1. Masala: Berilgan $f(x) = 3x - 7$ va $g(x) = 4x + 3$ funksiyalar uchun $g(f(x))$ ni toping va teskari funksiyasini aniqlang. Formulalar: Kompozitsiya funksiyasi $g(f(x)) = g(f(x))$ va teskari funksiya $f^{-1}(x)$ shunday funksiya bo'ladiki, $f(f^{-1}(x)) = x$. Hisoblash: $$g(f(x)) = 4(3x - 7) + 3 = 12x - 28 + 3 = 12x - 25$$ Teskari funksiya uchun $y = 12x - 25$ ni $x$ ga nisbatan yechamiz: $$y = 12x - 25 \Rightarrow 12x = y + 25 \Rightarrow x = \frac{y + 25}{12}$$ Demak, teskari funksiya: $$f^{-1}(x) = \frac{x + 25}{12}$$ Javob: D) $y = \frac{4x + 25}{12}$ emas, balki $y = \frac{x + 25}{12}$, lekin variantlar orasida eng yaqin D) variant. 2. Masala: $f(x + 1) + 2f(x) = 12$ va $f(2) = 7$ bo'lsa, $f(4)$ ni toping. Formulalar: Berilgan tenglama rekursiv ko'rinishda. Hisoblash: $f(3) + 2f(2) = 12 \Rightarrow f(3) + 14 = 12 \Rightarrow f(3) = -2$ $f(4) + 2f(3) = 12 \Rightarrow f(4) + 2(-2) = 12 \Rightarrow f(4) - 4 = 12 \Rightarrow f(4) = 16$ Javob: D) 16 3. Masala: $f(3x + 2) = x^2 + 2$ bo'lsa, $f(x)$ ni toping. Formulalar: $f(3x + 2) = x^2 + 2$ ni $t = 3x + 2$ deb olamiz, $x = \frac{t - 2}{3}$. Hisoblash: $$f(t) = \left(\frac{t - 2}{3}\right)^2 + 2 = \frac{(t - 2)^2}{9} + 2 = \frac{t^2 - 4t + 4}{9} + 2 = \frac{t^2 - 4t + 4 + 18}{9} = \frac{t^2 - 4t + 22}{9}$$ Javob: A) $\frac{x^2 - 4x + 22}{9}$ 4. Masala: $f(x) = x^2 - x + 2$ va $g(f(x)) = 2x^2 - 2x + 1$ bo'lsa, $g(x)$ ni toping. Formulalar: $g(f(x)) = g(y)$, $y = f(x)$. Hisoblash: $f(x) = x^2 - x + 2$, $g(f(x)) = 2x^2 - 2x + 1$ Faraz qilamiz $g(y) = ay + b$ $$g(f(x)) = a(x^2 - x + 2) + b = a x^2 - a x + 2a + b$$ Bu $2x^2 - 2x + 1$ ga teng bo'lishi kerak, shuning uchun: $$a = 2, -a = -2 \Rightarrow a = 2$$ $$2a + b = 1 \Rightarrow 4 + b = 1 \Rightarrow b = -3$$ Demak, $$g(x) = 2x - 3$$ Javob: A) $2x - 3$ 5. Masala: $\frac{f(x)}{x - 4} \cdot (x + 1) \leq 0$ tengsizlikni yeching, grafik asosida. Grafikdan ko'rinib turibdi: $f(x)$ ning ildizlari $x = -5$ va $x = -1$ atrofida, $f(x)$ $x=0$ da 0 ga teng. Tengsizlikni yechish uchun nol nuqtalarini va taqsimotini aniqlaymiz: Nol nuqtalari: $f(x) = 0$ da $x = -5, -1$ va $x - 4 = 0$ da $x=4$. Tengsizlik: $$\frac{f(x)}{x - 4} (x + 1) \leq 0$$ Bu ifoda $x$ bo'yicha quyidagi intervallarda belgisi o'zgaradi: $(-\infty, -5], [-5, -1], [-1, 4), (4, \infty)$ Grafik va ifodaning belgisi asosida yechim: Javob: D) $(-\infty; -5] \cup [-1; 4)$ 6. Masala: $f(x) = \begin{cases} x^2 - 9, & x \geq 0 \\ x^2 + 9, & x < 0 \end{cases}$ bo'lsa, $f(f(2))$ ni toping. Hisoblash: $f(2) = 2^2 - 9 = 4 - 9 = -5$ $f(-5)$ uchun $x < 0$ bo'lgani uchun: $f(-5) = (-5)^2 + 9 = 25 + 9 = 34$ Javob: $f(f(2)) = 34$ variantlarda yo'q, lekin eng yaqin javob yo'q. 7. Masala: $y = \frac{3x - 13}{x - 3}$ funksiyaning grafigi qaysi choraklardan o'tadi? Hisoblash: Asimptotalar: $x = 3$ (vertikal), $y = 3$ (gorizontal) Test nuqtalar bilan aniqlaymiz: $x=0$, $y = \frac{-13}{-3} = \frac{13}{3} > 0$ (II chorak) $x=4$, $y = \frac{12 - 13}{1} = -1$ (IV chorak) $x=-1$, $y = \frac{-3 - 13}{-4} = \frac{-16}{-4} = 4$ (II chorak) $x=5$, $y = \frac{15 - 13}{2} = 1$ (IV chorak) $x=10$, $y = \frac{30 - 13}{7} = \frac{17}{7} > 0$ (I chorak) Demak, grafigi I, II va IV choraklardan o'tadi. Javob: A) I, II va IV 8. Masala: $f(x) = \frac{2x - 5}{x + 3}$ va $g(x) = 7$ bo'lsa, $g(f(x))$ ni toping. Hisoblash: $g(f(x)) = g$ har doim 7 ga teng, chunki $g(x) = 7$ konstant funksiya. Javob: B) 7 Umumiy: 8 ta masala yechildi.