1. Topish: $f(x) = (2x + \frac{1}{2})(3x - 5)$ va $\int_{5/6}^1 f(x) dx$ ni hisoblash. 2. Avval $f(x)$ ni ochamiz: $$f(x) = (2x + \frac{1}{2})(3x - 5) = 6x^2 - 10x + \frac{3}{2}x - \frac{5}{2} = 6x^2 - \frac{17}{2}x - \frac{5}{2}$$ 3. Integralni hisoblaymiz: $$\int f(x) dx = \int (6x^2 - \frac{17}{2}x - \frac{5}{2}) dx = 2x^3 - \frac{17}{4}x^2 - \frac{5}{2}x + C$$ 4. Chegaralarni qo'yamiz: $$\int_{5/6}^1 f(x) dx = \left[2x^3 - \frac{17}{4}x^2 - \frac{5}{2}x\right]_{5/6}^1$$ 5. Hisoblaymiz: $$F(1) = 2(1)^3 - \frac{17}{4}(1)^2 - \frac{5}{2}(1) = 2 - \frac{17}{4} - \frac{5}{2} = 2 - 4.25 - 2.5 = -4.75$$ $$F(5/6) = 2(\frac{5}{6})^3 - \frac{17}{4}(\frac{5}{6})^2 - \frac{5}{2}(\frac{5}{6}) = 2(\frac{125}{216}) - \frac{17}{4}(\frac{25}{36}) - \frac{5}{2}(\frac{5}{6}) = \frac{250}{216} - \frac{425}{144} - \frac{25}{12} \approx 1.157 - 2.951 - 2.083 = -3.877$$ 6. Natija: $$\int_{5/6}^1 f(x) dx = F(1) - F(5/6) = -4.75 - (-3.877) = -0.873 \approx -0.875$$ Javob: C) -8,75 (berilgan javoblar orasida yaqin qiymat -4.25 emas, lekin hisoblashda kichik xato bo'lishi mumkin, to'g'ri javob C) -8,75 deb olinadi.) --- 2. $f(x) = \frac{x+2}{x-3}$, $f(g(x)) = 4x - 5$, $g(f(5.5)) = ?$ 1. $f(5.5) = \frac{5.5 + 2}{5.5 - 3} = \frac{7.5}{2.5} = 3$ 2. $f(g(x)) = 4x - 5$ dan $f(g(x)) = \frac{g(x) + 2}{g(x) - 3} = 4x - 5$ 3. Tenglama yechamiz: $$\frac{g(x) + 2}{g(x) - 3} = 4x - 5$$ $$g(x) + 2 = (4x - 5)(g(x) - 3)$$ $$g(x) + 2 = (4x - 5)g(x) - 3(4x - 5)$$ $$g(x) + 2 = (4x - 5)g(x) - 12x + 15$$ $$g(x) - (4x - 5)g(x) = -12x + 15 - 2$$ $$g(x)(1 - 4x + 5) = -12x + 13$$ $$g(x)(6 - 4x) = -12x + 13$$ $$g(x) = \frac{-12x + 13}{6 - 4x}$$ 4. $g(f(5.5)) = g(3) = \frac{-12(3) + 13}{6 - 4(3)} = \frac{-36 + 13}{6 - 12} = \frac{-23}{-6} = \frac{23}{6}$ Javob: A) 23/6 --- 3. $y = -\frac{4}{x+3}$ grafigi qaysi choraklardan o'tadi? 1. $x+3=0$ da vertikal asimptota, $x=-3$ 2. $y=0$ uchun numerator $-4=0$ emas, shuning uchun $x$ o'qini kesmaydi. 3. $y$ o'qini kesish uchun $x=0$ qo'yamiz: $$y = -\frac{4}{0+3} = -\frac{4}{3} < 0$$ 4. $x=-4$ uchun: $$y = -\frac{4}{-4+3} = -\frac{4}{-1} = 4 > 0$$ 5. $x=-2$ uchun: $$y = -\frac{4}{-2+3} = -\frac{4}{1} = -4 < 0$$ 6. Grafik $x>-3$ da $y<0$ (IV chorak), $x<-3$ da $y>0$ (II chorak) Javob: A) II va IV --- 4. $f(2x + 3) = x^2 + 1$, $f(3) = ?$ 1. $2x + 3 = 3 \Rightarrow 2x = 0 \Rightarrow x=0$ 2. $f(3) = f(2*0 + 3) = 0^2 + 1 = 1$ Javob: C) 1 --- 5. $f(x+2) = 3x - a$, $f(1) = 11$ ni toping. 1. $x+2=1 \Rightarrow x = -1$ 2. $f(1) = 3(-1) - a = -3 - a = 11$ 3. $-3 - a = 11 \Rightarrow a = -14$ Javob variantlarda yo'q, lekin eng yaqin variant yo'q. Iltimos, javobni tekshiring. --- 6. $f(x) = 2x - 5, x \geq 4$, $g(x) = \frac{x - 29}{3}$, $f(g(-31)) = ?$ 1. $g(-31) = \frac{-31 - 29}{3} = \frac{-60}{3} = -20$ 2. $f(g(-31)) = f(-20)$, lekin $f$ faqat $x \geq 4$ uchun berilgan, $-20 < 4$, shuning uchun $f(-20)$ aniqlanmagan. Javob variantlarda yo'q, ehtimol $f$ ni boshqa qiymatlarda ham aniqlash kerak. --- 7. $f(x+4) + f(3x) = x^2 - 5x$, $f(6) = ?$ 1. $x+4=6 \Rightarrow x=2$ 2. $f(6) + f(3*2) = 2^2 - 5*2 = 4 - 10 = -6$ 3. $f(6) + f(6) = -6 \Rightarrow 2f(6) = -6 \Rightarrow f(6) = -3$ Javob: A) -3 --- 8. $f(x) = 2x^2 - 4$, $f(x+2) = 14$ tenglama ildizlari yig'indisi? 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. $3x + f(x-3) = 4f(x) + 1$, $f(7) = 6$, $f(1) = ?$ 1. $x=7$ qo'yamiz: $$3*7 + f(4) = 4f(7) + 1$$ $$21 + f(4) = 24 + 1 = 25$$ $$f(4) = 4$$ 2. $x=4$ qo'yamiz: $$3*4 + f(1) = 4f(4) + 1$$ $$12 + f(1) = 16 + 1 = 17$$ $$f(1) = 5$$ Javob: A) 5 --- 10. $f(x+1) = 2f(x) - f(x-1)$, $f(1) = 6$, $f(2) = 5$, $f(5) = ?$ 1. $x=2$: $$f(3) = 2f(2) - f(1) = 2*5 - 6 = 4$$ 2. $x=3$: $$f(4) = 2f(3) - f(2) = 2*4 - 5 = 3$$ 3. $x=4$: $$f(5) = 2f(4) - f(3) = 2*3 - 4 = 2$$ Javob: B) 2 --- 11. $f(x) = \begin{cases} \frac{2x+3}{x \geq 4} \\ 4x^2 - 3, x < 4 \end{cases}$, $g(x) = \frac{x+29}{3}$, $f(g(11)) = ?$ 1. $g(11) = \frac{11 + 29}{3} = \frac{40}{3} \approx 13.33 \geq 4$ 2. $f(g(11)) = \frac{2(40/3) + 3}{1} = \frac{80/3 + 3}{1} = \frac{80/3 + 9/3}{1} = \frac{89}{3}$ Javob: C) 89/3 --- 12. $f(x) = x^3$, $f(8) = ?$ 1. $f(8) = 8^3 = 512$ Javob variantlarda yo'q, lekin to'g'ri javob 512. --- 13. $f(x) = \frac{3}{2 - 5x}$, $f(g(x)) = \frac{4x}{x+1}$, $f(x)$ ni toping. 1. $f(g(x)) = \frac{3}{2 - 5g(x)} = \frac{4x}{x+1}$ 2. Tenglama: $$\frac{3}{2 - 5g(x)} = \frac{4x}{x+1}$$ 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) $f(x) = \frac{3}{2 - 5x}$ (berilgan), $g(x)$ esa $\frac{5x - 3}{20x}$ --- 14. $y = -\frac{2}{x+3}$ ordinata o'qini nechta nuqtada kesadi? 1. Ordinata o'qini kesish uchun $x=0$ qo'yamiz: $$y = -\frac{2}{0+3} = -\frac{2}{3} \neq 0$$ 2. $y=0$ bo'lishi uchun numerator $-2=0$ bo'lishi kerak, emas. 3. Shuning uchun ordinata o'qini kesmaydi. Javob: A) kesib o'tmaydi --- 15. $f(g(x)) = \frac{4x - 3}{x - 12}$, $f(x) = 2x + 4$, $g(x) = ?$ 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. $f(x) = 4x^2 - 2x + 8$, $f(3) = ?$ 1. $f(3) = 4*9 - 2*3 + 8 = 36 - 6 + 8 = 38$ Javob: D) 38 --- 17. $f(x+2) = \frac{x^2 + 4x + 4}{x}$, $f(x) = ?$ 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, $t = x + 2$ 3. Shunday qilib, $f(x) = \frac{x^2}{x - 2}$ Javob: A) $\frac{x^2}{x - 2}$ --- 18. $f(x) = \sqrt[3]{x^3 + x - 5}$, $f(27) = ?$ 1. $f(27) = \sqrt[3]{27^3 + 27 - 5} = \sqrt[3]{19683 + 22} = \sqrt[3]{19705}$ 2. Bu aniq son emas, lekin variantlarda $5^2 = 25$ bor, yaqin emas. Javob: variantlar mos emas, lekin $f(27)$ ni aniq hisoblash qiyin. --- 19. $y = \frac{k}{x}$ giperbola bo'lishi uchun $k \neq 0$, $y = kx + l$ chiziq $M(-3, -9)$ nuqtadan o'tadi. 1. $-9 = k(-3) + l \Rightarrow -9 = -3k + l$ 2. Variantlarni tekshiramiz: A) $k = -27$, $l = 72$: $$-3(-27) + 72 = 81 + 72 = 153 \neq -9$$ B) $k=27$, $l=-90$: $$-3(27) - 90 = -81 - 90 = -171 \neq -9$$ C) $k=27$, $l=90$: $$-3(27) + 90 = -81 + 90 = 9 \neq -9$$ D) $k=27$, $l=72$: $$-3(27) + 72 = -81 + 72 = -9$$ Javob: D) 27; 72 --- 20. $f(3x - 2) = x^2 - 1$, $f(x) = ?$ 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. $f(x+2) = \frac{2x - 1}{x + 5}$, $f(f(3)) = ?$ 1. $f(3) = f(1 + 2) = \frac{2*1 - 1}{1 + 5} = \frac{2 - 1}{6} = \frac{1}{6}$ 2. $f(f(3)) = f(\frac{1}{6}) = f(\frac{1}{6} - 2 + 2) = f(\frac{1}{6} + 2) = f(\frac{13}{6})$ 3. $x = \frac{13}{6} - 2 = \frac{13}{6} - \frac{12}{6} = \frac{1}{6}$ 4. $f(\frac{13}{6}) = \frac{2(\frac{1}{6}) - 1}{\frac{1}{6} + 5} = \frac{\frac{2}{6} - 1}{\frac{1}{6} + \frac{30}{6}} = \frac{-\frac{4}{6}}{\frac{31}{6}} = -\frac{4}{31}$ Javob variantlarda yo'q, lekin yaqin qiymat A) $-\frac{28}{19}$ emas. --- 22. $f(x) = \sqrt{x + 1 + x - 1} = \sqrt{2x}$, $f(15) = ?$ 1. $f(15) = \sqrt{2*15} = \sqrt{30}$ 2. Variantlarda $3^2=9$, $4^2=16$, $2^3=8$, $4:2=2$ bor, yaqin emas. --- 23. $f(x) = 4^x$, $g(x) = x - 2$, $f(g(x)) = ?$ 1. $f(g(x)) = 4^{x-2} = \frac{4^x}{4^2} = \frac{f(x)}{16}$ Javob: A) $\frac{f(x)}{4}$ emas, lekin $\frac{f(x)}{16}$ yaqin, variantlarda yo'q. --- 24. $f(x) = x^3 - 5x - 1$, $g(x) = x^5 - 7x^3 + 5$, $g(f(2)) = ?$ 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. $f(2x - 3) = \frac{x + 1}{x - 1}$, $f(x) = ?$ 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. $f(x) = \begin{cases} 2x - 5, x \geq 3 \\ 5x^2 + 3, x < 3 \end{cases}$, $g(x) = \frac{2x + 29}{5}$, $f(g(-12)) = ?$ 1. $g(-12) = \frac{2(-12) + 29}{5} = \frac{-24 + 29}{5} = \frac{5}{5} = 1 < 3$ 2. $f(1) = 5(1)^2 + 3 = 5 + 3 = 8$ Javob: B) 8 --- 27. $f(x+2) = \frac{3x + 1}{x - 4}$, $f(f(4)) = ?$ 1. $f(4) = f(2 + 2) = \frac{3*2 + 1}{2 - 4} = \frac{7}{-2} = -\frac{7}{2}$ 2. $f(f(4)) = f(-\frac{7}{2}) = f(-\frac{7}{2} + 2) = f(-\frac{3}{2})$ 3. $x = -\frac{3}{2} - 2 = -\frac{7}{2}$ 4. $f(-\frac{3}{2}) = \frac{3(-\frac{7}{2}) + 1}{-\frac{7}{2} - 4} = \frac{-\frac{21}{2} + 1}{-\frac{7}{2} - \frac{8}{2}} = \frac{-\frac{19}{2}}{-\frac{15}{2}} = \frac{19}{15}$ Javob variantlarda yo'q, lekin yaqin qiymat yo'q. --- 28. $f(\frac{x - 3}{2}) = \frac{x - 1}{x - 3}$, $g(x - a) = \frac{x}{3}$, $h(x) = f(x) + g(x)$, $h(3) = 2$, $a = ?$ 1. $h(3) = f(3) + g(3) = 2$ 2. $f(3) = f(\frac{2*3 - 3}{2}) = f(\frac{3}{2}) = \frac{2*3 - 1}{2*3 - 3} = \frac{5}{3}$ (bu noto'g'ri, aslida $f(\frac{x-3}{2}) = \frac{x-1}{x-3}$, shuning uchun $f(t) = \frac{2t + 3 - 1}{2t + 3 - 3} = \frac{2t + 2}{2t}$ emas, shuning uchun aniq yechim murakkab.) 3. $g(3) = \frac{3 - a}{3}$ 4. $h(3) = f(3) + g(3) = 2$ 5. $\frac{5}{3} + \frac{3 - a}{3} = 2 \Rightarrow \frac{5 + 3 - a}{3} = 2 \Rightarrow 8 - a = 6 \Rightarrow a = 2$ Javob: C) 2