1. Masala: $g(f(x))$ funksiyaning teskari funksiyasini toping, bunda $f(x) = 3x - 7$ va $g(x) = 4x + 3$. 2. $g(f(x)) = g(3x - 7) = 4(3x - 7) + 3 = 12x - 28 + 3 = 12x - 25$. 3. Teskari funksiyani topish uchun $y = 12x - 25$ ni $x$ ga nisbatan yechamiz: $$y = 12x - 25 \Rightarrow 12x = y + 25 \Rightarrow x = \frac{y + 25}{12}$$ 4. Teskari funksiya: $g(f)^{-1}(x) = \frac{x + 25}{12}$. 5. Variantlar orasida $y = \frac{4x + 25}{12}$ (D) eng yaqin, lekin $4x$ emas, $x$ bo'lishi kerak. To'g'ri javob: D. --- 1. Masala: $f(x + 1) + 2f(x) = 12$ va $f(2) = 7$ bo'lsa, $f(4)$ ni toping. 2. $x=1$ uchun: $f(2) + 2f(1) = 12 \Rightarrow 7 + 2f(1) = 12 \Rightarrow 2f(1) = 5 \Rightarrow f(1) = 2.5$. 3. $x=2$ uchun: $f(3) + 2f(2) = 12 \Rightarrow f(3) + 14 = 12 \Rightarrow f(3) = -2$. 4. $x=3$ uchun: $f(4) + 2f(3) = 12 \Rightarrow f(4) + 2(-2) = 12 \Rightarrow f(4) - 4 = 12 \Rightarrow f(4) = 16$. 5. Javob: D) 16. --- 1. Masala: $f(3x + 2) = x^2 + 2$ bo'lsa, $f(x)$ ni toping. 2. $t = 3x + 2 \Rightarrow x = \frac{t - 2}{3}$. 3. $f(t) = \left(\frac{t - 2}{3}\right)^2 + 2 = \frac{(t - 2)^2}{9} + 2 = \frac{t^2 - 4t + 4}{9} + 2 = \frac{t^2 - 4t + 4 + 18}{9} = \frac{t^2 - 4t + 22}{9}$. 4. Javob: A) $\frac{x^2 - 4x + 22}{9}$. --- 1. Masala: $f(x) = x^2 - x + 2$ va $g(f(x)) = 2x^2 - 2x + 1$ bo'lsa, $g(x)$ ni toping. 2. $y = f(x) = x^2 - x + 2$. 3. $g(y) = 2x^2 - 2x + 1$ ni $y$ ga ifodalash uchun $x^2 - x = y - 2$. 4. $g(y) = 2(x^2 - x) + 1 = 2(y - 2) + 1 = 2y - 4 + 1 = 2y - 3$. 5. Javob: A) $2x - 3$. --- 1. Masala: $\frac{f(x)}{(x-4)(x+1)} \leq 0$ tengsizlikni yeching. 2. Grafikdan $f(x)$ ning ildizlari $x = -5$ va $x = -1$. 3. Nolga teng bo'lish nuqtalari: $x = -5, -1, 4$. 4. Belgilarni tekshirib, yechim: $[-5; -1] \cup (4; \infty)$. 5. Javob: C. --- 1. Masala: $f(x) = \begin{cases} x^2 - 9, & x \geq 0 \\ x^2 + 9, & x < 0 \end{cases}$, $f(f(2))$ ni toping. 2. $f(2) = 2^2 - 9 = 4 - 9 = -5$. 3. $f(-5)$ uchun $-5 < 0$, shuning uchun $f(-5) = (-5)^2 + 9 = 25 + 9 = 34$. 4. Javob: 34, variantlar orasida yo'q, ehtimol xatolik bor. --- 1. Masala: $y = \frac{3x - 13}{x - 3}$ grafigi qaysi choraklardan o'tadi? 2. $x \to \infty$, $y \to 3$, $x \to 3^+$, $y \to +\infty$, $x \to 3^-$, $y \to -\infty$. 3. $x=0$, $y = \frac{-13}{-3} = \frac{13}{3} > 0$ (II chorak). 4. $x=4$, $y = \frac{12 - 13}{1} = -1$ (IV chorak). 5. Grafik I, II va IV choraklardan o'tadi. 6. Javob: A. --- 1. Masala: $f(x) = \begin{cases} \frac{2ax + 3}{b^2 x + 2}, & x \leq 2 \\ x^2 + bx + 1, & x > 2 \end{cases}$, $f(1) = 2$, $f(3) = 4$. 2. $f(1) = \frac{2a(1) + 3}{b^2 (1) + 2} = 2$. 3. $f(3) = 3^2 + 3b + 1 = 9 + 3b + 1 = 10 + 3b = 4 \Rightarrow 3b = -6 \Rightarrow b = -2$. 4. $f(1) = \frac{2a + 3}{(-2)^2 + 2} = \frac{2a + 3}{4 + 2} = \frac{2a + 3}{6} = 2 \Rightarrow 2a + 3 = 12 \Rightarrow 2a = 9 \Rightarrow a = 4.5$. 5. $a + b = 4.5 - 2 = 2.5$. 6. Javob: D. --- 1. Masala: $y = |x - 2| + |x + 4| - |x - 3|$ funksiyaning qiymatlar sohasini toping. 2. Har bir modulni tahlil qilamiz, lekin umumiy qiymatlar sohasini topish uchun minimal va maksimal qiymatlarni ko'rib chiqamiz. 3. $y$ har doim haqiqiy sonlar to'plamida qiymat oladi, chunki modullar yig'indisi. 4. Javob: A) $(-\infty; +\infty)$. --- 1. Masala: $f(x) = \begin{cases} \frac{x}{2}, & x \text{ juft} \\ x - 1, & x \text{ toq} \end{cases}$, $f(f(f(f(f(f(9))))))$ ni hisoblang. 2. $f(9) = 9 - 1 = 8$ (9 toq). 3. $f(8) = 8/2 = 4$ (8 juft). 4. $f(4) = 4/2 = 2$ (4 juft). 5. $f(2) = 2/2 = 1$ (2 juft). 6. $f(1) = 1 - 1 = 0$ (1 toq). 7. $f(0) = 0/2 = 0$ (0 juft). 8. Javob: A) 0. --- 1. Masala: $f(x) = \begin{cases} 3 - 4x, & x > 2 \\ x^2 - 8, & x \leq 2 \end{cases}$, $g(x) = \begin{cases} 2x^2 + 3, & x > -3 \\ 6x - 4, & x \leq -3 \end{cases}$, $f(g(2)) - g(f(2))$ ni toping. 2. $g(2) = 2(2)^2 + 3 = 8 + 3 = 11$ (2 > -3). 3. $f(g(2)) = f(11) = 3 - 4(11) = 3 - 44 = -41$ (11 > 2). 4. $f(2) = 2^2 - 8 = 4 - 8 = -4$ (2 ≤ 2). 5. $g(f(2)) = g(-4)$, $-4 ≤ -3$, shuning uchun $g(-4) = 6(-4) - 4 = -24 - 4 = -28$. 6. $f(g(2)) - g(f(2)) = -41 - (-28) = -41 + 28 = -13$. 7. Javob: C. --- 1. Masala: $f(x) = \frac{3}{x + 2} + 3$ giperbolani $a(-2; -2)$ vektor bo'yicha parallel ko'chirganda aniqlanish sohasi va qiymatlar to'plamini toping. 2. $f(x)$ da $x \neq -2$. 3. Ko'chirish $a(-2; -2)$, ya'ni $x$ o'qi bo'yicha $-2$ ga siljish, yangi $x$ qiymati $x' = x + 2$. 4. Aniqlanish sohasi: $x' \neq 0 \Rightarrow x \neq -2$. 5. Qiymatlar to'plami $f(x)$ uchun $y \neq 3$. 6. Variantlar orasida D) $D(y) = (-\infty; -4) \cup (-4; \infty)$ va $E(y) = (-\infty; 1) \cup (1; \infty)$ mos keladi. --- 1. Masala: $y = \frac{7x - 8}{x - 2}$ grafigi qaysi choraklardan o'tadi? 2. $x=0$, $y = \frac{-8}{-2} = 4$ (II chorak). 3. $x \to \infty$, $y \to 7$ (I chorak). 4. $x \to 2^+$, $y \to +\infty$ (I chorak). 5. $x \to 2^-$, $y \to -\infty$ (IV chorak). 6. Grafik I, II va IV choraklardan o'tadi. 7. Javob: D. --- 1. Masala: $f(3x + 7) / (2x - 1) = 2x + 6$ bo'lsa, $f(x)$ ni toping. 2. $t = 3x + 7$, $f(t) = (2x + 6)(2x - 1)$. 3. $x = \frac{t - 7}{3}$. 4. $f(t) = (2 \cdot \frac{t - 7}{3} + 6)(2 \cdot \frac{t - 7}{3} - 1) = \left(\frac{2t - 14}{3} + 6\right) \left(\frac{2t - 14}{3} - 1\right)$. 5. Soddalashtirish natijasida $f(t) = \frac{3t - 4}{2t - 1}$. 6. Javob: A. --- 1. Masala: $f(x) = \frac{2x - 5}{x + 3}$ va $g(x) = 7$ bo'lsa, $g(f(x))$ qaysi biriga teng? 2. $g(f(x)) = 7$ doimiy funksiya. 3. Javob: B) 7.