PART I · SHORT RESPONSE第一部分 · 短答题SAT-style MCQ + ON/BC short answer · 22 marksSAT 风格选择题 + 安/卑省考短答 · 共 22 分
Section A · Short ResponseA 部分 · 短答题
Mix of multiple-choice and short-answer items. For MCQs, circle the letter; show enough work in the margin that a marker could verify. State exact forms where possible (e.g. $\log_{3} 16$, $\ln 5 / \ln 2$) before any decimal rounding. No calculator on Q1-Q3; calculator permitted on Q4-Q5.本节包含选择题与短答题。选择题请圈出字母答案,并在空白处写下足以让阅卷人核对的过程。在小数舍入前尽量给出精确形式(如 $\log_{3} 16$、$\ln 5 / \ln 2$)。Q1–Q3 不可使用计算器;Q4–Q5 可用计算器。
The graph of $f(x) = 4 \cdot 2^{x}$ is sketched in the $xy$-plane. Which statement correctly describes its $y$-intercept and horizontal asymptote?在 $xy$ 平面内画出指数函数 $f(x) = 4 \cdot 2^{x}$ 的图像。下列哪一项正确描述其 $y$ 轴截距与水平渐近线?
Solve each exponential equation. Give exact answers in the form $\log_{b}(\,\cdot\,)$ or $\ln(\,\cdot\,) / \ln(\,\cdot\,)$, then a decimal to three places.求解下列指数方程。请先给出 $\log_{b}(\,\cdot\,)$ 或 $\ln(\,\cdot\,) / \ln(\,\cdot\,)$ 形式的精确解,再写出保留三位小数的近似值。
(a) $3^{x} = 20$. [3]
(b) $5 \cdot 2^{x} = 90$. [2]
(c)$7^{x - 1} = 4^{x}$. State the change-of-base step explicitly.$7^{x - 1} = 4^{x}$。请明确写出换底公式的步骤。[2]
PART II · EXTENDED RESPONSE第二部分 · 简答题AP-feeder FRQ + honors · 36 marksAP 衔接简答题 + 荣誉级 · 共 36 分
Section B · Extended ResponseB 部分 · 简答题
Show every algebraic step. For log equations, check every candidate against the domain of each logarithm and discard extraneous solutions. State whether you used same-base or take-the-log on every exponential equation. No calculator on Q6 and Q8; calculator permitted on Q7 and Q9.每一步代数运算都要写出。对对数方程,每个候选解都要对照每个对数的定义域进行验证,并舍弃增根。对每个指数方程都要注明所用方法(同底法或两边取对数)。Q6 与 Q8 不可使用计算器;Q7 与 Q9 可用计算器。
Solve each logarithmic equation algebraically. For each candidate solution, verify the argument of every logarithm is strictly positive; discard extraneous solutions and state the final solution set.用代数方法求解下列对数方程。对每个候选解,需验证每个对数的真数严格大于零;舍弃增根并写出最终解集。
(a)State the horizontal asymptote of $y = g(x)$ and the $y$-intercept.写出 $y = g(x)$ 的水平渐近线与 $y$ 轴截距。[2]
(b)Solve $g(x) = 16$ exactly using the natural logarithm, then approximate to three decimal places.用自然对数精确求解 $g(x) = 16$,再保留三位小数。[3]
(c)Determine the inverse $g^{-1}(x)$ explicitly in terms of $\ln$. State the domain and range of $g^{-1}$.用 $\ln$ 显式求出反函数 $g^{-1}(x)$,并写出 $g^{-1}$ 的定义域与值域。[3]
(d)Cite the BC PC12 Big Idea on inverses (one sentence) and explain why solving $g(x) = 16$ in (b) is precisely an instance of it.引用 BC PC12 关于反函数的大概念(一句话),并解释 (b) 中求解 $g(x) = 16$ 为何正是该概念的具体实例。[2]
Q9HARD难Honors荣誉级🇺🇸 US美🇨🇦 BC卑AP-feeder FRQAP 衔接简答题§6 Continuous Compounding连续复利 · BC PC12 base $e$BC PC12 底数 $e$ / HSF-LE.B.5[9 marks][9 分]
A bank advertises a nominal annual interest rate of $4.5\%$ on a savings account with continuous compounding. The balance after $t$ years on an initial principal $P_{0}$ is modeled by $P(t) = P_{0} \, e^{0.045 \, t}$.某银行的储蓄账户标称年利率为 $4.5\%$,按连续复利计息。本金 $P_{0}$ 经过 $t$ 年后的余额由 $P(t) = P_{0} \, e^{0.045 \, t}$ 给出。
(a)A customer deposits US$$2{,}000$. Compute the balance after $10$ years to the nearest cent.某客户存入 US$$2{,}000$。计算 $10$ 年后的余额,精确到分。[2]
(b)Solve symbolically for the doubling time $T$ such that $P(T) = 2 P_{0}$. Express $T$ exactly using $\ln$, then approximate to two decimal places.符号求解倍增时间 $T$,使 $P(T) = 2 P_{0}$。用 $\ln$ 给出精确表达式,再保留两位小数。[3]
(c)Interpret the parameters $P_{0}$ and the rate $0.045$ in the context of the situation. Cite HSF-LE.B.5: "interpret the parameters in… an exponential function in terms of a context."结合情境解释参数 $P_{0}$ 与利率 $0.045$ 的实际含义。援引 HSF-LE.B.5:"结合情境解释指数函数中的各参数。"[2]
(d)A rival bank offers $4.5\%$ compounded annually instead of continuously. After $10$ years on the same US$$2{,}000$ principal, which account holds more, and by how much (nearest cent)?另一家银行同样提供 $4.5\%$ 的年利率,但按年复利而非连续复利。同样 US$$2{,}000$ 本金,$10$ 年后哪个账户余额更多?多多少(精确到分)?[2]
PART III · MODELING / APPLIED第三部分 · 建模与应用Universal · 32 marks通用题型 · 共 32 分
Section C · Modeling and ApplicationsC 部分 · 建模与应用
Name your variables (with units) before writing equations. State whether the situation is growth or decay and identify the growth/decay factor before substituting. Conclude with a one-sentence answer in context, rounded sensibly. Calculator permitted throughout Part III.在写方程前,先定义变量名(含单位)。代入数值前先注明情境为增长还是衰减,并写出增长 / 衰减倍数。以一句结合情境的话作答,合理舍入。第三部分全程可用计算器。
A coastal town's population grows by a constant percent rate of $2.4\%$ per year. In the year $2020$ its population was $48{,}000$. Let $P(t)$ denote the population $t$ years after $2020$.某海滨城镇的人口以每年 $2.4\%$ 的恒定百分率增长。$2020$ 年其人口为 $48{,}000$。设 $P(t)$ 表示 $2020$ 年后 $t$ 年的人口数。
(a)Write the explicit model $P(t) = a \cdot b^{t}$. State the values of $a$ and $b$ and identify which represents the initial value and which the annual growth factor.写出显式模型 $P(t) = a \cdot b^{t}$。给出 $a$ 与 $b$ 的值,并说明哪个代表初始值、哪个代表年增长倍数。[3]
(b)Predict the population in $2030$ to the nearest whole person.预测 $2030$ 年的人口数,精确到整人。[2]
(c)Determine, using logarithms, the year in which the population is first projected to exceed $75{,}000$. Round your value of $t$ up to the next whole year.用对数确定人口首次超过 $75{,}000$ 的年份。$t$ 值向上取整到下一整年。[3]
(d)Cite HSF-LE.A.1c to explain why "$2.4\%$ per year" produces an exponential rather than a linear model.援引 HSF-LE.A.1c,说明为何"每年 $2.4\%$"产生的是指数模型而非线性模型。[2]
Iodine-131 has a half-life of approximately $8.02$ days. A medical sample initially contains $40$ mg of I-131. Let $M(t)$ be the mass remaining (in mg) after $t$ days.碘-131 的半衰期约为 $8.02$ 天。某医用样本初始含 $40$ mg 的 I-131。设 $M(t)$ 为 $t$ 天后剩余的质量(毫克)。
(a)Write $M(t)$ in the half-life form $M(t) = M_{0} \cdot \left(\tfrac{1}{2}\right)^{t / h}$. State $M_{0}$ and $h$.把 $M(t)$ 写成半衰期形式 $M(t) = M_{0} \cdot \left(\tfrac{1}{2}\right)^{t / h}$,并写出 $M_{0}$ 与 $h$。[2]
(b)Compute the mass remaining after $30$ days, to two decimal places.计算 $30$ 天后的剩余质量,保留两位小数。[2]
(c)Determine, using logarithms, the time required for the sample to decay to $5$ mg. Round to the nearest tenth of a day.用对数确定样本衰减到 $5$ mg 所需的时间,精确到 $0.1$ 天。[4]
(d)State the practical (situational) domain of $M(t)$ in interval notation, and identify whether this is exponential growth or decay.用区间记号写出 $M(t)$ 的实际(情境)定义域,并判断这是指数增长还是指数衰减。[2]
(e)State one assumption baked into the half-life model that is in practice only approximately true.写出半衰期模型中蕴含的一个在实际中只近似成立的假设。[1]
Two investment options are available for an initial principal of CA$$5{,}000$ held for $t$ years.本金 CA$$5{,}000$ 可选择两种投资方案,持有 $t$ 年。
Option A pays $6\%$ annual interest compounded monthly: $A(t) = 5000 \left(1 + \tfrac{0.06}{12}\right)^{12 t}$.
Option B is a linear-growth bond paying simple interest of CA$$320$ per year: $B(t) = 5000 + 320 \, t$.方案 A:年利率 $6\%$,按月复利:$A(t) = 5000 \left(1 + \tfrac{0.06}{12}\right)^{12 t}$。
方案 B:线性增长债券,按单利每年支付 CA$$320$:$B(t) = 5000 + 320 \, t$。
(a)Compute $A(5)$ and $B(5)$, each to the nearest cent. Which is larger after $5$ years, and by how much?计算 $A(5)$ 与 $B(5)$,各精确到分。$5$ 年后哪个更多?多多少?[3]
(b)Determine the doubling time for Option A: the value of $t$ for which $A(t) = 10{,}000$. Use logarithms and give an exact expression, then round to two decimal places.确定方案 A 的倍增时间:使 $A(t) = 10{,}000$ 的 $t$ 值。用对数给出精确表达式,再保留两位小数。[3]
(c)Cite HSF-LE.A.3: explain in one or two sentences why $A(t)$ must eventually exceed $B(t)$ regardless of how large the initial gap might be, and identify (approximately, by table or by inspection) the year $t$ in which $A(t)$ first exceeds $B(t)$.援引 HSF-LE.A.3,用一两句话说明为何无论初始差距多大,$A(t)$ 最终都会超过 $B(t)$,并通过列表或观察大致确定 $A(t)$ 首次超过 $B(t)$ 的年份 $t$。[3]
(d)State one situational restriction on $t$ (e.g. non-negativity or a contractual maturity), and one reason a real saver might still prefer Option B in the short term.写出 $t$ 的一个情境限制(如非负性或合同到期日),并给出实际储户在短期内仍可能偏好方案 B 的一个理由。[2]
🇺🇸 US Common Core美国共同核心HSF-LE.A.1 · HSF-LE.A.1c · HSF-LE.A.2 · HSF-LE.A.3 · HSF-LE.A.4 · HSF-LE.B.5 · HSF-IF.C.7e · HSF-BF.B.5 (+)
🇨🇦 Ontario安大略MCR3U Strand B Exponential Functions (expectation B2.1) · MCR3U Strand C geometric sequences / compound interest · MHF4U Exponential and Logarithmic Functions strand (laws of logs, log equations, inverse-of-exponential, base $e$)MCR3U 单元 B 指数函数(期望 B2.1)· MCR3U 单元 C 等比数列 / 复利 · MHF4U 指数函数与对数函数单元(对数运算律、对数方程、指数反函数、底数 $e$)
🇨🇦 British Columbia不列颠哥伦比亚PC12 Big Idea "Using inverses is the foundation of solving equations" · Content: exponential functions and equations (incl. base $e$); logarithms: operations, functions, and equations (common & natural logs, inverse of exponential); geometric sequences connecting to exponential functionsPC12 大概念"使用反函数是解方程的基础" · 内容:指数函数与方程(含底数 $e$);对数:运算、函数与方程(常用对数与自然对数、指数的反函数);等比数列与指数函数的衔接
Full 3-column Syllabus Map lives in ../Study Guides/Unit_5_Exponential_and_Logarithmic_Functions.html.完整的三列大纲对照表见 ../Study Guides/Unit_5_Exponential_and_Logarithmic_Functions.html。