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 each transformation in words before drawing. When asked for the image of a point, give the coordinate transformation explicitly: $(x, y) \mapsto (\text{new } x, \text{new } y)$. Read inside-the-bracket operations as acting on the input (horizontal, reverse sign) and outside-the-bracket operations as acting on the output (vertical, direct sign). No calculator permitted.动笔前先用文字写出每个变换。求点的像时，显式写出坐标映射：$(x, y) \mapsto (\text{new } x, \text{new } y)$。括号内的运算作用于输入（横向，符号反转），括号外的运算作用于输出（纵向，符号直接）。不可使用计算器。
The graph of $y = f(x)$ is transformed to give the graph of $y = -2 f(x - 3) + 1$. Describe the sequence of four transformations applied to $y = f(x)$, in the correct order.把 $y = f(x)$ 的图像变换为 $y = -2 f(x - 3) + 1$。按正确顺序描述施加于 $y = f(x)$ 的四个变换。
(a)State the horizontal translation.写出横向平移。[1]
(b)State the vertical stretch factor.写出纵向伸缩系数。[1]
(c)State the reflection.写出反射。[1]
(d)State the vertical translation.写出纵向平移。[1]
Q2MEDIUMPaper 1ASL 2.11 Image of a Point[5 marks]
The graph of $y = f(x)$ passes through the point $P(2, 5)$.$y = f(x)$ 的图像过点 $P(2, 5)$。
(a)Find the image $P'$ of $P$ on the graph of $y = f(2 x) - 3$.求 $P$ 在 $y = f(2 x) - 3$ 上的像 $P'$。[2]
(b)Find the image of $P$ on the graph of $y = 3 f\!\left(\dfrac{x}{2}\right) + 4$.求 $P$ 在 $y = 3 f\!\left(\dfrac{x}{2}\right) + 4$ 上的像。[2]
(c)State, in one sentence, the general coordinate mapping $(x, y) \mapsto (\ldots, \ldots)$ that takes a point on $y = f(x)$ to its image on $y = a f(b x - c) + d$.用一句话写出把 $y = f(x)$ 上的点映射到 $y = a f(b x - c) + d$ 上对应点的通用坐标映射 $(x, y) \mapsto (\ldots, \ldots)$。[1]
Q3MEDIUMPaper 1ASL 2.11 Equation from Description[6 marks]
A curve $C$ has equation $y = f(x)$, where $f$ has range $[-1, 4]$ and is zero only at $x = 0$ and $x = 6$. The curve $C$ is then subjected, in order, to the following four transformations:曲线 $C$ 的方程为 $y = f(x)$，其中 $f$ 的值域为 $[-1, 4]$，零点仅在 $x = 0$ 与 $x = 6$。$C$ 按以下顺序经过四个变换：
a horizontal stretch with scale factor $\tfrac{1}{2}$;横向伸缩系数 $\tfrac{1}{2}$；
a translation by the vector $\binom{1}{0}$;按向量 $\binom{1}{0}$ 平移；
a reflection in the $x$-axis;关于 $x$ 轴反射；
a translation by the vector $\binom{0}{5}$.按向量 $\binom{0}{5}$ 平移。
(a)Write the equation of the resulting curve in the form $y = a f(b(x - h)) + k$. State $a, b, h, k$.把所得曲线的方程写成 $y = a f(b(x - h)) + k$ 的形式，并写出 $a, b, h, k$。[3]
(b)State the range of the resulting function.写出所得函数的值域。[2]
(c)Find the $x$-coordinates of the zeros of the resulting function.求所得函数零点的 $x$ 坐标。[1]
Q4HARDPaper 1AAHL 2.16 Outer Absolute Value (HL)[6 marks]
Consider the function $g(x) = |x^{2} - 4|$.考虑函数 $g(x) = |x^{2} - 4|$。
(a)Sketch the graph of $g$ on $-3 \le x \le 3$. Label the zeros, the corner points, and the local maximum produced by the absolute-value fold.在 $-3 \le x \le 3$ 上画出 $g$ 的图像，标出零点、尖点以及绝对值翻折产生的局部极大。[3]
(b)State the range of $g$ on $\mathbb{R}$.写出 $g$ 在 $\mathbb{R}$ 上的值域。[1]
(c)Find all real solutions of $g(x) = 3$.求 $g(x) = 3$ 的所有实数解。[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
For each sketch, mark the image of every zero, every vertex, and every $y$-intercept. Sketches without labelled key features score the M1 only. For absolute-value sketches, mark the fold axis ($x$-axis or $y$-axis) on the diagram.每张草图须标出每个零点、每个顶点与 $y$ 截距的像。未标关键特征的草图只能拿 M1。绝对值草图须在图上标出反射轴（$x$ 轴或 $y$ 轴）。
Let $f(x) = x^{2} - 2 x$. The graph of $y = f(x)$ has zeros at $x = 0$ and $x = 2$, and vertex at $(1, -1)$.设 $f(x) = x^{2} - 2 x$。$y = f(x)$ 的图像零点为 $x = 0$ 与 $x = 2$，顶点 $(1, -1)$。
(a)On separate axes, sketch the graphs of $y = f(x + 1)$ and $y = -f(x)$. For each, state the zeros and the vertex.在不同坐标系下分别画 $y = f(x + 1)$ 与 $y = -f(x)$ 的图像。各写出零点与顶点。[3]
(b)On a single set of axes, sketch the graph of $y = f(-x)$. State its zeros and vertex, and explain in one sentence why the transformation produces the same parabola as a reflection in the line $x = 1$.在一组坐标系下画 $y = f(-x)$ 的图像。写出零点与顶点，并用一句话解释为何该变换与"关于直线 $x = 1$ 反射"产生同一抛物线。[2]
(c)Sketch $y = |f(x)|$. State the zeros, the location of any corner points, and the local maximum produced by the fold.画 $y = |f(x)|$。写出零点、尖点位置以及由翻折产生的局部极大。[3]
(d)Sketch $y = f(|x|)$. State all zeros, all local minima, and the local maximum at $x = 0$.画 $y = f(|x|)$。写出所有零点、所有局部极小以及 $x = 0$ 处的局部极大。[3]
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. When asked to confirm two expressions are equivalent, you may overlay their graphs on the GDC and report what you observe (single curve, no second curve visible) but you must also justify algebraically. State the window you used.需要图形计算器（GDC）。要求确认两个表达式等价时，可在 GDC 上叠加两图并报告所见（单一曲线，看不到第二条），但须同时给出代数论证。须写出所用窗口。
(a)Graph $y_{1}$ and $y_{2}$ on the GDC on the window $-\pi \le x \le \pi$, $-1.5 \le y \le 1.5$. State what you observe and what this suggests about the relationship between the two expressions.在 GDC 上以窗口 $-\pi \le x \le \pi$、$-1.5 \le y \le 1.5$ 绘出 $y_{1}$ 与 $y_{2}$。写出观察结果及其暗示的两式关系。[2]
(b)Prove the equivalence algebraically by factoring the argument of $y_{1}$.通过对 $y_{1}$ 的自变量提因式，代数证明两者等价。[2]
(c)Hence describe the single sequence of two transformations that maps $y = \sin x$ to $y_{1}$.由此描述把 $y = \sin x$ 映射到 $y_{1}$ 的两步变换序列。[2]
(d)State the period and the phase shift of $y_{1}$ from $y = \sin x$.写出 $y_{1}$ 的周期，以及相对 $y = \sin x$ 的相位平移。[1]
Q7HARDPaper 2AHL 2.16 Reciprocal Sketch from Quadratic (HL)[9 marks]
Let $f(x) = x^{2} - 1$.设 $f(x) = x^{2} - 1$。
(a)Identify the zeros of $f$ and the coordinates of its vertex.写出 $f$ 的零点与顶点坐标。[1]
(b)For $y = \dfrac{1}{f(x)}$, state the equations of all vertical asymptotes and of the horizontal asymptote.对 $y = \dfrac{1}{f(x)}$，写出所有竖直渐近线与水平渐近线的方程。[2]
(c)Find the local extremum of $y = \dfrac{1}{f(x)}$ and state whether it is a local maximum or a local minimum.求 $y = \dfrac{1}{f(x)}$ 的局部极值，并说明是极大还是极小。[2]
(d)Sketch $y = \dfrac{1}{f(x)}$ on $-3 \le x \le 3$, $-3 \le y \le 3$. Mark the asymptotes, the extremum, and the sign of $y$ on each of the three intervals separated by the asymptotes.在 $-3 \le x \le 3$、$-3 \le y \le 3$ 上画 $y = \dfrac{1}{f(x)}$。标出渐近线、极值，以及被渐近线分成的三个区间上 $y$ 的符号。[3]
(e)Find all $x$ at which the graphs of $y = f(x)$ and $y = 1/f(x)$ cross.求 $y = f(x)$ 与 $y = 1/f(x)$ 图像所有交点的 $x$ 坐标。[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. When sketching nested transformations such as $y = |f(|x|)|$, draw the intermediate stage as a faint construction curve and label the final graph in bold. For fixed-point work, plot $y = f(x)$ together with $y = x$ on the GDC, read the intersections, and confirm by algebra.需要图形计算器（GDC）。方法分占主导。画 $y = |f(|x|)|$ 等嵌套变换时，把中间步骤画成淡色辅助曲线，终图加粗。处理不动点时，在 GDC 上同绘 $y = f(x)$ 与 $y = x$，读出交点并以代数验证。
Q8HARDPaper 3AHL 2.16 Nested Absolute Value + Composition (HL)[15 marks]
Let $f(x) = x^{2} - 2$ for all $x \in \mathbb{R}$.设 $f(x) = x^{2} - 2$，$x \in \mathbb{R}$。
(a)Sketch $y = f(x)$. State its zeros, its vertex, and its range.画 $y = f(x)$。写出零点、顶点与值域。[2]
(b)Sketch $y = f(|x|)$, $y = |f(x)|$, and $y = |f(|x|)|$ on separate axes. Note that $f$ is even, so $y = f(|x|)$ is identical to $y = f(x)$. Use this fact to deduce that $y = |f(|x|)| = |f(x)|$. State the zeros, the local extrema, and (for the absolute-value sketches) the corner points of the final graph.在不同坐标系下分别画 $y = f(|x|)$、$y = |f(x)|$ 与 $y = |f(|x|)|$。注意 $f$ 为偶函数，故 $y = f(|x|)$ 与 $y = f(x)$ 完全相同。由此推出 $y = |f(|x|)| = |f(x)|$。写出终图的零点、局部极值以及（绝对值草图的）尖点。[5]
(c)Find the fixed points of $f$, i.e., all real $x$ such that $f(x) = x$. Show the algebra and confirm using the GDC by sketching $y = f(x)$ and $y = x$ on the same axes.求 $f$ 的不动点，即所有满足 $f(x) = x$ 的实数 $x$。展示代数过程，并在 GDC 上同绘 $y = f(x)$ 与 $y = x$ 加以确认。[3]
(d)Define $h(x) = f(f(x))$. Show that $h(x) = x^{4} - 4 x^{2} + 2$ and hence find the four real solutions of $h(x) = x$. (Hint: every fixed point of $f$ is also a fixed point of $h$, so $(x - 2)(x + 1)$ divides the polynomial $h(x) - x$. Carry out the division to find the remaining two roots; give exact surd form.)定义 $h(x) = f(f(x))$。证明 $h(x) = x^{4} - 4 x^{2} + 2$，由此求 $h(x) = x$ 的四个实数解。（提示：$f$ 的每个不动点都是 $h$ 的不动点，故 $(x - 2)(x + 1)$ 整除多项式 $h(x) - x$。作多项式除法求其余两根；以精确根式给出。）[5]