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
Show every step. Optimisation answers must justify the maximum/minimum with a second-derivative or sign-change argument. Kinematics: state units, and never confuse displacement with distance. Areas: sketch the region first. No calculator permitted.写出每一步。最优化题须用二阶导或符号变化论证最大/最小值。运动学：注明单位，切勿把位移与路程混为一谈。面积：先画区域草图。不可使用计算器。
A rectangular pen is built against a long straight wall. The wall forms one side; the other three sides use $40$ m of fencing. Let $w$ m be the side perpendicular to the wall and $\ell$ m the side parallel to the wall.在一段长直墙边搭建矩形围栏。墙构成其中一条边；其余三条边共用 $40$ m 围栏。设 $w$ m 为垂直墙的边长，$\ell$ m 为平行墙的边长。
(a)Write $\ell$ in terms of $w$ using the fencing constraint, then express the enclosed area $A$ as a function of $w$ alone.用围栏约束把 $\ell$ 写成 $w$ 的函数，并把面积 $A$ 仅用 $w$ 表示。[2]
(b)Find the value of $w$ that maximises $A$, and state the maximum area. Justify that it is a maximum.求使 $A$ 取最大值的 $w$，并写出最大面积。须论证此为最大值。[2]
Q2MEDIUMPaper 1ASL 5.9 Kinematics (Times at Rest)[5 marks]
A particle moves on a line so that its displacement (in metres) from the origin at time $t \ge 0$ (in seconds) is $$ s(t) \;=\; t^{3} - 6t^{2} + 9t + 1. $$一质点沿直线运动，其在 $t \ge 0$（秒）时相对原点的位移（米）为 $$ s(t) \;=\; t^{3} - 6t^{2} + 9t + 1。 $$
(a)Find $v(t)$ and the values of $t$ at which the particle is instantaneously at rest.求 $v(t)$ 及质点瞬时静止的 $t$ 值。[2]
(b)Find $a(t)$ and determine whether the particle is speeding up or slowing down at $t = 2$.求 $a(t)$ 并判断 $t = 2$ 时质点是在加速还是减速。[2]
(c)State the displacement at each rest instant.写出每个静止瞬刻的位移。[1]
Q3MEDIUMPaper 1ASL 5.9 Displacement vs Distance[6 marks]
A particle has velocity $v(t) = t^{2} - 4t + 3$ (m s$^{-1}$) for $0 \le t \le 4$ (s).质点的速度为 $v(t) = t^{2} - 4t + 3$（m s$^{-1}$），$0 \le t \le 4$（秒）。
(a)Factor $v(t)$ and find the times at which $v = 0$ on $[0, 4]$.对 $v(t)$ 因式分解，并在 $[0, 4]$ 上求 $v = 0$ 的时刻。[1]
(b)Find the displacement $\displaystyle\int_{0}^{4} v(t)\,dt$ over $[0, 4]$.求 $[0, 4]$ 上的位移 $\displaystyle\int_{0}^{4} v(t)\,dt$。[2]
(c)Find the total distance $\displaystyle\int_{0}^{4} |v(t)|\,dt$ travelled over $[0, 4]$. Show the sign-split clearly.求 $[0, 4]$ 上的总路程 $\displaystyle\int_{0}^{4} |v(t)|\,dt$。须清楚显示分段处理。[3]
Q4HARDPaper 1ASL 5.11 Area Between Curves[5 marks]
Consider the region $R$ enclosed by the curves $y = x$ and $y = x^{3}$ for $0 \le x \le 1$.考虑由曲线 $y = x$ 与 $y = x^{3}$ 在 $0 \le x \le 1$ 上围成的区域 $R$。
(a)Justify which curve lies above the other on $(0, 1)$ by comparing values at $x = \tfrac{1}{2}$ and citing the sign of $x - x^{3}$.通过比较 $x = \tfrac{1}{2}$ 处的函数值并引用 $x - x^{3}$ 的符号，论证 $(0, 1)$ 上哪条曲线在上。[2]
(b)Compute the area of $R$ as $\displaystyle\int_{0}^{1} \bigl(\text{top} - \text{bottom}\bigr)\,dx$. Give the exact value.用 $\displaystyle\int_{0}^{1} \bigl(\text{上} - \text{下}\bigr)\,dx$ 计算 $R$ 的面积。给出精确值。[3]
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
In optimisation extended responses, set up the objective and constraint separately, eliminate the dependent variable, differentiate, solve, then verify the second-derivative sign. Carry units symbolically when the constants are themselves symbolic ($V_{0}$, $S_{0}$, etc.).长题最优化中，先分别写出目标与约束，消元，求导，求解，最后用二阶导符号验证。当常数本身是符号（如 $V_{0}$、$S_{0}$）时，单位也要符号化保留。
A closed right circular cylinder has fixed volume $V_{0} > 0$, radius $r > 0$, and height $h > 0$. Its surface area (including both lids) is $S = 2\pi r^{2} + 2\pi r h$.一闭合直圆柱具有固定体积 $V_{0} > 0$、半径 $r > 0$、高 $h > 0$。其表面积（含两底）为 $S = 2\pi r^{2} + 2\pi r h$。
(a)Use the volume constraint $V_{0} = \pi r^{2} h$ to express $S$ as a function of $r$ alone.用体积约束 $V_{0} = \pi r^{2} h$ 把 $S$ 仅用 $r$ 表示。[2]
(b)Find $\dfrac{dS}{dr}$ and solve $\dfrac{dS}{dr} = 0$ to show that the surface area is stationary when $r^{3} = \dfrac{V_{0}}{2\pi}$, i.e. $r = \left(\dfrac{V_{0}}{2\pi}\right)^{1/3}$.求 $\dfrac{dS}{dr}$ 并解 $\dfrac{dS}{dr} = 0$，证明表面积在 $r^{3} = \dfrac{V_{0}}{2\pi}$ 即 $r = \left(\dfrac{V_{0}}{2\pi}\right)^{1/3}$ 时取驻点。[3]
(c)Show that $\dfrac{d^{2}S}{dr^{2}} > 0$ at this $r$, so the stationary point is a minimum.证明在此 $r$ 处 $\dfrac{d^{2}S}{dr^{2}} > 0$，故驻点为最小值。[2]
(d)Substitute back into the constraint to express $h$ in terms of $r$, and hence show that the optimal cylinder satisfies $h = 2r$ (height equals diameter).代回约束，把 $h$ 用 $r$ 表示，并由此证明最优圆柱满足 $h = 2r$（高 $=$ 直径）。[4]
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. For volumes of revolution, decide whether the disc/washer formula uses $\pi \int y^{2}\,dx$ (about $x$-axis) or $\pi \int x^{2}\,dy$ (about $y$-axis) before integrating. Related rates: state every chain-rule link explicitly before plugging numbers.需要图形计算器（GDC）。旋转体体积须在积分前先决定盘片/垫圈公式用 $\pi \int y^{2}\,dx$（绕 $x$ 轴）还是 $\pi \int x^{2}\,dy$（绕 $y$ 轴）。相关变化率：代入数值前须显式写出每一条链式法则环节。
Q6MEDIUMPaper 2SL 5.11 Volume of Revolution ($x$-axis)[7 marks]
The region bounded by the curve $y = \sqrt{x}$, the $x$-axis, and the line $x = 4$ is rotated through $2\pi$ about the $x$-axis to form a solid of revolution $\Omega$.由曲线 $y = \sqrt{x}$、$x$ 轴及直线 $x = 4$ 所围区域绕 $x$ 轴旋转 $2\pi$ 角度，得到旋转体 $\Omega$。
(a)Sketch (or describe) the region and write the disc-formula integral for the volume of $\Omega$.作图（或描述）该区域，并写出 $\Omega$ 体积的圆盘公式积分式。[2]
(b)Evaluate the integral exactly. Give the answer in the form $k\pi$.精确计算该积分。结果写成 $k\pi$ 形式。[3]
(c)Confirm your exact answer numerically using your GDC and state both the symbolic and decimal forms.用 GDC 数值核对该精确值，写出符号与小数两种形式。[2]
Q7HARDPaper 2AHL 5.13 Related Rates (Conical Tank, HL)[10 marks]
An inverted right circular cone has its vertex pointing down. At every height the radius equals half the height: $r = \dfrac{h}{2}$. Water is being poured in at a constant rate of $2$ m$^{3}$ min$^{-1}$. Let $V(t)$ and $h(t)$ be the volume and depth of water at time $t$ (min).一倒置的直圆锥（顶点朝下），任意高度处半径为高度之半：$r = \dfrac{h}{2}$。以恒定速率 $2$ m$^{3}$ min$^{-1}$ 注水。设 $V(t)$、$h(t)$ 为时刻 $t$（分钟）水的体积与深度。
(a)Use $V = \tfrac{1}{3}\pi r^{2} h$ and the constraint $r = h/2$ to express $V$ as a function of $h$ alone.由 $V = \tfrac{1}{3}\pi r^{2} h$ 及约束 $r = h/2$，把 $V$ 仅用 $h$ 表示。[2]
(b)Differentiate implicitly with respect to $t$ to obtain $\dfrac{dV}{dt}$ in terms of $h$ and $\dfrac{dh}{dt}$.对 $t$ 隐式求导，得到 $\dfrac{dV}{dt}$ 关于 $h$ 与 $\dfrac{dh}{dt}$ 的表达式。[3]
(c)Find the exact value of $\dfrac{dh}{dt}$ at the instant when $h = 4$ m, given $\dfrac{dV}{dt} = 2$ m$^{3}$ min$^{-1}$.已知 $\dfrac{dV}{dt} = 2$ m$^{3}$ min$^{-1}$，求 $h = 4$ m 时 $\dfrac{dh}{dt}$ 的精确值。[3]
(d)Without further differentiation, explain in one sentence why $\dfrac{dh}{dt}$ decreases as $h$ grows, even though $\dfrac{dV}{dt}$ is constant.不再求导，用一句话解释：尽管 $\dfrac{dV}{dt}$ 恒定，为何 $\dfrac{dh}{dt}$ 随 $h$ 增大而减小。[2]
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. For a $y$-axis volume of revolution, rewrite the curve as $x = g(y)$ and integrate $\pi \int x^{2}\,dy$. Relate the inflow rate to the resulting volume formula via implicit differentiation.需要图形计算器（GDC）。方法分占主导。对绕 $y$ 轴旋转的体积，须先把曲线改写为 $x = g(y)$，再积分 $\pi \int x^{2}\,dy$。利用隐式求导把注入速率与所得体积公式相联系。
Let $R$ be the region in the first quadrant enclosed by the curve $y = x^{2}$, the line $y = 4$, and the $y$-axis. The region $R$ is rotated through $2\pi$ about the $y$-axis to form a bowl-shaped solid $\Omega$. Water is then poured into $\Omega$ from above at a constant rate of $\dfrac{dV}{dt} = 3$ cm$^{3}$ s$^{-1}$ (treat $\Omega$ as a hollow bowl whose interior is the solid; lengths in cm).设 $R$ 为第一象限内由曲线 $y = x^{2}$、直线 $y = 4$ 及 $y$ 轴所围区域。$R$ 绕 $y$ 轴旋转 $2\pi$ 得碗状立体 $\Omega$。以恒定速率 $\dfrac{dV}{dt} = 3$ cm$^{3}$ s$^{-1}$ 从上方注水入 $\Omega$（将 $\Omega$ 视为空心碗，其内腔为该立体；长度单位为 cm）。
(a)Sketch $R$ and identify the limits of $y$ on $\Omega$. Write the disc-formula integral $V(\Omega) = \pi \displaystyle\int x^{2}\,dy$ with $x = g(y)$ and appropriate limits.作 $R$ 的草图，写出 $\Omega$ 上 $y$ 的范围。写出圆盘公式 $V(\Omega) = \pi \displaystyle\int x^{2}\,dy$（$x = g(y)$，并给出积分限）。[3]
(b)Evaluate the integral in (a) exactly. Give the total volume of $\Omega$ in the form $k\pi$.精确计算 (a) 中积分，给出 $\Omega$ 的总体积，形式为 $k\pi$。[3]
(c)Let $h$ be the depth of water in $\Omega$ (so $0 \le h \le 4$). Show that the volume of water when the depth is $h$ is $V(h) = \dfrac{\pi h^{2}}{2}.$设 $\Omega$ 中水深为 $h$（$0 \le h \le 4$）。证明水深 $h$ 时水的体积为 $V(h) = \dfrac{\pi h^{2}}{2}$。[3]
(d)Hence find $\dfrac{dh}{dt}$ as a function of $h$, and evaluate it at $h = 2$ cm.由此求 $\dfrac{dh}{dt}$ 作为 $h$ 的函数，并在 $h = 2$ cm 处求值。[3]
(e)Find the time $T$ (in seconds) needed to fill the bowl completely. Compare $T$ with the bowl's total volume $V(\Omega)$ from (b) and verify your two answers are consistent.求注满该碗所需时间 $T$（秒）。用 (b) 中的 $V(\Omega)$ 比较并验证两答案相容。[3]