AP Precalculus · AI 押题冲刺AP Precalculus · AI Cram

2024 首考的新科目,真题少 — 但这正是 AI 押题最有价值的场景。覆盖 8 道公开真题 FRQ(2024 + 2025 北美) + 40 道 AI 押题 MCQ。New course (first administered 2024). Past papers are scarce — which is exactly where AI prediction shines. Covers 8 official released FRQs (2024 + 2025) + 40 AI-predicted MCQs.

📝 考试格式Exam Format:

Section I · 40 MCQ · 2 小时(Part A 28 无 calc · Part B 12 必 calc)· 占 62.5%Section I · 40 MCQ · 2 h (Part A 28 no-calc · Part B 12 calc) · 62.5%
Section II · 4 FRQ · 1 小时(Part A 2 calc · Part B 2 no calc)· 占 37.5%Section II · 4 FRQ · 1 h (Part A 2 calc · Part B 2 no-calc) · 37.5%
U1 多项式/有理 ~40% · U2 指对 ~27% · U3 三角/极坐标 ~33%(U4 不考)U1 Polynomial / Rational ~40% · U2 Exp / Log ~27% · U3 Trig / Polar ~33% (U4 not tested)

Part I · 押题分析Part I · Hot Slots

📌 三秒看懂:Precalc 4 道 FRQ 槽位都是固定题型3-Second Summary: All 4 FRQ slots are fixed archetypes

FRQ1 函数概念 / 表格分析Function Concepts / Table Analysis 100% calc
FRQ2 非周期建模(指 / 对 / 多项式)Non-Periodic Modeling (exp / log / poly) 100% calc
FRQ3 周期建模(三角 sin / cos)Periodic Modeling (sin / cos) 100% no-calc
FRQ4 符号变换(组合 / 反函数 / 等式 / 三角恒等)Symbolic Manipulation (composition / inverse / identity) 100% no-calc

因为 4 个槽位题型固定,这门课押题难度比 AP Calc 低。重点是模板熟练。Because all 4 slots are predictable, predicting Precalc is easier than AP Calc — focus on mastering templates.

Part II · 历年真题 FRQ + 评分Part II · Past FRQs + Scoring

2024 + 2025 北美卷,合计 8 道。点击展开看原题,二级展开看官方评分细则。2024 + 2025 North America forms, 8 FRQs total. Expand to view; nested toggle reveals official scoring guidelines.

2024 北美卷 · 4 道 FRQ2024 North America · 4 FRQs

2024 FRQ1 · 非周期建模(Function Concepts)Function Concepts

📋 查看评分细则 / Show scoring

2024 FRQ2 · 非周期建模(Modeling Non-Periodic)Modeling Non-Periodic

📋 查看评分细则 / Show scoring

2024 FRQ3 · 周期建模(Modeling Periodic)Modeling Periodic

📋 查看评分细则 / Show scoring

2024 FRQ4 · 符号变换(Symbolic Manipulation)Symbolic Manipulation

📋 查看评分细则 / Show scoring

2025 北美卷 · 4 道 FRQ2025 North America · 4 FRQs

2025 FRQ1 · 非周期建模(Function Concepts)Function Concepts

📋 查看评分细则 / Show scoring

2025 FRQ2 · 非周期建模(Modeling Non-Periodic)Modeling Non-Periodic

📋 查看评分细则 / Show scoring

2025 FRQ3 · 周期建模(Modeling Periodic)Modeling Periodic

📋 查看评分细则 / Show scoring

2025 FRQ4 · 符号变换(Symbolic Manipulation)Symbolic Manipulation

📋 查看评分细则 / Show scoring

Part III · MCQ 题库(64 题)Part III · MCQ Bank (64 Qs)

40 道 AI 押题 + 24 道 CED 官方样题(50 no-calc · 14 calc)。按单元 / 来源 / 计算器筛选,点击 A–D 即时判分。40 AI-predicted + 24 official CED samples (50 no-calc · 14 calc). Filter by unit / source / calc, then click A–D to grade.

MCQ 题库 — 64 题MCQ Bank — 64 questions

MCQ 1

U1 多项式 / 有理 / U1 Polynomial / Rational

AI 押题AI Pick

A polynomial function $p$ has the following end behavior: \[\lim_{x\to-\infty} p(x) = +\infty \quad\text{and}\quad \lim_{x\to+\infty} p(x) = -\infty.\] Which of the following statements about $p$ must be true?

(A) $p$ has even degree and a positive leading coefficient.
(B) $p$ has odd degree and a negative leading coefficient.
(C) $p$ has odd degree and a positive leading coefficient.
(D) $p$ has even degree and a negative leading coefficient.

📖 查看解题思路 / Show solution

Because the two ends of the graph go in opposite directions (up on the left, down on the right), $p$ must have odd degree. Since $p(x)\to-\infty$ as $x\to+\infty$, the leading coefficient is negative. Therefore $p$ has odd degree with a negative leading coefficient.

MCQ 2

U1 多项式 / 有理 / U1 Polynomial / Rational

AI 押题AI Pick

Let $p$ be the polynomial defined by \[p(x)=(x^2-9)(x+1)^3(x-2)^2.\] Counting each zero according to its multiplicity, how many real zeros does $p$ have?

(A) 4
(B) 5
(C) 6
(D) 7

📖 查看解题思路 / Show solution

Factoring $x^2-9=(x-3)(x+3)$ gives $p(x)=(x-3)(x+3)(x+1)^3(x-2)^2$. The real zeros are $x=3$ (multiplicity 1), $x=-3$ (multiplicity 1), $x=-1$ (multiplicity 3), and $x=2$ (multiplicity 2). The total, counted with multiplicity, is $1+1+3+2=7$.

MCQ 3

U1 多项式 / 有理 / U1 Polynomial / Rational

AI 押题AI Pick

Let $f(x) = \dfrac{x^2 - x - 6}{x^2 - 4x + 3}$. Which of the following correctly describes the vertical asymptote(s) and removable discontinuity (hole) of the graph of $f$?

(A) Vertical asymptotes at $x = 1$ and $x = 3$; no removable discontinuity.
(B) Vertical asymptote at $x = 3$; removable discontinuity at $x = 1$.
(C) Vertical asymptote at $x = 1$; removable discontinuity at $x = 3$.
(D) Vertical asymptotes at $x = 1$ and $x = 3$; removable discontinuity at $x = -2$.

📖 查看解题思路 / Show solution

Factor: $f(x) = \dfrac{(x-3)(x+2)}{(x-1)(x-3)}$. The factor $(x-3)$ cancels, producing a removable discontinuity at $x=3$. The remaining factor $(x-1)$ in the denominator does not cancel, so $x=1$ is a vertical asymptote. The value $x=-2$ is a zero of $f$, not a hole.

MCQ 4

U1 多项式 / 有理 / U1 Polynomial / Rational

AI 押题AI Pick

Let $f(x) = \dfrac{2x^3 - 3x^2 + x - 5}{x^2 + 1}$. Which of the following is the equation of the slant (oblique) asymptote of the graph of $f$?

(A) $y = 2x - 3$
(B) $y = 2x + 3$
(C) $y = 2x$
(D) $y = 2x - 1$

📖 查看解题思路 / Show solution

Because the degree of the numerator is exactly one more than the degree of the denominator, the graph has a slant asymptote given by the polynomial quotient. Performing the division $(2x^3 - 3x^2 + x - 5) \div (x^2 + 1)$: $2x^3 \div x^2 = 2x$, and $2x(x^2+1) = 2x^3 + 2x$, leaving $-3x^2 - x - 5$. Then $-3x^2 \div x^2 = -3$, and $-3(x^2+1) = -3x^2 - 3$, leaving remainder $-x - 2$. The quotient is $2x - 3$, so the slant asymptote is $y = 2x - 3$.

MCQ 5

U1 多项式 / 有理 / U1 Polynomial / Rational

AI 押题AI Pick

The polynomial function $f$ is defined by $f(x) = x^3 - 6x^2$. The function $g$ is defined by $g(x) = \dfrac{1}{2}\,f(x+1)$. Which of the following is an equivalent expression for $g(x)$?

(A) $\dfrac{1}{2}(x-1)^2(x-7)$
(B) $\dfrac{1}{2}(x+1)^2(x-5)$
(C) $2(x+1)^2(x-5)$
(D) $\dfrac{1}{2}(x+1)^2(x-6)$

📖 查看解题思路 / Show solution

Factor $f(x) = x^2(x-6)$, so $f(x+1) = (x+1)^2\big((x+1)-6\big) = (x+1)^2(x-5)$. Multiplying by $\tfrac{1}{2}$ gives $g(x) = \tfrac{1}{2}(x+1)^2(x-5)$.

MCQ 6

U1 多项式 / 有理 / U1 Polynomial / Rational

AI 押题AI Pick

Let $f(x) = x^3 - 2x^2 + 5$. What is the average rate of change of $f$ on the interval $[-1, 3]$?

(A) $3$
(B) $6$
(C) $-3$
(D) $2$

📖 查看解题思路 / Show solution

Compute $f(3) = 27 - 18 + 5 = 14$ and $f(-1) = -1 - 2 + 5 = 2$. The average rate of change is $\dfrac{f(3) - f(-1)}{3 - (-1)} = \dfrac{14 - 2}{4} = 3$.

MCQ 7

U1 多项式 / 有理 / U1 Polynomial / Rational

AI 押题AI Pick

The polynomial function $p$ is defined by $p(x) = (3x+5)(x-2)^3(4-x)$. What is the sum of the distinct real zeros of $p$?

(A) $\dfrac{23}{3}$
(B) $-\dfrac{11}{3}$
(C) $\dfrac{13}{3}$
(D) $\dfrac{25}{3}$

📖 查看解题思路 / Show solution

Setting each factor equal to zero: $3x+5=0 \Rightarrow x=-\tfrac{5}{3}$; $(x-2)^3=0 \Rightarrow x=2$; $4-x=0 \Rightarrow x=4$. The distinct real zeros are $-\tfrac{5}{3},\,2,\,4$, and their sum is $-\tfrac{5}{3}+2+4=\tfrac{13}{3}$.

MCQ 8

U1 多项式 / 有理 / U1 Polynomial / Rational

AI 押题AI Pick

What are all real solutions to the equation \[\dfrac{x}{x-2} + \dfrac{3}{x+1} = \dfrac{6}{x^{2}-x-2}\,?\]

(A) $x = 2$ and $x = -6$
(B) $x = 2$ only
(C) $x = 6$ only
(D) $x = -6$ only

📖 查看解题思路 / Show solution

Since $x^{2}-x-2=(x-2)(x+1)$, multiplying both sides by $(x-2)(x+1)$ gives $x(x+1)+3(x-2)=6$, which simplifies to $x^{2}+4x-12=0$, or $(x+6)(x-2)=0$. Thus $x=-6$ or $x=2$, but $x=2$ makes the original denominators zero and is extraneous. The only solution is $x=-6$.

MCQ 9

U1 多项式 / 有理 / U1 Polynomial / Rational

AI 押题AI Pick

Let $f$ be the polynomial function defined by $f(x) = x^3 - 6x^2 + 5$. On which of the following intervals is the graph of $f$ concave up?

(A) $(2, \infty)$
(B) $(-\infty, 2)$
(C) $(4, \infty)$
(D) $(0, 4)$

📖 查看解题思路 / Show solution

The graph of $f$ is concave up on intervals where the rate of change of $f$ is increasing. Computing successive average rates of change of $f$ over unit intervals starting at $x=0$ gives $-5, -11, -11, -5, 7, 25,\dots$; the second differences are $-6, 0, 6, 12, 18$, turning positive once $x>2$. Equivalently, the rate of change $3x^2-12x$ is itself increasing where $6x-12>0$, i.e., on $(2,\infty)$.

MCQ 10

U1 多项式 / 有理 / U1 Polynomial / Rational

AI 押题AI Pick

Let $f$ be the piecewise function defined by \[f(x)=\begin{cases} x^{2}-kx & \text{for } x<-2 \\ kx+12 & \text{for } x\ge -2 \end{cases}\] where $k$ is a real constant. For what value of $k$ is $f$ continuous at $x=-2$?

(A) $k=2$
(B) $k=-2$
(C) $k=4$
(D) $k=8$

📖 查看解题思路 / Show solution

Continuity at $x=-2$ requires the two pieces to agree there. From the left: $(-2)^2-k(-2)=4+2k$. From the right: $k(-2)+12=-2k+12$. Setting $4+2k=-2k+12$ gives $4k=8$, so $k=2$.

MCQ 11

U1 多项式 / 有理 / U1 Polynomial / Rational

AI 押题AI Pick

The polynomial $p(x)$ has degree 4 and real coefficients. Given that $2-3i$ and $-1$ are zeros of $p(x)$, which of the following must also be a zero of $p(x)$?

(A) $-2+3i$
(B) $2+3i$
(C) $1$
(D) $-2-3i$

📖 查看解题思路 / Show solution

By the Complex Conjugate Root Theorem, nonreal zeros of a polynomial with real coefficients occur in conjugate pairs. Since $2-3i$ is a zero, its conjugate $2+3i$ must also be a zero of $p(x)$.

MCQ 12

U1 多项式 / 有理 / U1 Polynomial / Rational

AI 押题AI Pick

The height of a young tree, in centimeters, is modeled by the function $H(t) = -0.3t^3 + 4t^2 + 2t + 50$, where $t$ is the time in weeks since the tree was planted. What is the average rate of change of the tree's height, in centimeters per week, over the interval from $t = 2$ to $t = 6$?

(A) $12.267$ centimeters per week
(B) $-18.400$ centimeters per week
(C) $18.400$ centimeters per week
(D) $73.600$ centimeters per week

📖 查看解题思路 / Show solution

Compute $H(2) = -0.3(8) + 4(4) + 2(2) + 50 = 67.6$ and $H(6) = -0.3(216) + 4(36) + 2(6) + 50 = 141.2$. The average rate of change is $\dfrac{H(6) - H(2)}{6 - 2} = \dfrac{141.2 - 67.6}{4} = \dfrac{73.6}{4} = 18.4$ centimeters per week.

MCQ 13

U1 多项式 / 有理 / U1 Polynomial / Rational

AI 押题AI Pick

The table below gives values of a function $f$ at selected values of $x$. \[ \begin{array}{|c|c|}\hline x & f(x) \\\hline 1 & -1.3 \\ 2 & 0.4 \\ 3 & 4.5 \\ 4 & 13.4 \\ 5 & 29.5 \\ 6 & 55.2 \\\hline \end{array} \] A cubic regression model of the form $y = ax^3 + bx^2 + cx + d$ is fit to the data. What is the value of $a$?

(A) $0.4$
(B) $-1.2$
(C) $2.5$
(D) $-3.0$

📖 查看解题思路 / Show solution

Entering the six data points into a calculator and performing cubic regression returns $y = 0.4x^3 - 1.2x^2 + 2.5x - 3$. The leading coefficient is $a = 0.4$. The other choices are the values of $b$, $c$, and $d$, respectively.

MCQ 14

U1 多项式 / 有理 / U1 Polynomial / Rational

AI 押题AI Pick

A bacterial culture is treated with an antibiotic at time $t = 0$ hours. For $t \geq 0$, the number of bacteria in the culture, measured in thousands, is modeled by \[ N(t) = \dfrac{15t + 60}{t + 12}. \] Based on the model, which of the following is the best interpretation of the long-run behavior of the bacteria population?

(A) The number of bacteria approaches 5,000 as $t$ increases without bound.
(B) The number of bacteria approaches 15 as $t$ increases without bound.
(C) The number of bacteria approaches 15,000 as $t$ increases without bound.
(D) The number of bacteria decreases without bound as $t$ increases.

📖 查看解题思路 / Show solution

Since $N(t)$ is a rational function whose numerator and denominator have the same degree, the horizontal asymptote is the ratio of the leading coefficients: $y = \tfrac{15}{1} = 15$. Because $N$ is measured in thousands of bacteria, $N(t) \to 15$ thousand, i.e., 15,000 bacteria, as $t \to \infty$.

MCQ 15

U1 多项式 / 有理 / U1 Polynomial / Rational

AI 押题AI Pick

The profit, in thousands of dollars, of a small company $t$ months after launching a new product is modeled by \[P(t)=-0.05t^{3}+0.9t^{2}-2t-5\] for $0\le t\le 16$. The company "breaks even" when $P(t)=0$. To the nearest thousandth, what is the largest value of $t$ at which the company breaks even during this time interval?

(A) $t\approx 4.612$
(B) $t\approx 10.760$
(C) $t\approx 11.250$
(D) $t\approx 14.854$

📖 查看解题思路 / Show solution

Using a graphing/numerical solver on $P(t)=-0.05t^{3}+0.9t^{2}-2t-5=0$ over $[0,16]$ gives two real zeros, approximately $t\approx 4.612$ and $t\approx 14.854$. The largest zero in the interval is $t\approx 14.854$. (Choice A is the smaller zero; choice B is the time of maximum profit found from $P'(t)=0$; choice C is the maximum profit value, not a zero.)

MCQ 16

U1 多项式 / 有理 / U1 Polynomial / Rational

AI 押题AI Pick

The graph of the polynomial function $p$ has $x$-intercepts at $x = -2$, $x = 1$, and $x = 4$, each of multiplicity 1. The graph passes through the point $(0, 8)$, and the end behavior of $p$ is given by $\displaystyle \lim_{x \to -\infty} p(x) = -\infty$ and $\displaystyle \lim_{x \to \infty} p(x) = \infty$. Which of the following is the solution set to the inequality $p(x) \le 0$?

(A) $(-\infty, -2) \cup (1, 4)$
(B) $(-\infty, -2] \cup [1, 4]$
(C) $[-2, 1] \cup [4, \infty)$
(D) $(-\infty, -2] \cup [4, \infty)$

📖 查看解题思路 / Show solution

Because $p$ is a cubic with positive leading coefficient (from the end behavior) and simple roots at $-2, 1, 4$, the sign of $p(x)$ alternates: negative on $(-\infty,-2)$, positive on $(-2,1)$, negative on $(1,4)$, and positive on $(4,\infty)$. Including the zeros (since the inequality is non-strict), $p(x)\le 0$ on $(-\infty,-2] \cup [1,4]$.

MCQ 17

U1 多项式 / 有理 / U1 Polynomial / Rational

CED 样题CED Sample

The polynomial function $p$ is given by $p(x) = -4x^3 + 3x^2 + 1$. Which of the following statements about the end behavior of $p$ is true?

(A) The sign of the leading term of $p$ is positive, and the degree of the leading term of $p$ is even; therefore, $\lim_{x \to -\infty} p(x) = \infty$ and $\lim_{x \to \infty} p(x) = \infty$.
(B) The sign of the leading term of $p$ is negative, and the degree of the leading term of $p$ is odd; therefore, $\lim_{x \to -\infty} p(x) = \infty$ and $\lim_{x \to \infty} p(x) = -\infty$.
(C) The sign of the leading term of $p$ is positive, and the degree of the leading term of $p$ is odd; therefore, $\lim_{x \to -\infty} p(x) = -\infty$ and $\lim_{x \to \infty} p(x) = \infty$.
(D) The sign of the leading term of $p$ is negative, and the degree of the leading term of $p$ is odd; therefore, $\lim_{x \to -\infty} p(x) = -\infty$ and $\lim_{x \to \infty} p(x) = \infty$.

MCQ 18

U1 多项式 / 有理 / U1 Polynomial / Rational

CED 样题CED Sample

The depth of water, in feet, at a certain place in a lake is modeled by a function $W$. The graph of $y = W(t)$ is shown for $0 \leq t \leq 30$, where $t$ is the number of days since the first day of a month. What are all intervals of $t$ on which the depth of water is increasing at a decreasing rate?

(A) $(3, 6)$ only
(B) $(3, 12)$
(C) $(0, 3)$ and $(18, 30)$ only
(D) $(0, 6)$ and $(18, 30)$

MCQ 19

U1 多项式 / 有理 / U1 Polynomial / Rational

CED 样题CED Sample

Which of the following functions has a zero at $x = 3$ and has a graph in the $xy$-plane with a vertical asymptote at $x = 2$ and a hole at $x = 1$?

(A) $h(x) = \dfrac{x^2 - 4x + 3}{x^2 - 3x + 2}$
(B) $j(x) = \dfrac{x^2 - 5x + 6}{x^2 - 3x + 2}$
(C) $k(x) = \dfrac{x - 3}{x^2 - 3x + 2}$
(D) $m(x) = \dfrac{x - 3}{x^2 - 4x + 3}$

MCQ 20

U1 多项式 / 有理 / U1 Polynomial / Rational

CED 样题CED Sample

The polynomial function $p$ is an odd function. If $p(3) = -4$ is a relative maximum of $p$, which of the following statements about $p(-3)$ must be true?

(A) $p(-3) = 4$ is a relative maximum.
(B) $p(-3) = -4$ is a relative maximum.
(C) $p(-3) = 4$ is a relative minimum.
(D) $p(-3) = -4$ is a relative minimum.

MCQ 21

U1 多项式 / 有理 / U1 Polynomial / Rational

CED 样题CED Sample

The function $g$ is given by $g(x) = x^3 - 3x^2 - 18x$, and the function $h$ is given by $h(x) = x^2 - 2x - 35$. Let $k$ be the function given by $k(x) = \dfrac{h(x)}{g(x)}$. What is the domain of $k$?

(A) all real numbers $x$ where $x \neq 0$
(B) all real numbers $x$ where $x \neq -5, x \neq 7$
(C) all real numbers $x$ where $x \neq -3, x \neq 0, x \neq 6$
(D) all real numbers $x$ where $x \neq -5, x \neq -3, x \neq 0, x \neq 6, x \neq 7$

MCQ 22

U1 多项式 / 有理 / U1 Polynomial / Rational

CED 样题CED Sample

The figure shown is the graph of a polynomial function $g$. Which of the following could be an expression for $g(x)$?

(A) $0.25(x - 5)(x - 1)(x + 8)$
(B) $0.25(x + 5)(x + 1)(x - 8)$
(C) $0.25(x - 5)^2(x - 1)(x + 8)$
(D) $0.25(x + 5)^2(x + 1)(x - 8)$

MCQ 23

U1 多项式 / 有理 / U1 Polynomial / Rational

CED 样题CED Sample

The table gives values for a polynomial function $f$ at selected values of $x$. Let $g(x) = af(bx) + c$, where $a$, $b$, and $c$ are positive constants. In the $xy$-plane, the graph of $g$ is constructed by applying three transformations to the graph of $f$ in this order: a horizontal dilation by a factor of $2$, a vertical dilation by a factor of $3$, and a vertical translation by $5$ units. What is the value of $g(-4)$?

$x$	$-8$	$-4$	$-2$	$-1$	$0$	$3$
$f(x)$	$87$	$55$	$5$	$-4$	$-7$	$20$

(A) $266$
(B) $170$
(C) $28$
(D) $20$

MCQ 24

U1 多项式 / 有理 / U1 Polynomial / Rational

CED 样题CED Sample

A food vendor developed a new sandwich type for sale. The vendor made estimates about the sales of the new sandwich type over time. A linear regression was used to develop a model for the sales over time. The figure shows a graph of the residuals of the linear regression. Which of the following statements about the linear regression is true?

📊 Residuals plot showing points scattered around $y = 0$ with a curved pattern, with residuals ranging from approximately $-20$ to $20$ over time.

(A) The linear model is not appropriate, because there is a clear pattern in the graph of the residuals.
(B) The linear model is not appropriate, because the graph of the residuals has more points above $0$ than below $0$.
(C) The linear model is appropriate, because there is a clear pattern in the graph of the residuals.
(D) The linear model is appropriate, because the positive residual farthest from $0$ and the negative residual farthest from $0$ are about the same distance, although more points are above than below $0$.

MCQ 25

U1 多项式 / 有理 / U1 Polynomial / Rational

CED 样题CED Sample

The temperature, in degrees Celsius (°C), in a city on a particular day is modeled by the function $T$ defined by $T(t) = \dfrac{75t^3 - 836t^2 + 3100t - 4185}{14t^2 + 10t - 35}$, where $t$ is measured in hours from 12 P.M. for $2 \le t \le 9$. Based on the model, how many hours did it take for the temperature to increase from $0°C$ to $5°C$?

(A) 7.701
(B) 5.420
(C) 4.114
(D) 2.280

MCQ 26

U2 指数 / 对数 / U2 Exponential / Logarithmic

AI 押题AI Pick

The expression $\left(3^{2x+1}\right)^{2} \cdot 3^{\,x-3}$ is equivalent to which of the following?

(A) $3^{5x-1}$
(B) $3^{3x-1}$
(C) $3^{\,4x^{2}-10x-6}$
(D) $9^{5x-1}$

📖 查看解题思路 / Show solution

Apply the power rule: $\left(3^{2x+1}\right)^{2}=3^{2(2x+1)}=3^{4x+2}$. Then use the product rule: $3^{4x+2}\cdot 3^{x-3}=3^{(4x+2)+(x-3)}=3^{5x-1}$.

MCQ 27

U2 指数 / 对数 / U2 Exponential / Logarithmic

AI 押题AI Pick

Which of the following is equivalent to $\log_b\left(\dfrac{x^{3}\sqrt{y}}{z^{2}}\right)$, where $x>0$, $y>0$, and $z>0$?

(A) $3\log_b x + \dfrac{1}{2}\log_b y + 2\log_b z$
(B) $\dfrac{3\log_b x + \frac{1}{2}\log_b y}{2\log_b z}$
(C) $\big(3\log_b x\big)\big(\tfrac{1}{2}\log_b y\big) - 2\log_b z$
(D) $3\log_b x + \dfrac{1}{2}\log_b y - 2\log_b z$

📖 查看解题思路 / Show solution

Apply the quotient property: $\log_b\left(\dfrac{x^{3}\sqrt{y}}{z^{2}}\right)=\log_b(x^{3}\sqrt{y})-\log_b(z^{2})$. Then the product property gives $\log_b(x^{3})+\log_b(y^{1/2})-\log_b(z^{2})$, and the power property yields $3\log_b x+\tfrac{1}{2}\log_b y-2\log_b z$.

MCQ 28

U2 指数 / 对数 / U2 Exponential / Logarithmic

AI 押题AI Pick

What is the solution to the equation $\log_2(x+1) + \log_2(x-1) = 3$?

(A) $x = 4$
(B) $x = -3$
(C) $x = 3$
(D) $x = \pm 3$

📖 查看解题思路 / Show solution

Combine the logs: $\log_2[(x+1)(x-1)] = 3$, so $(x+1)(x-1) = 2^3$, giving $x^2 - 1 = 8$ and $x^2 = 9$. The candidates are $x = \pm 3$, but the original equation requires $x - 1 > 0$, so $x = -3$ is extraneous. The only valid solution is $x = 3$.

MCQ 29

U2 指数 / 对数 / U2 Exponential / Logarithmic

AI 押题AI Pick

Let $f(x) = 3^{\,x-2} + 5$. Which of the following is an expression for $f^{-1}(x)$?

(A) $f^{-1}(x) = \log_{3}(x-5) + 2$
(B) $f^{-1}(x) = \log_{3}(x+5) + 2$
(C) $f^{-1}(x) = \log_{3}(x-5) - 2$
(D) $f^{-1}(x) = \log_{3}(x-2) + 5$

📖 查看解题思路 / Show solution

Set $y = 3^{\,x-2} + 5$ and solve for $x$: $y - 5 = 3^{\,x-2}$, so $\log_{3}(y-5) = x - 2$, giving $x = \log_{3}(y-5) + 2$. Swapping variables yields $f^{-1}(x) = \log_{3}(x-5) + 2$.

MCQ 30

U2 指数 / 对数 / U2 Exponential / Logarithmic

AI 押题AI Pick

Let $f(x) = 2^x$ and $g(x) = \log_{2}(x - 1) + 3$. Which of the following is an expression for $(f \circ g)(x)$ on the domain $x > 1$?

(A) $x + 7$
(B) $(x-1)^{3}$
(C) $8x - 8$
(D) $8x + 8$

📖 查看解题思路 / Show solution

We have $(f\circ g)(x) = 2^{\log_{2}(x-1) + 3} = 2^{\log_{2}(x-1)} \cdot 2^{3} = (x-1)\cdot 8 = 8x - 8$. Choice A reflects converting the sum in the exponent to a sum of powers, choice B reflects converting the sum in the exponent to a product, and choice D reflects a sign error on $-1$.

MCQ 31

U2 指数 / 对数 / U2 Exponential / Logarithmic

AI 押题AI Pick

An exponential function $f$ is defined by $f(x)=ab^{x}$, where $a$ and $b$ are positive constants. If $f(1)=6$ and $f(4)=162$, which of the following is an expression for $f(x)$?

(A) $\dfrac{2}{9}\cdot 27^{x}$
(B) $6\cdot 3^{x}$
(C) $2\cdot 3^{x-1}$
(D) $2\cdot 3^{x}$

📖 查看解题思路 / Show solution

Form the ratio $\dfrac{f(4)}{f(1)}=\dfrac{ab^{4}}{ab}=b^{3}=\dfrac{162}{6}=27$, so $b=3$. Then $ab=6$ gives $a=2$. Therefore $f(x)=2\cdot 3^{x}$.

MCQ 32

U2 指数 / 对数 / U2 Exponential / Logarithmic

AI 押题AI Pick

Which of the following is equivalent to $\left(\log_{3} 8\right)\left(\log_{2} 9\right)$?

(A) $\log_{6} 72$
(B) $5$
(C) $6$
(D) $12$

📖 查看解题思路 / Show solution

Apply the power rule: $\log_{3} 8 = 3\log_{3} 2$ and $\log_{2} 9 = 2\log_{2} 3$. Then $\left(3\log_{3} 2\right)\left(2\log_{2} 3\right) = 6\,(\log_{3} 2)(\log_{2} 3)$. Since $(\log_{3} 2)(\log_{2} 3) = 1$ by the change-of-base identity, the product equals $6$.

MCQ 33

U2 指数 / 对数 / U2 Exponential / Logarithmic

AI 押题AI Pick

The table below shows values of a function $f$ believed to be exponential. \[\begin{array}{|c|c|}\hline x & f(x) \\\hline 0 & 24 \\ 1 & 36 \\ 2 & 54 \\ 3 & 81 \\\hline\end{array}\] Assuming $f$ is exponential, what is the predicted value of $f(6)$, rounded to the nearest thousandth?

(A) 96.000
(B) 216.000
(C) 273.375
(D) 410.063

📖 查看解题思路 / Show solution

The common ratio is $\dfrac{36}{24}=\dfrac{54}{36}=\dfrac{81}{54}=1.5$, and the initial value is $f(0)=24$, so $f(x)=24(1.5)^x$. Then $f(6)=24(1.5)^6=24(11.390625)=273.375$.

MCQ 34

U2 指数 / 对数 / U2 Exponential / Logarithmic

AI 押题AI Pick

An investment of \$5{,}000 is deposited into an account that earns an annual interest rate of $4.5\%$ compounded continuously. To the nearest tenth of a year, how long will it take for the value of the investment to triple?

(A) $24.4$ years
(B) $15.4$ years
(C) $10.6$ years
(D) $2.4$ years

📖 查看解题思路 / Show solution

Using $A = Pe^{rt}$ with $A/P = 3$ and $r = 0.045$, solve $3 = e^{0.045 t}$. Then $t = \dfrac{\ln 3}{0.045} \approx \dfrac{1.0986}{0.045} \approx 24.4$ years. Distractor (C) uses $\log_{10}$ instead of $\ln$; (B) solves for doubling; (D) misplaces the decimal in $r$.

MCQ 35

U2 指数 / 对数 / U2 Exponential / Logarithmic

AI 押题AI Pick

A scientist plots a data set as the points $(x, \log_{10} y)$ on a coordinate grid (a semi-log plot). The plotted points lie on a line that passes through $(0, 1)$ and $(4, 3)$. Which of the following functions best models $y$ as a function of $x$?

(A) $y = 0.5x + 10$
(B) $y = 10 \cdot 10^{-0.5x}$
(C) $y = 10 \cdot 10^{0.5x}$
(D) $y = 10^{0.5x}$

📖 查看解题思路 / Show solution

Since $(x, \log_{10} y)$ is linear, the line has slope $\dfrac{3-1}{4-0} = 0.5$ and $y$-intercept $1$, so $\log_{10} y = 0.5x + 1$. Exponentiating gives $y = 10^{0.5x+1} = 10^{1} \cdot 10^{0.5x} = 10 \cdot 10^{0.5x}$. Linearity on a semi-log plot indicates an exponential relationship.

MCQ 36

U2 指数 / 对数 / U2 Exponential / Logarithmic

AI 押题AI Pick

A radioactive isotope decays exponentially. A laboratory sample initially has a mass of 80 milligrams, and after 12 years the sample has a mass of 50 milligrams. Based on this model, what is the half-life of the isotope, to the nearest hundredth of a year?

(A) $8.14$ years
(B) $12.00$ years
(C) $16.00$ years
(D) $17.70$ years

📖 查看解题思路 / Show solution

Model the mass as $M(t)=80\left(\tfrac{1}{2}\right)^{t/h}$, where $h$ is the half-life. Setting $M(12)=50$ gives $\left(\tfrac{1}{2}\right)^{12/h}=\tfrac{50}{80}=\tfrac{5}{8}$. Solving, $h=\dfrac{12\ln(1/2)}{\ln(5/8)}\approx 17.70$ years.

MCQ 37

U2 指数 / 对数 / U2 Exponential / Logarithmic

CED 样题CED Sample

Let $k$, $w$, and $z$ be positive constants. Which of the following is equivalent to $\log_{10}\left(\dfrac{kz}{w^2}\right)$?

(A) $\log_{10}(k + z) - \log_{10}(2w)$
(B) $\log_{10} k + \log_{10} z - 2\log_{10} w$
(C) $\log_{10} k + \log_{10} z - \dfrac{1}{2}\log_{10} w$
(D) $\log_{10} k - \log_{10} z + 2\log_{10} w$

MCQ 38

U2 指数 / 对数 / U2 Exponential / Logarithmic

CED 样题CED Sample

Values of the terms of a geometric sequence $g_n$ are graphed in the figure. Which of the following is an expression for the $nth$ term of the geometric sequence?

(A) $g_n = 4\left(\dfrac{1}{2}\right)^{(n-2)}$
(B) $g_n = 8(2)^{(n-1)}$
(C) $g_n = 8\left(\dfrac{1}{2}\right)^n$
(D) $g_n = 16\left(\dfrac{1}{2}\right)^{(n-1)}$

MCQ 39

U2 指数 / 对数 / U2 Exponential / Logarithmic

CED 样题CED Sample

The table gives values of the function $g$ for selected values of $x$. The function $f$ is given by $f(x) = 3^x + x^2$. What is the value of $f(g(3))$?

$x$	$g(x)$
$-2$	$4$
$0$	$\dfrac{1}{2}$
$3$	$-2$
$4$	$3$
$36$	$9$

(A) $-72$
(B) $\dfrac{37}{9}$
(C) $9$
(D) $97$

MCQ 40

U2 指数 / 对数 / U2 Exponential / Logarithmic

CED 样题CED Sample

The value, in millions of dollars, of transactions processed by an online payment platform is modeled by the function $M$. The value is expected to increase by $6.1\%$ each quarter of a year. At time $t = 0$ years, 54 million dollars of transactions were processed. If $t$ is measured in years, which of the following is an expression for $M(t)$? (Note: A quarter is one fourth of a year.)

(A) $54(0.061)^{(t/4)}$
(B) $54(0.061)^{(4t)}$
(C) $54(1.061)^{(t/4)}$
(D) $54(1.061)^{(4t)}$

MCQ 41

U2 指数 / 对数 / U2 Exponential / Logarithmic

CED 样题CED Sample

Iodine-131 has a half-life of 8 days. In a particular sample, the amount of iodine-131 remaining after $d$ days can be modeled by the function $h$ given by $h(d) = A_0(0.5)^{(d/8)}$, where $A_0$ is the amount of iodine-131 in the sample at time $d = 0$. Which of the following functions $k$ models the amount of iodine-131 remaining after $t$ hours, where $A_0$ is the amount of iodine-131 in the sample at time $t = 0$? (There are 24 hours in a day, so $t = 24d$.)

(A) $k(t) = A_0(0.5)^{(t/24)}$
(B) $k(t) = A_0\left(0.5^{(1/24)}\right)^{(8t)}$
(C) $k(t) = A_0\left(0.5^{(24)}\right)^{(t/8)}$
(D) $k(t) = A_0\left(0.5^{(1/192)}\right)^t$

MCQ 42

U2 指数 / 对数 / U2 Exponential / Logarithmic

CED 样题CED Sample

What are all values of $x$ for which $\ln\left(x^3\right) - \ln x = 4$?

(A) $x = -2$ and $x = 2$
(B) $x = -e^2$ and $x = e^2$
(C) $x = e^2$ only
(D) $x = e^4$

MCQ 43

U2 指数 / 对数 / U2 Exponential / Logarithmic

CED 样题CED Sample

The table presents values for a function $f$ at selected values of $x$. An exponential regression $y = ab^x$ is used to model these data. What is the value of $f(1.5)$ predicted by the exponential function model?

$x$	$f(x)$
$-2$	10
$-1$	15
1	40
2	56

(A) 46.767
(B) 47.342
(C) 47.800
(D) 47.917

MCQ 44

U3 三角 / 极坐标 / U3 Trigonometric / Polar

AI 押题AI Pick

What is the exact value of $\cos\left(\dfrac{2\pi}{3}\right) + \sin\left(\dfrac{7\pi}{4}\right)$?

(A) $\dfrac{\sqrt{2}-1}{2}$
(B) $-\dfrac{1+\sqrt{2}}{2}$
(C) $\dfrac{1-\sqrt{2}}{2}$
(D) $\dfrac{1+\sqrt{2}}{2}$

📖 查看解题思路 / Show solution

The angle $\tfrac{2\pi}{3}$ lies in Quadrant II with reference angle $\tfrac{\pi}{3}$, so $\cos(\tfrac{2\pi}{3}) = -\tfrac{1}{2}$. The angle $\tfrac{7\pi}{4}$ lies in Quadrant IV with reference angle $\tfrac{\pi}{4}$, so $\sin(\tfrac{7\pi}{4}) = -\tfrac{\sqrt{2}}{2}$. Adding gives $-\tfrac{1}{2} - \tfrac{\sqrt{2}}{2} = -\tfrac{1+\sqrt{2}}{2}$.

MCQ 45

U3 三角 / 极坐标 / U3 Trigonometric / Polar

AI 押题AI Pick

The function $f$ is defined by $f(x) = -3\cos\!\left(\dfrac{\pi}{2}x - \pi\right) + 4$. Which of the following correctly identifies the amplitude, period, and midline of the graph of $f$?

(A) Amplitude $-3$, period $\dfrac{\pi}{2}$, midline $y = 4$
(B) Amplitude $3$, period $\dfrac{\pi}{2}$, midline $y = 4$
(C) Amplitude $3$, period $4$, midline $y = 4$
(D) Amplitude $3$, period $4$, midline $y = -4$

📖 查看解题思路 / Show solution

For $f(x) = a\cos(bx + c) + d$, the amplitude is $|a| = 3$, the period is $\dfrac{2\pi}{|b|} = \dfrac{2\pi}{\pi/2} = 4$, and the midline is $y = d = 4$. Thus the amplitude is $3$, the period is $4$, and the midline is $y = 4$.

MCQ 46

U3 三角 / 极坐标 / U3 Trigonometric / Polar

AI 押题AI Pick

The function $f$ is defined by $f(x) = 2\sin(3x + \tfrac{\pi}{2}) - 1$. Which of the following correctly describes the phase shift and vertical shift of the graph of $f$ compared to the graph of $y = \sin x$?

(A) Phase shift of $\dfrac{\pi}{6}$ units to the left and vertical shift of $1$ unit down
(B) Phase shift of $\dfrac{\pi}{2}$ units to the left and vertical shift of $1$ unit down
(C) Phase shift of $\dfrac{\pi}{6}$ units to the right and vertical shift of $1$ unit down
(D) Phase shift of $\dfrac{\pi}{6}$ units to the left and vertical shift of $1$ unit up

📖 查看解题思路 / Show solution

Factor the argument: $3x + \tfrac{\pi}{2} = 3\left(x + \tfrac{\pi}{6}\right)$, so $f(x) = 2\sin\!\left(3\left(x + \tfrac{\pi}{6}\right)\right) - 1$. The $+\tfrac{\pi}{6}$ inside indicates a phase shift of $\tfrac{\pi}{6}$ units to the left, and the $-1$ outside indicates a vertical shift of $1$ unit down.

MCQ 47

U3 三角 / 极坐标 / U3 Trigonometric / Polar

AI 押题AI Pick

Which of the following is equivalent to $\dfrac{\sec^2 x - 1}{\sin^2 x}$ for all values of $x$ in the domain of the expression?

(A) $\tan^2 x$
(B) $\csc^2 x$
(C) $1 - \cos^2 x$
(D) $\sec^2 x$

📖 查看解题思路 / Show solution

By the Pythagorean identity, $\sec^2 x - 1 = \tan^2 x = \dfrac{\sin^2 x}{\cos^2 x}$. Therefore $\dfrac{\sec^2 x - 1}{\sin^2 x} = \dfrac{\sin^2 x/\cos^2 x}{\sin^2 x} = \dfrac{1}{\cos^2 x} = \sec^2 x.$

MCQ 48

U3 三角 / 极坐标 / U3 Trigonometric / Polar

AI 押题AI Pick

How many solutions does the equation $2\sin(3x) + 1 = 0$ have on the interval $[0, 2\pi]$?

(A) 2
(B) 6
(C) 3
(D) 12

📖 查看解题思路 / Show solution

Rewrite as $\sin(3x) = -\dfrac{1}{2}$. Let $u = 3x$; since $x \in [0, 2\pi]$, we have $u \in [0, 6\pi]$. On any interval of length $2\pi$, $\sin(u) = -\dfrac{1}{2}$ has exactly two solutions ($u = \dfrac{7\pi}{6}$ and $u = \dfrac{11\pi}{6}$). Across three full periods in $[0, 6\pi]$, there are $2 \times 3 = 6$ solutions.

MCQ 49

U3 三角 / 极坐标 / U3 Trigonometric / Polar

AI 押题AI Pick

What is the exact value of $\cos\!\left(2\arcsin\!\left(-\dfrac{1}{3}\right)\right)$?

(A) $-\dfrac{7}{9}$
(B) $\dfrac{7}{9}$
(C) $\dfrac{8}{9}$
(D) $\dfrac{1}{3}$

📖 查看解题思路 / Show solution

Let $\theta=\arcsin(-\tfrac{1}{3})$, so $\sin\theta=-\tfrac{1}{3}$. Using the double-angle identity, $\cos(2\theta)=1-2\sin^{2}\theta=1-2\cdot\tfrac{1}{9}=\tfrac{7}{9}$. The sign of $\sin\theta$ does not affect the result since it is squared.

MCQ 50

U3 三角 / 极坐标 / U3 Trigonometric / Polar

AI 押题AI Pick

Suppose $\theta$ is an angle such that $\sec\theta = -\dfrac{5}{3}$ and $\tan\theta > 0$. What is the value of $\cot\theta$?

(A) $-\dfrac{4}{3}$
(B) $\dfrac{4}{3}$
(C) $-\dfrac{3}{4}$
(D) $\dfrac{3}{4}$

📖 查看解题思路 / Show solution

From $\sec\theta=-\dfrac{5}{3}$, we have $\cos\theta=-\dfrac{3}{5}$. Since $\tan\theta>0$ and $\cos\theta<0$, we need $\sin\theta<0$; using $\sin^2\theta+\cos^2\theta=1$ gives $\sin\theta=-\dfrac{4}{5}$. Therefore $\cot\theta=\dfrac{\cos\theta}{\sin\theta}=\dfrac{-3/5}{-4/5}=\dfrac{3}{4}$.

MCQ 51

U3 三角 / 极坐标 / U3 Trigonometric / Polar

AI 押题AI Pick

Without using a calculator, evaluate $\cos\left(\dfrac{7\pi}{6}\right) - \sin\left(\dfrac{5\pi}{4}\right)$.

(A) $\dfrac{\sqrt{2}-\sqrt{3}}{2}$
(B) $\dfrac{\sqrt{3}-\sqrt{2}}{2}$
(C) $\dfrac{\sqrt{2}+\sqrt{3}}{2}$
(D) $-\dfrac{\sqrt{2}+\sqrt{3}}{2}$

📖 查看解题思路 / Show solution

The angle $\tfrac{7\pi}{6}$ lies in Quadrant III with reference angle $\tfrac{\pi}{6}$, so $\cos\tfrac{7\pi}{6}=-\tfrac{\sqrt{3}}{2}$. The angle $\tfrac{5\pi}{4}$ lies in Quadrant III with reference angle $\tfrac{\pi}{4}$, so $\sin\tfrac{5\pi}{4}=-\tfrac{\sqrt{2}}{2}$. Therefore the expression equals $-\tfrac{\sqrt{3}}{2}-\left(-\tfrac{\sqrt{2}}{2}\right)=\dfrac{\sqrt{2}-\sqrt{3}}{2}$.

MCQ 52

U3 三角 / 极坐标 / U3 Trigonometric / Polar

AI 押题AI Pick

The point $(r, \theta) = \left(-4, \dfrac{5\pi}{6}\right)$ is given in polar coordinates. Which of the following is the rectangular form $(x, y)$ of this point?

(A) $(-2\sqrt{3},\ 2)$
(B) $(-2\sqrt{3},\ -2)$
(C) $(2\sqrt{3},\ 2)$
(D) $(2\sqrt{3},\ -2)$

📖 查看解题思路 / Show solution

Using $x = r\cos\theta$ and $y = r\sin\theta$ with $r=-4$ and $\theta=\dfrac{5\pi}{6}$: $x = -4\cos\!\left(\dfrac{5\pi}{6}\right) = -4\!\left(-\dfrac{\sqrt{3}}{2}\right) = 2\sqrt{3}$, and $y = -4\sin\!\left(\dfrac{5\pi}{6}\right) = -4\!\left(\dfrac{1}{2}\right) = -2$. Thus $(x,y) = (2\sqrt{3},\ -2)$.

MCQ 53

U3 三角 / 极坐标 / U3 Trigonometric / Polar

AI 押题AI Pick

Which of the following expressions is equivalent to $\dfrac{\sin(2\theta)}{1+\cos(2\theta)}$ for all values of $\theta$ where the expression is defined?

(A) $\cot\theta$
(B) $\tan\theta$
(C) $\sin\theta\cos\theta$
(D) $\tan(2\theta)$

📖 查看解题思路 / Show solution

Apply the double-angle identities: $\sin(2\theta) = 2\sin\theta\cos\theta$ and $1+\cos(2\theta) = 2\cos^{2}\theta$. Then $\dfrac{\sin(2\theta)}{1+\cos(2\theta)} = \dfrac{2\sin\theta\cos\theta}{2\cos^{2}\theta} = \dfrac{\sin\theta}{\cos\theta} = \tan\theta$.

MCQ 54

U3 三角 / 极坐标 / U3 Trigonometric / Polar

AI 押题AI Pick

A researcher records the height of ocean water (in feet) at a coastal station and finds that on a particular day, high tide first occurs at $t = 2$ hours after midnight with a height of $8.4$ feet, and the next low tide occurs at $t = 8$ hours with a height of $1.2$ feet. Assuming the water height $H$ is well modeled by a sinusoidal function of time $t$ (in hours), which of the following best models $H(t)$?

(A) $H(t) = 4.8 + 7.2\cos\!\left(\dfrac{\pi}{6}(t-2)\right)$
(B) $H(t) = 4.8 + 3.6\cos\!\left(\dfrac{\pi}{6}(t-2)\right)$
(C) $H(t) = 4.8 + 3.6\cos\!\left(\dfrac{\pi}{3}(t-2)\right)$
(D) $H(t) = 4.8 + 3.6\sin\!\left(\dfrac{\pi}{6}(t-2)\right)$

📖 查看解题思路 / Show solution

The midline is $\tfrac{8.4+1.2}{2}=4.8$ and the amplitude is $\tfrac{8.4-1.2}{2}=3.6$. Since consecutive high and low tides are half a period apart, the period is $2(8-2)=12$, giving $b=\tfrac{2\pi}{12}=\tfrac{\pi}{6}$. Because the maximum occurs at $t=2$, a cosine with horizontal shift $2$ fits: $H(t)=4.8+3.6\cos\!\left(\tfrac{\pi}{6}(t-2)\right)$.

MCQ 55

U3 三角 / 极坐标 / U3 Trigonometric / Polar

AI 押题AI Pick

Consider the polar function defined by $ r = 3 - 6\cos\theta $ for $ 0 \le \theta \le 2\pi $. Which of the following statements about the graph of this function is true?

(A) The graph is a circle of radius 3 centered at the point with rectangular coordinates $(-3, 0)$.
(B) The graph is a limaçon with an inner loop, and it is symmetric about the polar axis (the $x$-axis).
(C) The graph is a cardioid that is symmetric about the line $\theta = \dfrac{\pi}{2}$ (the $y$-axis).
(D) The graph is a four-petaled rose whose maximum distance from the origin is $6$.

📖 查看解题思路 / Show solution

A polar function of the form $r = a + b\cos\theta$ is a limaçon. Because $|a| = 3 < 6 = |b|$, the limaçon has an inner loop ($r$ changes sign, equaling $0$ at $\theta = \pi/3, 5\pi/3$). Since the function depends only on $\cos\theta$, replacing $\theta$ with $-\theta$ leaves $r$ unchanged, so the graph is symmetric about the polar axis. The maximum distance from the origin is $|3 - 6(-1)| = 9$, not 6, ruling out (D); it is not a circle or a cardioid (which would require $|a| = |b|$).

MCQ 56

U3 三角 / 极坐标 / U3 Trigonometric / Polar

AI 押题AI Pick

The depth of water at a coastal dock, in feet, is modeled by the function \[ d(t) = 8 + 6\cos\!\left(\dfrac{\pi}{6}(t-6)\right), \] where $t$ is the number of hours after midnight. According to this model, at what time after 6 A.M. is the depth of the water first equal to 5 feet?

(A) 8:00 A.M.
(B) 9:00 A.M.
(C) 2:00 P.M.
(D) 10:00 A.M.

📖 查看解题思路 / Show solution

Set $8 + 6\cos\!\left(\frac{\pi}{6}(t-6)\right) = 5$, giving $\cos\!\left(\frac{\pi}{6}(t-6)\right) = -\tfrac{1}{2}$. The smallest positive solution is $\frac{\pi}{6}(t-6) = \frac{2\pi}{3}$, so $t - 6 = 4$ and $t = 10$. Thus the depth first reaches 5 feet at 10:00 A.M.

MCQ 57

U3 三角 / 极坐标 / U3 Trigonometric / Polar

CED 样题CED Sample

Let $f(x) = 1 + 3\sec x$ and $g(x) = -5$. In the $xy$-plane, what are the $x$-coordinates of the points of intersection of the graphs of $f$ and $g$ for $0 \le x < 2\pi$?

(A) $x = \dfrac{\pi}{3}$ and $x = \dfrac{5\pi}{3}$
(B) $x = \dfrac{\pi}{6}$ and $x = \dfrac{5\pi}{6}$
(C) $x = \dfrac{2\pi}{3}$ and $x = \dfrac{4\pi}{3}$
(D) $x = \dfrac{7\pi}{6}$ and $x = \dfrac{11\pi}{6}$

MCQ 58

U3 三角 / 极坐标 / U3 Trigonometric / Polar

CED 样题CED Sample

The figure shows the graph of a sinusoidal function $g$. What are the values of the period and amplitude of $g$?

(A) The period is 4, and the amplitude is 3.
(B) The period is 8, and the amplitude is 3.
(C) The period is 4, and the amplitude is 6.
(D) The period is 8, and the amplitude is 6.

MCQ 59

U3 三角 / 极坐标 / U3 Trigonometric / Polar

CED 样题CED Sample

Which of the following is the graph of the polar function $r = f(\theta)$, where $f(\theta) = 3\cos\theta + 2$, in the polar coordinate system for $0 \leq \theta \leq 2\pi$?

📊 Two polar graphs labeled (A) and (B) showing different rose-like curves on polar coordinate grids with radial markings from 1 to 5.

(A) Graph (A)
(B) Graph (B)
(C)
(D)

MCQ 60

U3 三角 / 极坐标 / U3 Trigonometric / Polar

CED 样题CED Sample

What are all values of $\theta$, $-\pi \le \theta \le \pi$, for which $2\cos\theta > -1$ and $\sin\theta > \sqrt{3}$?

(A) $-\dfrac{5\pi}{6} < \theta < \dfrac{5\pi}{6}$
(B) $\dfrac{\pi}{6} < \theta < \dfrac{5\pi}{6}$ only
(C) $-\dfrac{2\pi}{3} < \theta < \dfrac{2\pi}{3}$ only
(D) $\dfrac{\pi}{3} < \theta < \dfrac{2\pi}{3}$ only

MCQ 61

U3 三角 / 极坐标 / U3 Trigonometric / Polar

CED 样题CED Sample

A polar function is given by $r = f(\theta) = -1 + \sin\theta$. As $\theta$ increases on the interval $0 < \theta < \dfrac{\pi}{2}$, which of the following is true about the points on the graph of $r = f(\theta)$ in the $xy$-plane?

(A) The points on the graph are above the $x$-axis and are getting closer to the origin.
(B) The points on the graph are above the $x$-axis and are getting farther from the origin.
(C) The points on the graph are below the $x$-axis and are getting closer to the origin.
(D) The points on the graph are below the $x$-axis and are getting farther from the origin.

MCQ 62

U3 三角 / 极坐标 / U3 Trigonometric / Polar

CED 样题CED Sample

The number of minutes of daylight per day for a certain city can be modeled by the function $D$ given by $D(t) = 160\cos\left(\frac{2\pi}{365}(t-172)\right) + 729$, where $t$ is the day of the year for $1 \le t \le 365$. Which of the following best describes the behavior of $D(t)$ on day 150?

(A) The number of minutes of daylight per day is increasing at a decreasing rate.
(B) The number of minutes of daylight per day is decreasing at a decreasing rate.
(C) The number of minutes of daylight per day is increasing at an increasing rate.
(D) The number of minutes of daylight per day is decreasing at an increasing rate.

MCQ 63

U3 三角 / 极坐标 / U3 Trigonometric / Polar

CED 样题CED Sample

The function $g$ is given by $g(x) = \sin x - \cos x$ and has a period of $2\pi$. In order to define the inverse function of $g$, which of the following specifies a restricted domain for $g$ and provides a rationale for why $g$ is invertible on that domain?

(A) $0 \le x \le \pi$, because all possible values of $g(x)$ occur without repeating on this interval.
(B) $-\frac{\pi}{4} \le x \le \frac{3\pi}{4}$, because all possible values of $g(x)$ occur without repeating on this interval.
(C) $0 \le x \le \pi$, because the length of this interval is half of the period.
(D) $-\frac{\pi}{4} \le x \le \frac{3\pi}{4}$, because the length of this interval is half of the period.

MCQ 64

U3 三角 / 极坐标 / U3 Trigonometric / Polar

CED 样题CED Sample

A theme park thrill ride involves a tower and a carriage that rapidly moves passengers up and down along a vertical axis, as shown in the figure. The carriage is lifted to the top of the tower, then released to move down the tower. The ride involves 10 controlled bounces from the highest point to the lowest point, and back to the highest point. A point $X$ is located on the bottom of the carriage. The height of $X$ above the ground, in feet, can be modeled by a periodic function $H$. At time $t = 0$ seconds, $X$ is at its highest point of 120 feet. The lowest point for $X$ is at a height of 20 feet. The next time $X$ is at its highest point is at time $t = 8$ seconds, which is the end of the first bounce. Which of the following can be an expression for $H(t)$, where $t$ is the time in seconds?

📊 Figure showing a vertical tower with a carriage that moves up and down. Point $X$ is marked on the bottom of the carriage.

(A) $50\sin\left(\dfrac{\pi}{4}t\right) + 70$
(B) $50\cos\left(\dfrac{\pi}{4}t\right) + 70$
(C) $50\sin\left(\dfrac{\pi}{8}t\right) + 70$
(D) $50\cos\left(\dfrac{\pi}{8}t\right) + 70$

Part V · 考前 24 小时 Cheat SheetPart V · 24-Hour Cheat Sheet

必背公式卡Must-Know Formula Cards

多项式 End BehaviorPolynomial End Behavior

次数 odd, 首项 + → ↙↗（左下右上）Odd degree, leading + → ↙↗ (down left, up right)

次数 odd, 首项 − → ↗↙（左上右下）Odd degree, leading − → ↗↙ (up left, down right)

次数 even, 首项 + → ↗↗（两端向上）Even degree, leading + → ↗↗ (both up)

次数 even, 首项 − → ↙↙（两端向下）Even degree, leading − → ↙↙ (both down)

零点 odd 重 → 穿越 x 轴；even 重 → 接触不穿越Odd multiplicity → crosses x-axis; even → bounces off

有理函数 Rational FunctionRational Function

先因式分解再约分！Factor and cancel first!

Hole: 能约去的因子对应点Hole: location of cancelled factor

VA: 约简后分母 = 0 的点VA: remaining denominator = 0

HA: 分子次 < 分母次 → $y=0$；相等 → $y=$ 首项系数比；分子次 > 分母次 → 无 HAHA: deg(num) < deg(den) → $y=0$; equal → $y=$ ratio of leading coefficients; num > den → no HA

指数函数 ExponentialExponential Functions

$y = ab^x$：a = initial value, b = growth factor

$b>1$ → 增长；$0<b<1$ → 衰减$b>1$ → growth; $0<b<1$ → decay

$y = ae^{kx}$：$k>0$ 增，$k<0$ 减$k>0$ growth, $k<0$ decay

必写解释：当 $x=0$ 时 $y=a$ 代表初始量Must write: "When $x=0$, $y=a$ represents the initial…"

对数函数 LogarithmicLogarithmic Functions

$\log_b(MN)=\log_b M+\log_b N$

$\log_b(M/N)=\log_b M-\log_b N$

$\log_b(M^n)=n\log_b M$

换底：Change of base: $\log_b a = \dfrac{\ln a}{\ln b}$

Domain: 参数 > 0argument > 0

三角/周期函数 SinusoidalSinusoidal Functions

$y = A\sin(B(x-C))+D$

Amplitude $=|A|$ （振幅，永远为正）(always positive)

Period $= \dfrac{2\pi}{B}$

Phase shift $= C$ （向右移 C）(shift right by C)

Midline $y=D$

$\sin^2\theta+\cos^2\theta=1$

极坐标 PolarPolar Coordinates

$x=r\cos\theta,\quad y=r\sin\theta$

$r^2=x^2+y^2$

$\theta=\arctan(y/x)$

圆：Circle: $r=a$

Rose: $r=a\cos(n\theta)$

Limaçon: $r=a+b\cos\theta$

反函数 & 复合 Inverse & CompositionInverse & Composition

$(f\circ g)(x)=f(g(x))$ 先算 $g$ 再算 $f$apply $g$ first, then $f$

$f^{-1}$ 存在 ↔ $f$ 是 one-to-oneexists ↔ $f$ is one-to-one

$f(f^{-1}(x))=x$, $f^{-1}(f(x))=x$

Domain of $f^{-1}$ = Range of $f$

FRQ 4 槽位速记4-Slot Quick Reference

Q1 Function Concepts：从表格读 rate of change，判断 concavityFunction Concepts: rate of change from table, concavity

Q2 Modeling：$y=ab^t$，必须在 context 中解释 $a$ 和 $b$Modeling: $y=ab^t$, interpret $a$ and $b$ in context

Q3 Trig：必写全部 4 参数 — amplitude · period · shift · midlineTrig: write all 4 — amplitude · period · shift · midline

Q4 Algebra：inverse / composition / trig identity，每步写出Algebra: inverse / composition / identity, show every step

考场 5 条纪律5 Exam-Room Rules

每问写单位（meters, degrees, dollars…）没单位 = 丢点，无论答案是否正确。Write units on every sub-part (meters, degrees, dollars…). Missing units = lost points regardless of the answer.
写解释句：每个 context 答案后加一句 "This represents…" 或 "This means that at time $t=…$"Write an interpretation sentence: follow every contextual answer with "This represents…" or "This means that at $t=…$"
FRQ3 必写 4 要素：amplitude, period, phase shift, midline — 每个都是独立给分点，一个不能漏。FRQ3: state all 4 sinusoidal parameters — amplitude, period, phase shift, midline — each is a separate scoring point.
Calculator 题（FRQ1, 2）：先写 setup 表达式再写数值结果，不能只写一个数字。积分和方程求根交给计算器。Calculator sections (FRQ1, 2): write the setup expression first, then the numerical answer. Use the calculator for definite integrals and equation solving.
College Board 按步骤给分，不看最终答案。过程对即使算错也能得分；只写答案不写过程 = 零分。College Board grades by process, not the final answer. Correct steps with a wrong final value still earn points; a bare answer with no work earns zero.

9 大高频陷阱9 High-Frequency Traps

End behavior 由 degree（奇/偶）和 leading coefficient（正/负）共同决定，两个条件缺一不可。End behavior requires BOTH the degree (odd/even) AND the sign of the leading coefficient — never just one.
有理函数：先因式分解再判断。能约掉 = hole；约不掉在分母 = vertical asymptote。两者不能混淆。Rational functions: factor first. Cancelled factor → hole; remaining factor in denominator → vertical asymptote. Never confuse the two.
指数建模中 $b$ 是 growth factor，不是 rate。Growth rate $= b-1$（例：$b=1.08$ → 8% 增长率）。In exponential models, $b$ is the growth factor, not the rate. Growth rate $=b-1$ (e.g., $b=1.08$ → 8% rate).
$\log(x+y)\neq\log x+\log y$！Product rule 只适用于 log 内的乘法，不是相加。这是最常见的对数错误。$\log(x+y)\neq\log x+\log y$! The product rule applies only to multiplication inside the log, not addition — the most common log mistake.
Sinusoidal period $=\dfrac{2\pi}{B}$，不是 $2\pi B$ 也不是 $B$。从 context 找到 period 后再倒推 $B = \dfrac{2\pi}{\text{period}}$。Sinusoidal period $=\dfrac{2\pi}{B}$, not $2\pi B$ or $B$. Find the period from context, then solve $B=\dfrac{2\pi}{\text{period}}$.
Amplitude $=|A|$，永远是正数。题目给出 max/min 时：$A=\dfrac{\text{max}-\text{min}}{2}$，Midline $D=\dfrac{\text{max}+\text{min}}{2}$。Amplitude $=|A|$, always positive. Given max/min: $A=\dfrac{\text{max}-\text{min}}{2}$, Midline $D=\dfrac{\text{max}+\text{min}}{2}$.
复合函数 $(f\circ g)(x)$：先算 $g(x)$ 再算 $f$ — 从右往左读，不是从左往右。Composition $(f\circ g)(x)$: apply $g$ first, then $f$ — read right to left, not left to right.
$f^{-1}$ 存在 ↔ $f$ 必须是 one-to-one（过 Horizontal Line Test）。若不是，必须限制定义域再求反函数。$f^{-1}$ exists ↔ $f$ must be one-to-one (passes the Horizontal Line Test). If not, restrict the domain first.
零点重数 odd → 图像穿越 $x$ 轴；even → 接触 $x$ 轴后弹回，不穿越。Odd multiplicity zero → graph crosses the $x$-axis; even multiplicity → touches and bounces back, does not cross.

\(x\)	\(g(x)\)
\(-2\)	\(4\)
\(0\)	\(\dfrac{1}{2}\)
\(3\)	\(-2\)
\(4\)	\(3\)
\(36\)	\(9\)

\(x\)	\(f(x)\)
\(-2\)	10
\(-1\)	15
1	40
2	56