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 · 2026 AI 押题 MCQ(40 题)Part III · 2026 AI Predicted MCQ (40)

按 CED 权重生成(28 no-calc + 12 calc)。点击 A-D 即时判分,展开看解题。Generated per CED weights (28 no-calc + 12 calc). Click A-D for instant grading; expand to see the solution.

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

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 13

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 14

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 15

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 16

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 17

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 18

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 19

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 20

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 21

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 22

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 23

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 24

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 25

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 26

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 27

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 28

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 29

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 30

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 31

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 32

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 33

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 34

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 35

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 36

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 37

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 38

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 39

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 40

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.

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

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