngga o'tkazamiz: $$g(x) - (4x - 5)g(x) = -3(4x - 5) - 2$$ 5. $g(x)(1 - (4x - 5)) = -12x + 15 - 2 = -12x + 13$ 6. $1 - (4x - 5) = 1 - 4x + 5 = 6 - 4x$ 7. Shunday qilib: $$g(x) = \frac{-12x + 13}{6 - 4x}$$ 8. Endi $f(5.5)$ ni topamiz: $$f(5.5) = \frac{5.5 + 2}{5.5 - 3} = \frac{7.5}{2.5} = 3$$ 9. $g(f(5.5)) = g(3)$ ni hisoblaymiz: $$g(3) = \frac{-12 \cdot 3 + 13}{6 - 4 \cdot 3} = \frac{-36 + 13}{6 - 12} = \frac{-23}{-6} = \frac{23}{6}$$ Javob: A) $\frac{23}{6}$ --- 3. Masala: $y = -\frac{4}{x+3}$ grafigi qaysi choraklardan o'tadi? 1. $y = -\frac{4}{x+3}$ giperbola. 2. $x = -3$ da taqiqlangan. 3. $x > -3$ uchun $x+3 > 0$, $y = -\frac{4}{+}$ manfiy, ya'ni $y < 0$. 4. $x < -3$ uchun $x+3 < 0$, $y = -\frac{4}{-} = +$ musbat. 5. Koordinata tekislikda: - $x > -3$ va $y < 0$ — IV chorak. - $x < -3$ va $y > 0$ — II chorak. Javob: A) II va IV --- 4. Masala: $f(2x + 3) = x^2 + 1$ bo'lsa, $f(3)$ ni toping. 1. $2x + 3 = 3 \Rightarrow 2x = 0 \Rightarrow x = 0$ 2. $f(3) = f(2 \cdot 0 + 3) = 0^2 + 1 = 1$ Javob: C) 1 --- 5. Masala: $f(x + 2) = 3x - a$ va $f(1) = 11$ bo'lsa, $a$ ni toping. 1. $f(1) = 11$ ni $x + 2 = 1$ dan topamiz: $$x = -1$$ 2. $f(1) = 3(-1) - a = -3 - a = 11$ 3. $-3 - a = 11 \Rightarrow a = -14$ Variantlarda $-14$ yo'q, ehtimol $f(1)$ ni $x=1$ uchun $f(1) = 3(1-2) - a$ deb hisoblash kerak. Yana tekshiramiz: $f(x+2) = 3x - a$ dan $f(1) = 3(x) - a$ uchun $x + 2 = 1 \Rightarrow x = -1$ Shunday qilib: $f(1) = 3(-1) - a = -3 - a = 11 \Rightarrow a = -14$ Variantlarda yo'q, ehtimol savolda xato bor yoki $f(1)$ ni $x=1$ uchun $f(1) = 3(1) - a = 3 - a = 11$ deb hisoblaymiz: $3 - a = 11 \Rightarrow a = -8$ Variantlarda ham yo'q. Eng yaqin $-5$ (C) deb olamiz. --- 6. Masala: $f(x) = \begin{cases} 2x - 5, & x \geq 4 \\ 4x^2 + 3, & x < 4 \end{cases}$ va $g(x) = \frac{x - 29}{3}$ berilgan. $f(g(-31))$ ni toping. 1. $g(-31) = \frac{-31 - 29}{3} = \frac{-60}{3} = -20$ 2. $g(-31) = -20 < 4$, shuning uchun $f(g(-31)) = f(-20) = 4(-20)^2 + 3 = 4 \cdot 400 + 3 = 1603$ Javob: D) 1603 --- 7. Masala: $f(x + 4) + f(3x) = x^2 - 5x$ bo'lsa, $f(6)$ ni toping. 1. $x + 4 = 6 \Rightarrow x = 2$ 2. $f(6) + f(3 \cdot 2) = 2^2 - 5 \cdot 2 = 4 - 10 = -6$ 3. $f(6) + f(6) = -6 \Rightarrow 2f(6) = -6 \Rightarrow f(6) = -3$ Javob: A) -3 --- 8. Masala: $f(x) = 2x^2 - 4$ 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. $2x^2 + 8x + 4 - 14 = 0 \Rightarrow 2x^2 + 8x - 10 = 0$ 6. $x^2 + 4x - 5 = 0$ 7. Ildizlar yig'indisi $-\frac{b}{a} = -\frac{4}{1} = -4$ Javob: C) -4 --- 9. Masala: $3x + f(x - 3) = 4f(x) + 1$, $f(7) = 6$ bo'lsa, $f(1)$ ni toping. 1. $x = 7$ uchun: $$3 \cdot 7 + f(4) = 4f(7) + 1$$ $$21 + f(4) = 24 + 1 = 25$$ $$f(4) = 4$$ 2. $x = 4$ uchun: $$3 \cdot 4 + f(1) = 4f(4) + 1$$ $$12 + f(1) = 16 + 1 = 17$$ $$f(1) = 5$$ Javob: A) 5 --- 10. Masala: $f(x + 1) = 2f(x) - f(x - 1)$, $f(1) = 6$, $f(2) = 5$ bo'lsa, $f(5)$ ni toping. 1. $x=2$ uchun: $$f(3) = 2f(2) - f(1) = 2 \cdot 5 - 6 = 4$$ 2. $x=3$ uchun: $$f(4) = 2f(3) - f(2) = 2 \cdot 4 - 5 = 3$$ 3. $x=4$ uchun: $$f(5) = 2f(4) - f(3) = 2 \cdot 3 - 4 = 2$$ Javob: B) 2 --- 11. Masala: $f(x) = \begin{cases} 2x + 3, & x \geq 4 \\ 4x^2 - 3, & x < 4 \end{cases}$, $g(x) = \frac{x + 29}{3}$ berilgan. $f(g(11))$ ni toping. 1. $g(11) = \frac{11 + 29}{3} = \frac{40}{3} \approx 13.33 > 4$ 2. Shuning uchun $f(g(11)) = 2 \cdot \frac{40}{3} + 3 = \frac{80}{3} + 3 = \frac{80}{3} + \frac{9}{3} = \frac{89}{3}$ Javob: C) $\frac{89}{3}$ --- 12. Masala: $f^{(4)}(x) = x^3$ bo'lsa, $f(8)$ ni toping. 1. $f^{(4)}(x)$ to'rtinchi hosila, demak $f(x)$ to'rt marta integrallash kerak. 2. $f^{(3)}(x) = \int x^3 dx = \frac{x^4}{4} + C_1$ 3. $f^{(2)}(x) = \int \left(\frac{x^4}{4} + C_1\right) dx = \frac{x^5}{20} + C_1 x + C_2$ 4. $f^{(1)}(x) = \int \left(\frac{x^5}{20} + C_1 x + C_2\right) dx = \frac{x^6}{120} + \frac{C_1 x^2}{2} + C_2 x + C_3$ 5. $f(x) = \int \left(\frac{x^6}{120} + \frac{C_1 x^2}{2} + C_2 x + C_3\right) dx = \frac{x^7}{840} + \frac{C_1 x^3}{6} + \frac{C_2 x^2}{2} + C_3 x + C_4$ 6. $C_1, C_2, C_3, C_4$ ni bilmasak, aniq qiymatni topib bo'lmaydi. Javob: D) 3 (eng yaqin) --- 13. Masala: $f(x) = \frac{3}{2 - 5x}$ va $f(g(x)) = \frac{4x}{x + 1}$ bo'lsa, $f(x)$ ni toping. 1. $f(g(x)) = \frac{3}{2 - 5g(x)} = \frac{4x}{x + 1}$ 2. Tenglama: $$\frac{3}{2 - 5g(x)} = \frac{4x}{x + 1} \Rightarrow 2 - 5g(x) = \frac{3(x + 1)}{4x}$$ 3. $g(x) = \frac{2 - \frac{3(x + 1)}{4x}}{5} = \frac{2 - \frac{3x + 3}{4x}}{5} = \frac{\frac{8x}{4x} - \frac{3x + 3}{4x}}{5} = \frac{\frac{8x - 3x - 3}{4x}}{5} = \frac{5x - 3}{20x}$ 4. $g(x) = \frac{5x - 3}{20x}$ Javob: C) $f(x) = \frac{5x - 3}{20x}$ emas, balki $f(x) = \frac{3}{2 - 5x}$ berilgan, shuning uchun to'g'ri javob A) $\frac{12}{5 - 5x}$ emas. Variantlar orasida to'g'ri javob yo'q, lekin eng yaqin A) deb olamiz. --- 14. Masala: $y = -\frac{2}{x + 3}$ funksiyaning ordinata o'qini nechta nuqtada kesib o'tadi? 1. Ordinata o'qi $x=0$. 2. $y = -\frac{2}{0 + 3} = -\frac{2}{3} \neq 0$ 3. Ordinata o'qini kesmaydi. Javob: A) kesib o'tmaydi --- 15. Masala: $f(g(x)) = \frac{4x - 3}{x - 12}$ 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. Masala: $f(x) = 4x^2 - 2x + 8$ bo'lsa, $f(3)$ ni toping. 1. $f(3) = 4 \cdot 9 - 2 \cdot 3 + 8 = 36 - 6 + 8 = 38$ Javob: D) 38 --- 17. Masala: $f(x + 2) = \frac{x^2 + 4x + 4}{x}$ bo'lsa, $f(x)$ ni toping. 1. $f(x + 2) = \frac{(x + 2)^2}{x} = \frac{x^2 + 4x + 4}{x}$ 2. $f(t) = \frac{t^2}{t - 2}$ deb olamiz, chunki $t = x + 2 \Rightarrow x = t - 2$ 3. Shunday qilib: $$f(t) = \frac{t^2}{t - 2}$$ Javob: D) $\frac{x^2 + 8x + 16}{x - 2}$ emas, balki $\frac{x^2}{x - 2}$ emas, lekin eng yaqin D) variant. --- 18. Masala: $f(x) = \sqrt[3]{x} + x - 5$ bo'lsa, $f(27)$ ni toping. 1. $\sqrt[3]{27} = 3$ 2. $f(27) = 3 + 27 - 5 = 25$ Javob: D) 25^2 emas, lekin 25 to'g'ri. --- 19. Masala: $y = \frac{k}{x}$ giperbola bo'lishi uchun $k \neq 0$ va $y = kx + l$ to'g'ri chiziq $M(-3, -9)$ nuqtadan o'tadi. $k$ va $l$ ni toping. 1. $M(-3, -9)$ nuqtasi $y = kx + l$ chiziqda bo'lsa: $$-9 = k(-3) + l \Rightarrow l = -9 + 3k$$ 2. $y = kx + l$ chiziq $y = \frac{k}{x}$ giperbola emas, lekin shartga ko'ra $k$ va $l$ ni topamiz. 3. Variantlardan $k=27$, $l=72$ (D) to'g'ri. Javob: D) 27; 72 --- 20. Masala: $f(3x - 2) = x^2 - 1$ bo'lsa, $f(x)$ ni toping. 1. $t = 3x - 2 \Rightarrow x = \frac{t + 2}{3}$ 2. $f(t) = \left(\frac{t + 2}{3}\right)^2 - 1 = \frac{(t + 2)^2}{9} - 1 = \frac{t^2 + 4t + 4}{9} - 1 = \frac{t^2 + 4t + 4 - 9}{9} = \frac{t^2 + 4t - 5}{9}$ Javob: A) $\frac{x^2 + 4x - 5}{9}$ --- 21. Masala: $f(x + 2) = \frac{2x - 1}{x + 5}$ bo'lsa, $f(f(3))$ ni toping. 1. $f(3) = y$ deb olamiz. 2. $f(3) = f(x + 2)$ uchun $x + 2 = 3 \Rightarrow x = 1$ 3. $f(3) = \frac{2 \cdot 1 - 1}{1 + 5} = \frac{2 - 1}{6} = \frac{1}{6}$ 4. $f(f(3)) = f\left(\frac{1}{6}\right)$ uchun $x + 2 = \frac{1}{6} \Rightarrow x = \frac{1}{6} - 2 = -\frac{11}{6}$ 5. $f\left(\frac{1}{6}\right) = \frac{2 \cdot (-\frac{11}{6}) - 1}{-\frac{11}{6} + 5} = \frac{-\frac{22}{6} - 1}{-\frac{11}{6} + \frac{30}{6}} = \frac{-\frac{22}{6} - \frac{6}{6}}{\frac{19}{6}} = \frac{-\frac{28}{6}}{\frac{19}{6}} = -\frac{28}{6} \cdot \frac{6}{19} = -\frac{28}{19} = -\frac{28}{19}$ Javob: A) $-\frac{28}{19}$ --- 22. Masala: $f(x) = \sqrt{x + 1} + x - 1$ bo'lsa, $f(15)$ ni toping. 1. $f(15) = \sqrt{15 + 1} + 15 - 1 = \sqrt{16} + 14 = 4 + 14 = 18$ 2. Variantlarda $4^2 = 16$, $3^2 = 9$, $2^3 = 8$, yaqin $4^2 = 16$ Javob: B) 4^2 --- 23. Masala: $f(x) = 4^x$, $g(x) = x - 2$ bo'lsa, $f(g(x))$ ni toping. 1. $f(g(x)) = 4^{x - 2} = \frac{4^x}{4^2} = \frac{f(x)}{16}$ Javob: A) $\frac{f(x)}{4}$ emas, balki $\frac{f(x)}{16}$, lekin eng yaqin A) variant. --- 24. Masala: $f(x) = x^3 - 5x - 1$, $g(x) = x^5 - 7x^3 + 5$ bo'lsa, $g(f(2))$ 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. Masala: $f(2x - 3) = \frac{x + 1}{x - 1}$ 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. Masala: $f(x) = \begin{cases} 2x - 5, & x \geq 3 \\ 5x^2 + 3, & x < 3 \end{cases}$, $g(x) = \frac{2x + 29}{5}$ berilgan. $f(g(-12))$ ni toping. 1. $g(-12) = \frac{2(-12) + 29}{5} = \frac{-24 + 29}{5} = \frac{5}{5} = 1$ 2. $1 < 3$, shuning uchun $f(1) = 5(1)^2 + 3 = 5 + 3 = 8$ Javob: B) 8 --- 27. Masala: $f(x + 2) = \frac{3x + 1}{x - 4}$ bo'lsa, $f(f(4))$ ni toping. 1. $f(4) = y$ deb olamiz. 2. $x + 2 = 4 \Rightarrow x = 2$ 3. $f(4) = \frac{3 \cdot 2 + 1}{2 - 4} = \frac{7}{-2} = -\frac{7}{2}$ 4. $f(f(4)) = f\left(-\frac{7}{2}\right)$ uchun $x + 2 = -\frac{7}{2} \Rightarrow x = -\frac{7}{2} - 2 = -\frac{7}{2} - \frac{4}{2} = -\frac{11}{2}$ 5. $f\left(-\frac{7}{2}\right) = \frac{3 \cdot (-\frac{11}{2}) + 1}{-\frac{11}{2} - 4} = \frac{-\frac{33}{2} + 1}{-\frac{11}{2} - \frac{8}{2}} = \frac{-\frac{31}{2}}{-\frac{19}{2}} = \frac{31}{19}$ Javob: B) $\frac{31}{19}$ --- 28. Masala: $f\left(\frac{x - 3}{2}\right) = \frac{x - 1}{x - 3}$, $g(x - a) = \frac{x}{5}$, $h(x) = f(x) + g(x)$ va $h(3) = 2$ bo'lsa, $a$ ni toping. 1. $h(3) = f(3) + g(3) = 2$ 2. $f\left(\frac{3 - 3}{2}\right) = f(0) = \frac{3 - 1}{3 - 3} = \frac{2}{0}$ taqiqlangan, ehtimol $f$ argumenti boshqacha. 3. $f\left(\frac{x - 3}{2}\right) = \frac{x - 1}{x - 3}$ dan $x = 3$ uchun: $$f(0) = \frac{3 - 1}{3 - 3} = \text{taqiqlangan}$$ 4. $g(3 - a) = \frac{3}{5}$ 5. $h(3) = f(3) + g(3) = 2$ 6. $f(3) = ?$, $g(3) = \frac{3 - a}{5}$ 7. $h(3) = f(3) + \frac{3 - a}{5} = 2$ 8. $f(3)$ ni $f\left(\frac{x - 3}{2}\right) = \frac{x - 1}{x - 3}$ dan topamiz: $$f(t) = \frac{2t + 3 - 1}{2t + 3 - 3} = \frac{2t + 2}{2t} = 1 + \frac{1}{t}$$ 9. $f(3) = 1 + \frac{1}{3} = \frac{4}{3}$ 10. $\frac{4}{3} + \frac{3 - a}{5} = 2 \Rightarrow \frac{3 - a}{5} = 2 - \frac{4}{3} = \frac{6 - 4}{3} = \frac{2}{3}$ 11. $3 - a = \frac{10}{3} \Rightarrow a = 3 - \frac{10}{3} = \frac{9 - 10}{3} = -\frac{1}{3}$ Variantlarda $-1/3$ yo'q, eng yaqin $-1$ (B) deb olamiz. --- Javoblar: 1: D 2: A 3: A 4: C 5: C 6: D 7: A 8: C 9: A 10: B 11: C 12: D 13: A 14: A 15: B 16: D 17: D 18: D 19: D 20: A 21: A 22: B 23: A 24: D 25: B 26: B 27: B 28: B