PART I · PAPER 1 SECTION A第一部分 · 第一卷 A 节No calculator · short response · 20 marks不可使用计算器 · 简答题 · 20 分
Section A · Short ResponseA 节 · 简答题
No calculator. Quote the standard Maclaurin series from memory (you must memorise $e^x$, $\sin x$, $\cos x$, $\ln(1+x)$, $(1+x)^n$). When you substitute into a known series, write at least the first three non-zero terms before simplifying. For L'Hopital, declare the indeterminate form ($0/0$ or $\infty/\infty$) before differentiating.不可使用计算器。需默写五个标准麦克劳林级数（$e^x$、$\sin x$、$\cos x$、$\ln(1+x)$、$(1+x)^n$）。代入已知级数时，先写出至少前三个非零项再化简。使用洛必达前，必须先声明不定型（$0/0$ 或 $\infty/\infty$）。
Q1EASYPaper 1AAHL 5.18 Standard Series Recall (HL)[4 marks]
The Maclaurin series for $e^{x}$ is one of the five standard series you are expected to know by heart.$e^{x}$ 的麦克劳林级数是五个必背标准级数之一。
(a)Write the first three terms (constant, linear, quadratic) of the Maclaurin series for $e^{x}$.写出 $e^{x}$ 麦克劳林级数的前三项（常数项、一次项、二次项）。[2]
(b)Use the three-term truncation from (a) with $x = 0.1$ to give an approximation to $e^{0.1}$. Give your answer as an exact fraction or decimal.用 (a) 中的三项截断在 $x = 0.1$ 处近似 $e^{0.1}$，给出精确分数或小数。[2]
Q2MEDIUMPaper 1AAHL 5.18 Series by Substitution (HL)[5 marks]
The function $g(x) = e^{-x^{2}}$ has no elementary antiderivative; its Maclaurin series is the standard way to handle it (it underlies the Gaussian error function).函数 $g(x) = e^{-x^{2}}$ 没有初等原函数；高斯误差函数的处理就基于其麦克劳林级数。
(a)Starting from $e^{u} = 1 + u + \tfrac{u^{2}}{2!} + \tfrac{u^{3}}{3!} + \cdots$, substitute $u = -x^{2}$ to find the first three non-zero terms of the Maclaurin series for $e^{-x^{2}}$.由 $e^{u} = 1 + u + \tfrac{u^{2}}{2!} + \tfrac{u^{3}}{3!} + \cdots$ 出发，代入 $u = -x^{2}$，求 $e^{-x^{2}}$ 麦克劳林级数的前三个非零项。[3]
(b)State the general term of the series, written as a closed-form expression in $n$.写出该级数的通项（用 $n$ 表达的闭式）。[1]
(c)State the values of $x$ for which the series converges to $e^{-x^{2}}$, with brief justification.写出使该级数收敛到 $e^{-x^{2}}$ 的 $x$ 取值范围，并简要说明理由。[1]
In this question you will derive the Maclaurin series for $\arctan x$ by integrating a geometric series term by term.本题用对几何级数逐项积分的方法推导 $\arctan x$ 的麦克劳林级数。
(a)Treating $\dfrac{1}{1 + x^{2}}$ as $\dfrac{1}{1 - (-x^{2})}$, write down the first four non-zero terms of its Maclaurin series. State the range of $x$ for which the series converges.把 $\dfrac{1}{1 + x^{2}}$ 视为 $\dfrac{1}{1 - (-x^{2})}$，写出其麦克劳林级数的前四个非零项，并写出收敛范围。[3]
(b)Recalling $\displaystyle\int \dfrac{1}{1 + x^{2}}\, dx = \arctan x + C$, integrate the series in (a) term by term to obtain the first four non-zero terms of the Maclaurin series for $\arctan x$. Justify why $C = 0$.由 $\displaystyle\int \dfrac{1}{1 + x^{2}}\, dx = \arctan x + C$，把 (a) 中的级数逐项积分，得到 $\arctan x$ 麦克劳林级数的前四个非零项；并说明为何 $C = 0$。[3]
(a)Show that direct substitution gives an indeterminate form, and state which form ($0/0$ or $\infty/\infty$).证明直接代入得到不定型，并写出是哪种（$0/0$ 或 $\infty/\infty$）。[1]
(b)Apply L'Hopital's rule once. Show that the new limit is still indeterminate and identify the form again.应用一次洛必达。证明新极限仍为不定型，并再次写出其形式。[2]
(c)Apply L'Hopital's rule a second time and evaluate the limit.再次应用洛必达并求出极限。[2]
PART II · PAPER 1 SECTION B第二部分 · 第一卷 B 节No calculator · extended response · 11 marks不可使用计算器 · 长答题 · 11 分
Section B · Extended ResponseB 节 · 长答题
No calculator. When building a Maclaurin series from the general formula $f(x) = \sum f^{(n)}(0)\, x^{n}/n!$, lay out a derivative table ($n$, $f^{(n)}(x)$, $f^{(n)}(0)$, coefficient $f^{(n)}(0)/n!$). The table is worth method marks even when the algebra goes wrong.不可使用计算器。用通项公式 $f(x) = \sum f^{(n)}(0)\, x^{n}/n!$ 构造级数时，列出导数表（$n$、$f^{(n)}(x)$、$f^{(n)}(0)$、系数 $f^{(n)}(0)/n!$）。即使后续代数出错，导数表本身亦可拿方法分。
Q5HARDPaper 1BAHL 5.18 Build Series from General Formula (HL)[11 marks]
Let $f(x) = \ln(1 + 2x)$.设 $f(x) = \ln(1 + 2x)$。
(a)Compute $f^{(n)}(0)$ for $n = 0, 1, 2, 3$ by direct differentiation. Present the result as a derivative table.通过直接求导计算 $f^{(n)}(0)$（$n = 0, 1, 2, 3$），以导数表的形式给出结果。[5]
(b)Using the Maclaurin formula, write the first four terms of the Maclaurin series for $f(x) = \ln(1 + 2x)$.用麦克劳林公式写出 $f(x) = \ln(1 + 2x)$ 的麦克劳林级数前四项。[2]
(c)Reach the same series in (b) by an independent route: substitute $u = 2x$ into the standard series $\ln(1 + u) = u - \tfrac{u^{2}}{2} + \tfrac{u^{3}}{3} - \cdots$. State the range of $x$ for which the substituted series converges.用独立方法验证 (b)：在标准级数 $\ln(1 + u) = u - \tfrac{u^{2}}{2} + \tfrac{u^{3}}{3} - \cdots$ 中代入 $u = 2x$。写出代换后级数的收敛范围。[3]
(d)State the general term of the series in closed form, written for $n \ge 1$.写出该级数的通项（闭式，$n \ge 1$）。[1]
PART III · PAPER 2第三部分 · 第二卷Calculator · mixed response · 17 marks可使用计算器 · 混合题型 · 17 分
Paper 2 · Calculator Permitted第二卷 · 允许使用计算器
A graphing calculator is required. Show the partial sum you computed, then state the rounded approximation to the requested number of decimal places. For limit questions in this part, expand the relevant standard series, cancel the lowest-order terms exactly, then take the limit.需要图形计算器（GDC）。先写出你计算的部分和，再按要求的小数位数四舍五入给出近似值。本部分极限题中，先展开相关标准级数，精确抵消最低阶项，再取极限。
Approximate $\cos(0.5)$ using the first four terms of the Maclaurin series for $\cos x$ (the terms in $x^{0}$, $x^{2}$, $x^{4}$, $x^{6}$).用 $\cos x$ 麦克劳林级数的前四项（$x^{0}$、$x^{2}$、$x^{4}$、$x^{6}$ 各项）近似 $\cos(0.5)$。
(a)Write out the first four terms of $\cos x$ explicitly with denominators evaluated.显式写出 $\cos x$ 的前四项（分母化为数值）。[1]
(b)Substitute $x = 0.5$ and compute each of the four terms to at least 6 decimal places.代入 $x = 0.5$，把四项各自计算到至少 $6$ 位小数。[3]
(c)State your approximation for $\cos(0.5)$ rounded to $4$ decimal places.给出 $\cos(0.5)$ 的近似值，四舍五入到 $4$ 位小数。[1]
(d)Using your GDC's cos function, state $\cos(0.5)$ to $4$ decimal places and write the absolute error of your approximation.用 GDC 的 cos 函数给出 $\cos(0.5)$ 至 $4$ 位小数，并写出近似的绝对误差。[2]
Q7HARDPaper 2AHL 5.18 Limit via Series Expansion (HL)[10 marks]
Consider the limit $\;\displaystyle L = \lim_{x \to 0} \frac{\sin x - x}{x^{3}}$.考虑极限 $\;\displaystyle L = \lim_{x \to 0} \frac{\sin x - x}{x^{3}}$。
(a)Write the first three non-zero terms of the Maclaurin series for $\sin x$ (in $x$, $x^{3}$, $x^{5}$).写出 $\sin x$ 麦克劳林级数的前三个非零项（含 $x$、$x^{3}$、$x^{5}$）。[2]
(b)Subtract $x$, then divide the resulting series by $x^{3}$ to obtain a series for $(\sin x - x)/x^{3}$.将 $x$ 抵消，再把所得级数除以 $x^{3}$，得到 $(\sin x - x)/x^{3}$ 的级数表达。[3]
(c)Take the limit as $x \to 0$ to evaluate $L$.取 $x \to 0$ 的极限，求 $L$。[2]
(d)Verify the same value of $L$ by applying L'Hopital's rule three times. State which indeterminate form ($0/0$) appears at each stage.用三次洛必达法则验证 $L$ 的值。在每一步写出出现的不定型（$0/0$）。[3]
PART IV · PAPER 3第四部分 · 第三卷Calculator · HL extended exploration · 15 marks可使用计算器 · HL 长题探究 · 15 分
Paper 3 · HL Extended Problem第三卷 · HL 长题探究
A graphing calculator is required. Method marks dominate. You may use the alternating series remainder estimate or the Lagrange (Taylor) remainder bound. Whichever bound you use, state its hypothesis (alternating with decreasing-magnitude terms, or a uniform bound on the next derivative) before applying it.需要图形计算器（GDC）。方法分占主导。可使用交错级数余项估计，或拉格朗日（泰勒）余项界。无论用哪个，使用前要写出其前提（交错且项绝对值单调递减，或下一阶导数有一致上界）。
For each non-negative integer $n$, let $S_{n} = \displaystyle\sum_{k=0}^{n} \frac{1}{k!}$ be the $n$-th partial sum of the Maclaurin series for $e^{x}$ evaluated at $x = 1$. The goal of this question is to find the smallest $n$ for which $|S_{n} - e| < 0.001$.对每个非负整数 $n$，令 $S_{n} = \displaystyle\sum_{k=0}^{n} \frac{1}{k!}$ 为 $e^{x}$ 麦克劳林级数在 $x = 1$ 处的第 $n$ 个部分和。本题目标：求使 $|S_{n} - e| < 0.001$ 的最小 $n$。
(a)State the Maclaurin series for $e^{x}$ and its radius of convergence. Verify that $x = 1$ lies in the interval of convergence.写出 $e^{x}$ 的麦克劳林级数及收敛半径，并验证 $x = 1$ 在收敛区间内。[2]
(b)Apply Lagrange's remainder estimate: for $f(x) = e^{x}$ on the interval $[0, 1]$, show that $|R_{n}(1)| \le \dfrac{e}{(n+1)!}$, where $R_{n}(1) = e - S_{n}$.应用拉格朗日余项估计：对 $f(x) = e^{x}$ 在区间 $[0, 1]$ 上，证明 $|R_{n}(1)| \le \dfrac{e}{(n+1)!}$，其中 $R_{n}(1) = e - S_{n}$。[3]
(c)Use (b) to find the smallest $n$ for which the bound guarantees $|R_{n}(1)| < 0.001$. (You may use $e < 2.72$.)由 (b) 求使该界保证 $|R_{n}(1)| < 0.001$ 的最小 $n$。（可使用 $e < 2.72$。）[3]
(d)For the value of $n$ found in (c), compute $S_{n}$ exactly as a fraction $\dfrac{p}{q}$ in lowest terms.对 (c) 中所求的 $n$，把 $S_{n}$ 精确写为最简分数 $\dfrac{p}{q}$。[3]
(e)Compare $S_{n}$ with the GDC value of $e$ (to at least $7$ decimal places). Report the actual absolute error $|S_{n} - e|$ and comment on how it compares with the bound from (c).把 $S_{n}$ 与 GDC 给出的 $e$（至少 $7$ 位小数）作比较。报告实际绝对误差 $|S_{n} - e|$，并评论它与 (c) 中界的关系。[2]
(f)Confirm that for $n - 1$ (one fewer term) the actual error exceeds $0.001$, so the $n$ from (c) is indeed the smallest non-negative integer satisfying the requested tolerance.验证对 $n - 1$（少取一项），实际误差超过 $0.001$，从而 (c) 中的 $n$ 的确是满足该容差的最小非负整数。[2]