PART I · PAPER 1 SECTION A第一部分 · 第一卷 A 节No calculator · short response · 21 marks不可使用计算器 · 简答题 · 21 分
Section A · Short ResponseA 节 · 简答题
State the domain (and range, when asked) for every function you write or derive. When inverting, swap $x$ and $y$ then solve; the domain of $f^{-1}$ equals the range of $f$. No calculator permitted.写出或求得任一函数时，给出定义域（题目要求时也给值域）。求反函数：交换 $x, y$ 后解出；$f^{-1}$ 的定义域 $=$ $f$ 的值域。不可使用计算器。
Q1EASYPaper 1A2.2 Domain (Largest)[4 marks]
Find the largest (real) domain of $f(x) = \dfrac{\sqrt{x - 1}}{x - 3}$.求 $f(x) = \dfrac{\sqrt{x - 1}}{x - 3}$ 的最大（实数）定义域。
(a)State the condition the radical $\sqrt{x - 1}$ imposes on $x$.写出根号 $\sqrt{x - 1}$ 对 $x$ 的限制条件。[1]
(b)State the condition the denominator $x - 3$ imposes.写出分母 $x - 3$ 的限制条件。[1]
(c)Combine both conditions and write the domain in set notation.合并两条件，用集合记号写出定义域。[2]
Q2MEDIUMPaper 1A2.1 Perpendicular Line[5 marks]
Find the equation of the straight line through $(1, -2)$ that is perpendicular to $3x - 2y = 6$. Give the answer in slope-intercept form.求过点 $(1, -2)$ 且垂直于 $3x - 2y = 6$ 的直线方程。以斜截式给出答案。
(a)Rewrite $3x - 2y = 6$ in slope-intercept form and state its slope $m_{1}$.把 $3x - 2y = 6$ 化为斜截式，并写出其斜率 $m_{1}$。[2]
(b)Compute the slope $m_{2}$ of any line perpendicular to it.求与之垂直的任一直线的斜率 $m_{2}$。[1]
(c)Use point-slope through $(1, -2)$ and simplify.用过 $(1, -2)$ 的点斜式并化简。[2]
Q3MEDIUMPaper 1A2.5 Inverse of a Rational[6 marks]
Let $f(x) = \dfrac{2x - 1}{x + 3}$ for $x \in \mathbb{R},\; x \ne -3$.设 $f(x) = \dfrac{2x - 1}{x + 3}$，$x \in \mathbb{R}$，$x \ne -3$。
(a)Write $y = f(x)$ and swap $x$ and $y$.写 $y = f(x)$ 后交换 $x$ 与 $y$。[1]
(b)Solve for $y$ to obtain $f^{-1}(x)$ in simplified form.对 $y$ 解出，化简得到 $f^{-1}(x)$。[3]
(c)State the domain of $f^{-1}$, and explain how it is read off the form of $f^{-1}(x)$.写出 $f^{-1}$ 的定义域，并说明如何从 $f^{-1}(x)$ 的形式中读出。[2]
Let $f(x) = x^{2} + 1$ for $x \in \mathbb{R}$, and $g(x) = \sqrt{x - 1}$ for $x \ge 1$.设 $f(x) = x^{2} + 1$，$x \in \mathbb{R}$；$g(x) = \sqrt{x - 1}$，$x \ge 1$。
(a)Find $(f \circ g)(x)$ in simplified form and state its domain.求 $(f \circ g)(x)$ 的化简式，并写出其定义域。[3]
(b)Find $(g \circ f)(x)$ in simplified form (use $|x|$ where appropriate) and state its domain.求 $(g \circ f)(x)$ 的化简式（必要时用 $|x|$），并写出其定义域。[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
For "show that" parts, every algebraic step must be visible. When solving $f(f(x)) = x$ for general parameters, equate coefficients of like powers of $x$ on both sides; do not divide by an expression that could be zero without justification."证明"题须显式写出每一步代数运算。对一般参数解 $f(f(x)) = x$ 时，两边按 $x$ 的同次幂比较系数；未论证非零前不可除以某表达式。
Let $f(x) = \dfrac{a x + b}{c x + d}$, where $a, b, c, d \in \mathbb{R}$ with $c \ne 0$ and $a d - b c \ne 0$.设 $f(x) = \dfrac{a x + b}{c x + d}$，其中 $a, b, c, d \in \mathbb{R}$，$c \ne 0$ 且 $a d - b c \ne 0$。
(a)Show that $f(f(x)) = \dfrac{(a^{2} + b c)\,x + b(a + d)}{c(a + d)\,x + (b c + d^{2})}.$证明 $f(f(x)) = \dfrac{(a^{2} + b c)\,x + b(a + d)}{c(a + d)\,x + (b c + d^{2})}$。[3]
(b)Hence prove that $f$ is self-inverse (that is, $f = f^{-1}$) if and only if $a + d = 0$.由此证明 $f$ 自逆（即 $f = f^{-1}$）当且仅当 $a + d = 0$。[4]
(c)Apply the criterion to decide which of the following are self-inverse: (i) $g(x) = \dfrac{2x + 3}{x - 2}$; (ii) $h(x) = \dfrac{3x - 1}{x + 3}$.用此判据判断下列是否自逆：(i) $g(x) = \dfrac{2x + 3}{x - 2}$；(ii) $h(x) = \dfrac{3x - 1}{x + 3}$。[2]
(d)State the geometric meaning of the condition $a + d = 0$ in terms of the graph of $f$.用 $f$ 图像的语言陈述条件 $a + d = 0$ 的几何意义。[2]
PART III · PAPER 2第三部分 · 第二卷Calculator · mixed response · 16 marks可使用计算器 · 混合题型 · 16 分
Paper 2 · Calculator Permitted第二卷 · 允许使用计算器
A graphing calculator is required. Give graphical-intersection answers to at least 3 significant figures unless an exact value is asked. Always sketch the viewing window you used and label intersection points.需要图形计算器（GDC）。图像交点答案保留至少 3 位有效数字（除非要求精确值）。要画出所用的取窗，并标出交点。
Solve the equation $e^{x} = 4 - x^{2}$ for $x \in [-3, 3]$ using a graphical method on the GDC.用 GDC 的图像法求方程 $e^{x} = 4 - x^{2}$ 在 $[-3, 3]$ 上的解。
(a)Sketch (or describe) the graphs of $y = e^{x}$ and $y = 4 - x^{2}$ on $[-3, 3]$; mark intersection points.在 $[-3, 3]$ 上画出（或描述）$y = e^{x}$ 与 $y = 4 - x^{2}$，并标出交点。[2]
(b)Use the GDC intersect feature to find both solutions to $3$ significant figures.用 GDC 的 intersect 功能求出两个解（保留 $3$ 位有效数字）。[3]
(c)Hence write the solution set of the inequality $e^{x} > 4 - x^{2}$ on $[-3, 3]$ as a union of intervals.由此把不等式 $e^{x} > 4 - x^{2}$ 在 $[-3, 3]$ 上的解集写为区间并集。[2]
Q7HARDPaper 22.5 + 2.10 Inverse via Graph (HL)[9 marks]
Let $f(x) = x^{3} + x + 1$ for $x \in \mathbb{R}$.设 $f(x) = x^{3} + x + 1$，$x \in \mathbb{R}$。
(a)Show algebraically that $f$ is strictly increasing on $\mathbb{R}$, and explain why $f^{-1}$ exists.代数证明 $f$ 在 $\mathbb{R}$ 上严格递增，并说明为何 $f^{-1}$ 存在。[2]
(b)Use the GDC to find $f^{-1}(3)$ correct to $3$ significant figures. (Hint: solve $x^{3} + x + 1 = 3$ graphically.)用 GDC 求 $f^{-1}(3)$（保留 $3$ 位有效数字）。（提示：图像法求 $x^{3} + x + 1 = 3$。）[3]
(c)The graphs of $y = f(x)$ and $y = f^{-1}(x)$ intersect on the line $y = x$. Find the $x$-coordinate of the intersection to $3$ significant figures (solve $f(x) = x$).$y = f(x)$ 与 $y = f^{-1}(x)$ 的图像交点位于直线 $y = x$ 上。求交点的 $x$ 坐标（解 $f(x) = x$，保留 $3$ 位有效数字）。[3]
(d)State why the symmetry of $f$ and $f^{-1}$ about $y = x$ guarantees that any real solution of $f(x) = f^{-1}(x)$ also satisfies $f(x) = x$.说明 $f$ 与 $f^{-1}$ 关于 $y = x$ 对称为何保证 $f(x) = f^{-1}(x)$ 的实数解也满足 $f(x) = x$。[1]
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. State each "claim" before its proof, and conclude with a clean classification statement.需要图形计算器（GDC）。方法分占主导。每个"断言"在证明前显式陈述，最后给出干净的分类结论。
A function $f : \mathbb{R} \to \mathbb{R}$ is called self-inverse if $f \circ f = \mathrm{id}$, that is, $f(f(x)) = x$ for every $x \in \mathbb{R}$.称函数 $f : \mathbb{R} \to \mathbb{R}$ 为自逆函数，若 $f \circ f = \mathrm{id}$，即对每个 $x \in \mathbb{R}$，$f(f(x)) = x$。
(a)Let $f(x) = m x + k$, $m, k \in \mathbb{R}$. Show that $f$ is self-inverse if and only if either (i) $m = 1$ and $k = 0$, or (ii) $m = -1$ and $k$ is arbitrary. Conclude: the non-trivial linear self-inverse functions on $\mathbb{R}$ are exactly $f(x) = c - x$ for $c \in \mathbb{R}$.设 $f(x) = m x + k$，$m, k \in \mathbb{R}$。证明 $f$ 自逆当且仅当 (i) $m = 1$ 且 $k = 0$，或 (ii) $m = -1$、$k$ 任意。结论：$\mathbb{R}$ 上非平凡的线性自逆函数恰为 $f(x) = c - x$（$c \in \mathbb{R}$）。[5]
(b)Show that every self-inverse function on $\mathbb{R}$ has graph symmetric about the line $y = x$, and that its graph meets $y = x$ at every fixed point of $f$. (A fixed point is an $x$ with $f(x) = x$.)证明 $\mathbb{R}$ 上每个自逆函数的图像关于直线 $y = x$ 对称；且其图像与 $y = x$ 交于 $f$ 的每个不动点。（不动点指满足 $f(x) = x$ 的 $x$。）[3]
(c)For $f(x) = c - x$, find all fixed points in terms of $c$, and verify your answer agrees with (b).对 $f(x) = c - x$，用 $c$ 表示所有不动点，并验证与 (b) 一致。[2]
(d)Define $g(x) = a - x$ and $h(x) = b - x$ on $\mathbb{R}$, with $a, b \in \mathbb{R}$. Compute $(g \circ h)(x)$ and $(h \circ g)(x)$, and decide whether either composite is self-inverse. (Justify, do not just compute one value.)在 $\mathbb{R}$ 上定义 $g(x) = a - x$ 与 $h(x) = b - x$，$a, b \in \mathbb{R}$。计算 $(g \circ h)(x)$ 与 $(h \circ g)(x)$，并判断两复合函数是否自逆。（须论证，不要只验一个数值。）[3]
(e)Find a non-linear self-inverse function on $\mathbb{R} \setminus \{0\}$ (other than the linear family above), and verify it satisfies $f(f(x)) = x$.在 $\mathbb{R} \setminus \{0\}$ 上举出一个非线性的自逆函数（不要上述线性族中的函数），并验证 $f(f(x)) = x$。[2]