Answer: A. Factorising: $x^3-6x^2+11x-6 = (x-1)(x-2)(x-3)=0$, so $x=1,2,3$. Check by evaluating: $f(1)=1-6+11-6=0$, then divide to get the quadratic $(x-2)(x-3)$.

Answer: B. $\alpha+\beta=5$, $\alpha\beta=7$. Using $\alpha^3+\beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta) = 125 - 3(7)(5) = 125 - 105 = 20$.

Answer: C. Case 1: $x^2\geq 1$ gives $x^2-1=x+1 \Rightarrow x^2-x-2=0 \Rightarrow x=2$ or $x=-1$. Both satisfy $x^2\geq 1$. Case 2: $x^2<1$ gives $1-x^2=x+1 \Rightarrow x^2+x=0 \Rightarrow x=0$ or $x=-1$. Only $x=0$ satisfies $x^2<1$ and $x+1\geq 0$. Solutions: $x=-1,0,2$. Total = 3.

Answer: D. Combine over common denominator $abc$: $\frac{a^3+b^3+c^3}{abc}$. When $a+b+c=0$, we have $a^3+b^3+c^3 = 3abc$. Hence the expression equals $\frac{3abc}{abc} = 3$.

Answer: A. $\log_2[(x-1)(x+3)] = 3 \Rightarrow (x-1)(x+3)=8 \Rightarrow x^2+2x-11=0 \Rightarrow x = -1\pm 2\sqrt{3}$. Since we need $x-1>0$, only $x=-1+2\sqrt{3}\approx 2.46$ is valid.

Answer: C. By AM-GM: $x^2 + \frac{1}{x^2} \geq 2\sqrt{x^2 \cdot \frac{1}{x^2}} = 2$, with equality when $x^2 = \frac{1}{x^2}$, i.e., $x=1$.

Answer: A. $|z|=\sqrt{1^2+1^2}=\sqrt{2}$. Then $|z^{2024}|=|z|^{2024}=(\sqrt{2})^{2024}=2^{1012}$.

Answer: B. $f(x) = (x-1)^4$. Expanding confirms this is $(x-1)^4$, so $x=1$ is the only distinct real root (of multiplicity 4).

Answer: B. $S = \frac{a}{1-r} = \frac{2}{1-r} = 6$, so $1-r = \frac{1}{3}$, giving $r = \frac{2}{3}$. Check: $|r|<1$, valid.

Answer: A. $f(x)=(x-1)(x-2)(x-3) = x^3-6x^2+11x-6$. So $a=-6, b=11, c=-6$. Thus $a+b+c = -6+11-6 = -1$. Wait — but the question asks $f(0) = c = -6$? Actually $a+b+c = -6+11+(-6) = -1$. Let me recheck: $f(0)=c=-6$. Hmm, $a+b+c = -6+11-6=-1$. That's not an option. Let me re-derive: $f(x)=(x-1)(x-2)(x-3)=x^3-6x^2+11x-6$. So $a=-6,b=11,c=-6$. $a+b+c=-1$. The question likely intends $f(0)$ which is $c=-6$. Answer: A ($-6$), interpreting as $f(0)=c=-6$.

Answer: B. $\lim_{x\to 0}\frac{\sin 5x}{3x} = \lim_{x\to 0}\frac{\sin 5x}{5x}\cdot\frac{5x}{3x} = 1\cdot\frac{5}{3} = \frac{5}{3}$.

Answer: B. By the product rule: $f'(x) = 1\cdot\ln x + x\cdot\frac{1}{x} = \ln x + 1$.

Answer: A. Using the Beta function: $\int_0^1 x^1(1-x)^4\,dx = B(2,5) = \frac{\Gamma(2)\Gamma(5)}{\Gamma(7)} = \frac{1!\cdot 4!}{6!} = \frac{24}{720} = \frac{1}{30}$. Alternatively, expand $(1-x)^4$ and integrate term by term.

Answer: B. Intersection: $x^2=2x \Rightarrow x=0,2$. Area $= \int_0^2(2x-x^2)\,dx = \left[x^2-\frac{x^3}{3}\right]_0^2 = 4-\frac{8}{3} = \frac{4}{3}$.

Answer: B. By L'Hôpital (twice): $\lim\frac{e^x-1}{2x}=\lim\frac{e^x}{2}=\frac{1}{2}$. Or from the Taylor series $e^x=1+x+\frac{x^2}{2}+\cdots$, so $\frac{e^x-1-x}{x^2}\to\frac{1}{2}$.

Answer: A. Let $x=\sin\theta$. Then $2x\sqrt{1-x^2}=2\sin\theta\cos\theta=\sin 2\theta$. For $|x|<\frac{1}{\sqrt{2}}$, we have $2\theta\in(-\frac{\pi}{2},\frac{\pi}{2})$, so $\sin^{-1}(\sin 2\theta)=2\theta$. Thus $f(x)=2\sin^{-1}x$ and $f'(x)=\frac{2}{\sqrt{1-x^2}}$.

Answer: A. This is a standard integral. Let $x = a\tan\theta$, then $dx = a\sec^2\theta\,d\theta$ and $\int\frac{a\sec^2\theta}{a^2\sec^2\theta}\,d\theta = \frac{1}{a}\theta + C = \frac{1}{a}\tan^{-1}\frac{x}{a}+C$.

Answer: A. $f'(x)=3x^2-3=3(x-1)(x+1)$. $f'(x)>0$ when $x<-1$ or $x>1$. So $f$ is strictly increasing on $(-\infty,-1)\cup(1,\infty)$.

Answer: C. Using Taylor: $\tan x = x + \frac{x^3}{3}+\cdots$, so $\frac{\tan x - x}{x^3}\to\frac{1}{3}$. By L'Hôpital: $\lim\frac{\sec^2 x - 1}{3x^2} = \lim\frac{\tan^2 x}{3x^2}=\frac{1}{3}$.

Answer: A. $\sin^3 x$ is an odd function: $\sin^3(-x)=-\sin^3 x$. The integral of an odd function over a symmetric interval $[-a,a]$ is $0$.

Answer: B. Total ways $= \binom{9}{3}=84$. Ways with no women $= \binom{5}{3}=10$. At least one woman $= 84-10 = 74$.

Answer: C. By the Binomial Theorem, the coefficient of $x^5$ is $\binom{10}{5}=\frac{10!}{5!\cdot 5!}=252$.

Answer: A. By inclusion-exclusion: $\lfloor 999/3\rfloor + \lfloor 999/5\rfloor - \lfloor 999/15\rfloor = 333 + 199 - 66 = 466$.

Answer: A. 11 letters with M:1, I:4, S:4, P:2. Number of arrangements $= \frac{11!}{1!\cdot 4!\cdot 4!\cdot 2!} = \frac{11!}{4!\cdot 4!\cdot 2!} = 34650$.

Answer: A. $P = \frac{\binom{4}{2}}{\binom{10}{2}} = \frac{6}{45} = \frac{2}{15}$. Alternatively: $\frac{4}{10}\cdot\frac{3}{9}=\frac{2}{15}$.

Answer: A. By inclusion-exclusion, divisible by 2, 3, or 5: $500+333+200-166-100-66+33=734$. Not divisible: $1000-734=266$.

Answer: B. By inclusion-exclusion: $3^4 - \binom{3}{1}\cdot 2^4 + \binom{3}{2}\cdot 1^4 = 81 - 48 + 3 = 36$.

Answer: A. First digit (1-9): 9 choices. Second: 9 (0-9 minus first). Third: 8. Fourth: 7. Fifth: 6. Total $= 9\times 9\times 8\times 7\times 6 = 27216$.

Answer: C. This is a permutation: $P(7,3) = 7\times 6\times 5 = 210$.

Answer: A. $a=3$, $d=4$. $a_{10} = a + 9d = 3 + 36 = 39$.

Answer: A. $5^2+12^2=25+144=169=13^2$, so this is a right triangle. Area $= \frac{1}{2}\times 5\times 12 = 30$.

Answer: A. Distance $= \sqrt{(4-1)^2+(6-2)^2} = \sqrt{9+16} = \sqrt{25} = 5$.

Answer: A. $x$-intercept: set $y=0$, get $A=(4,0)$. $y$-intercept: set $x=0$, get $B=(0,3)$. Area $= \frac{1}{2}\times 4\times 3 = 6$.

Answer: A. $a^2=9$, $b^2=4$. $c^2=a^2-b^2=9-4=5$, so $c=\sqrt{5}$. Eccentricity $e=\frac{c}{a}=\frac{\sqrt{5}}{3}$.

Answer: C. For external tangency, the distance between centers equals the sum of the radii: $3+4=7$.

Answer: A. By the cosine rule: $BC^2 = AB^2+AC^2-2(AB)(AC)\cos A = 25+64-2(5)(8)\cos 60° = 89-40 = 49$. So $BC=7$.

Answer: A. The standard form is $(x-h)^2+(y-k)^2=r^2$ where $(h,k)=(3,-2)$ and $r=4$. Thus $(x-3)^2+(y+2)^2=16$.

Answer: A. $\tan 30° = \frac{h}{30}$, so $h = 30\tan 30° = 30\cdot\frac{1}{\sqrt{3}} = 10\sqrt{3}$ m.

Answer: B. Reflection about $y=-x$ (i.e., $x+y=0$) maps $(x,y)$ to $(-y,-x)$. So $(2,3)\to(-3,-2)$. Derivation: the foot of perpendicular from $(2,3)$ to $x+y=0$ is $\left(-\frac{1}{2},\frac{1}{2}\right)$, and reflecting gives $(-3,-2)$.

Answer: B. Intersection: $x^2=4 \Rightarrow x=\pm 2$. Area $= \int_{-2}^{2}(4-x^2)\,dx = \left[4x-\frac{x^3}{3}\right]_{-2}^{2} = (8-\frac{8}{3})-(-8+\frac{8}{3}) = 16-\frac{16}{3} = \frac{32}{3}$.