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Unit B5 · SolutionsUnit B5 · 解析

Transformations of Graphs and Functions · Solutions图像与函数的变换 · 解析

Companion to the IB-Style Practice SetIB 风格练习题的解析配套

EASY MEDIUM HARD Paper 1A Paper 1B Paper 2 Paper 3

Syllabus SL 2.11 · AHL 2.16考纲 SL 2.11 · AHL 2.16AA HL

PART I  ·  PAPER 1 SECTION A · SOLUTIONS第一部分  ·  第一卷 A 节 · 解析No calculator · 21 marks不可使用计算器 · 21 分

Section A · Worked SolutionsA 节 · 详细解析

Q1EASYPaper 1ASL 2.11 Composite Description[4 marks]

Describe the four transformations that take $y = f(x)$ to $y = -2 f(x - 3) + 1$, in order.按顺序描述把 $y = f(x)$ 变为 $y = -2 f(x - 3) + 1$ 的四个变换。

Answers:答案：  (a) translation $+3$ in $x$沿 $x$ 平移 $+3$  ·  (b) vertical stretch, factor $2$纵向伸缩 $2$  ·  (c) reflection in $x$-axis关于 $x$ 轴反射  ·  (d) translation $+1$ in $y$沿 $y$ 平移 $+1$

(a) Horizontal translation A1

Inside the bracket, $x \mapsto x - 3$. This is the input being replaced, so the graph shifts in the reverse direction of the sign: a translation by $+3$ in the positive $x$-direction (i.e. right by $3$ units).

(b) Vertical stretch A1

Outside, multiplication by $|-2| = 2$ stretches the output: vertical stretch with scale factor $2$ from the $x$-axis.

(c) Reflection A1

The minus sign in $-2 f(\cdot)$ acts on the output: reflection in the $x$-axis.

(d) Vertical translation A1

The $+1$ outside is added to the output: translation by $+1$ in the positive $y$-direction (up by $1$).
Inside-out vs outside-in, and the canonical order. Operations inside the bracket (acting on $x$) are "horizontal and reversed": $x \mapsto x - 3$ shifts the graph right by $3$, not left; $x \mapsto 2x$ compresses horizontally by factor $2$, not stretches. Operations outside (acting on $y$) are "vertical and direct": $y \mapsto 2y$ stretches vertically by $2$; $y \mapsto y + 1$ shifts up by $1$. The IB-preferred order when describing a composite $y = a f(b(x - h)) + k$ is: horizontal stretch ($b$), horizontal shift ($h$), vertical stretch and reflection ($a$), vertical shift ($k$). Many students reverse "right" and "left" for inside-the-bracket shifts; train the muscle by computing the image of one specific point and check.

(a) 横向平移 A1

括号内 $x \mapsto x - 3$。这是输入的替换，图像沿与符号相反方向平移：沿正 $x$ 方向平移 $+3$（右移 $3$）。

(b) 纵向伸缩 A1

括号外乘以 $|-2| = 2$ 作用于输出：以 $x$ 轴为基准的纵向伸缩，系数 $2$。

(c) 反射 A1

$-2 f(\cdot)$ 中的负号作用于输出：关于 $x$ 轴反射。

(d) 纵向平移 A1

括号外的 $+1$ 加在输出上：沿正 $y$ 方向平移 $+1$（上移 $1$）。
"内反外正"与标准顺序。括号内（作用于 $x$）的运算是"横向且方向相反"：$x \mapsto x - 3$ 是右移 $3$，不是左移；$x \mapsto 2x$ 是横向压缩 $2$ 倍，不是拉伸。括号外（作用于 $y$）的运算是"纵向且方向直接"：$y \mapsto 2y$ 纵向拉伸 $2$；$y \mapsto y + 1$ 上移 $1$。描述 $y = a f(b(x - h)) + k$ 这类复合变换时 IB 偏好的顺序是：横向伸缩（$b$）、横向平移（$h$）、纵向伸缩与反射（$a$）、纵向平移（$k$）。许多学生在括号内平移时把"左""右"弄反；取一个具体点算像再核对，是练肌肉记忆的最快办法。
Q2MEDIUMPaper 1ASL 2.11 Image of a Point[5 marks]

$P(2, 5)$ lies on $y = f(x)$. Find its image on (a) $y = f(2x) - 3$, (b) $y = 3 f(x/2) + 4$, and (c) the general rule for $y = a f(bx - c) + d$.$P(2, 5)$ 在 $y = f(x)$ 上。求其在 (a) $y = f(2x) - 3$、(b) $y = 3 f(x/2) + 4$ 上的像，及 (c) $y = a f(bx - c) + d$ 的通用映射。

Answers:答案：  (a) $P' = (1, 2)$  ·  (b) $(4, 19)$  ·  (c) $(x, y) \mapsto \left(\dfrac{x + c}{b},\; a y + d\right)$

(a) Image under $y = f(2x) - 3$ M1·A1

Inside-the-bracket factor $b = 2$ compresses $x$ by factor $2$: $x \mapsto x/2$. Outside, $-3$ shifts $y$ down by $3$: $y \mapsto y - 3$. So $(2, 5) \mapsto (2/2,\; 5 - 3) = (1, 2)$. Check: $y\bigl|_{x = 1} = f(2 \cdot 1) - 3 = f(2) - 3 = 5 - 3 = 2$. $\checkmark$

(b) Image under $y = 3 f(x/2) + 4$ M1·A1

Inside, $b = 1/2$ stretches $x$ by factor $2$: $x \mapsto 2x$. Outside, $3 \cdot y + 4$: $y \mapsto 3y + 4$. So $(2, 5) \mapsto (2 \cdot 2,\; 3 \cdot 5 + 4) = (4, 19)$.

(c) General mapping A1

Write the inside as $b x - c = b(x - c/b)$. The transformations are: horizontal stretch by $1/b$ then shift right by $c/b$ (combined: $x \mapsto (x + c)/b$ if we trace $x$-coordinates forward), and vertical stretch by $a$ then shift up by $d$ (combined: $y \mapsto a y + d$). Hence $$ (x, y) \;\longmapsto\; \left(\frac{x + c}{b},\; a y + d\right). $$
Trace coordinates forward, not the formula. A common trap on image-of-a-point questions: students "do the inverse" because they remember $x \mapsto x - 3$ shifts right, which feels backwards. The cleanest mental model is to trace the original point's coordinates forward through the transformation list: if the new graph has equation $y_{\text{new}} = a f(b(x_{\text{new}} - h)) + k$, then a point $(x_{0}, y_{0})$ on $y = f(x)$ has image $(x_{\text{new}}, y_{\text{new}})$ where $b(x_{\text{new}} - h) = x_{0}$ (so $x_{\text{new}} = x_{0}/b + h$) and $y_{\text{new}} = a y_{0} + k$. Always sanity-check by plugging $x_{\text{new}}$ into $y_{\text{new}} = a f(b(x_{\text{new}} - h)) + k$ and confirming you recover $a y_{0} + k$.

(a) $y = f(2x) - 3$ 下的像 M1·A1

括号内 $b = 2$ 把 $x$ 压缩 $2$ 倍：$x \mapsto x/2$。括号外 $-3$ 把 $y$ 下移 $3$：$y \mapsto y - 3$。故 $(2, 5) \mapsto (2/2,\; 5 - 3) = (1, 2)$。验：$y\bigl|_{x = 1} = f(2) - 3 = 5 - 3 = 2$。$\checkmark$

(b) $y = 3 f(x/2) + 4$ 下的像 M1·A1

括号内 $b = 1/2$ 把 $x$ 拉伸 $2$ 倍：$x \mapsto 2x$。括号外 $3 \cdot y + 4$：$y \mapsto 3y + 4$。故 $(2, 5) \mapsto (2 \cdot 2,\; 3 \cdot 5 + 4) = (4, 19)$。

(c) 通用映射 A1

把括号内写作 $b x - c = b(x - c/b)$。变换为：横向伸缩 $1/b$ 再右移 $c/b$（合并 $x \mapsto (x + c)/b$），纵向伸缩 $a$ 再上移 $d$（合并 $y \mapsto a y + d$）。故 $$ (x, y) \;\longmapsto\; \left(\frac{x + c}{b},\; a y + d\right). $$
追踪坐标前进，而不是套公式。求像题的常见陷阱：因为 $x \mapsto x - 3$ "感觉像左移"，学生反而去"做反操作"。最干净的心理模型是把原点的坐标正向经过变换链：若新图方程为 $y_{\text{new}} = a f(b(x_{\text{new}} - h)) + k$，则原点 $(x_{0}, y_{0})$ 的像 $(x_{\text{new}}, y_{\text{new}})$ 满足 $b(x_{\text{new}} - h) = x_{0}$（即 $x_{\text{new}} = x_{0}/b + h$），$y_{\text{new}} = a y_{0} + k$。把 $x_{\text{new}}$ 代回新方程验回 $a y_{0} + k$ 即可。
Q3MEDIUMPaper 1ASL 2.11 Equation from Description[6 marks]

$y = f(x)$ has range $[-1, 4]$, zeros at $x = 0, 6$. Apply: (i) horizontal stretch $\tfrac{1}{2}$; (ii) shift by $\binom{1}{0}$; (iii) reflect in $x$-axis; (iv) shift by $\binom{0}{5}$. Find (a) the equation $y = a f(b(x - h)) + k$, (b) new range, (c) new zeros.$y = f(x)$ 值域 $[-1, 4]$，零点 $x = 0, 6$。施加 (i) 横向伸缩 $\tfrac{1}{2}$、(ii) 平移 $\binom{1}{0}$、(iii) 关于 $x$ 轴反射、(iv) 平移 $\binom{0}{5}$。求 (a) 方程 $y = a f(b(x - h)) + k$、(b) 新值域、(c) 新零点。

Answers:答案：  (a) $y = -f(2(x - 1)) + 5$, $(a, b, h, k) = (-1, 2, 1, 5)$  ·  (b) $[1, 6]$  ·  (c) no zeros (range $[1, 6]$ excludes 0); the original zeros map to $x = 1, 4$无零点（值域 $[1, 6]$ 不含 0）；原零点的像在 $x = 1, 4$

(a) Build the equation step by step M1·A1·A1

Start with $y = f(x)$.
  • (i) Horizontal stretch factor $\tfrac{1}{2}$ means $x$-coordinates are halved, i.e. $x \mapsto 2x$ inside: $y = f(2x)$.
  • (ii) Shift by $\binom{1}{0}$ (right $1$): $y = f(2(x - 1))$.
  • (iii) Reflect in $x$-axis: $y = -f(2(x - 1))$.
  • (iv) Shift by $\binom{0}{5}$ (up $5$): $y = -f(2(x - 1)) + 5$.
So $a = -1$, $b = 2$, $h = 1$, $k = 5$.

(b) Range M1·A1

Vertical operations only: original range $[-1, 4]$ under $y \mapsto -y$ becomes $[-4, 1]$, then $y \mapsto y + 5$ gives $[1, 6]$.

(c) Zeros A1

Zeros of the new function require $-f(2(x - 1)) + 5 = 0$, i.e. $f(2(x - 1)) = 5$. The range of $f$ is $[-1, 4]$, so $5 \notin \mathrm{range}\,f$: no real zeros. This is consistent with part (b): the new range $[1, 6]$ does not contain $0$. The original zeros of $f$ (at $x = 0, 6$) map under (i)+(ii) to $x = 0/2 + 1 = 1$ and $x = 6/2 + 1 = 4$, but at those $x$-values the new $y = -0 + 5 = 5$, not $0$ — so $x = 1, 4$ are images of the old zeros, not zeros of the new function.
"Zeros of the original" vs "zeros of the new function" — name them. This question is a Paper 1 favourite. The phrase "the zeros of the resulting function" sounds unambiguous but can refer to (i) the images of the original zeros under the transformation (the $x$-coordinates where the old zeros landed), or (ii) the genuine zeros of the new equation (where new $y = 0$). They coincide only when the vertical translation is zero. Here (ii) is empty because $[1, 6]$ excludes zero. Always state which you computed and verify with the new range: a function with range $[1, 6]$ obviously has no zeros, period. Marks tend to reward the image-of-zeros interpretation when the new function clearly has no real zeros, but explicit reasoning protects both.

(a) 一步步搭出方程 M1·A1·A1

从 $y = f(x)$ 起：
  • (i) 横向伸缩 $\tfrac{1}{2}$ 即 $x$ 坐标减半，括号内 $x \mapsto 2x$：$y = f(2x)$。
  • (ii) 按 $\binom{1}{0}$ 平移（右 $1$）：$y = f(2(x - 1))$。
  • (iii) 关于 $x$ 轴反射：$y = -f(2(x - 1))$。
  • (iv) 按 $\binom{0}{5}$ 平移（上 $5$）：$y = -f(2(x - 1)) + 5$。
$a = -1$、$b = 2$、$h = 1$、$k = 5$。

(b) 值域 M1·A1

只有纵向操作：原值域 $[-1, 4]$ 在 $y \mapsto -y$ 下变为 $[-4, 1]$，再 $y \mapsto y + 5$ 得 $[1, 6]$。

(c) 零点 A1

新函数零点须 $-f(2(x - 1)) + 5 = 0$，即 $f(2(x - 1)) = 5$。但 $5 \notin \mathrm{range}\,f = [-1, 4]$，故新函数无零点。注：原零点（$f = 0$ 处）的像在 $x = 1, 4$，但那里新 $y = -0 + 5 = 5$，非 $0$。最终：原零点的像在 $x = 1, 4$；新函数本身无零点（值域 $[1, 6]$ 不含 $0$）。
"原函数零点"与"新函数零点"——必须命名。这是 Paper 1 的常见绊脚石。"所得函数的零点"乍听明确，实可指 (i) 原零点在变换下的像（落点的 $x$ 坐标），或 (ii) 新方程的真零点（新 $y = 0$）。仅当纵向平移为零时两者相同。这里 (ii) 为空，因 $[1, 6]$ 不含零。务必声明计算哪一种，并用新值域核对：值域 $[1, 6]$ 的函数显然无零点。当新函数显然无实零点时，分数倾向"原零点的像"，但显式论证最稳。
Q4HARDPaper 1AAHL 2.16 Outer Absolute Value (HL)[6 marks]

$g(x) = |x^{2} - 4|$. Sketch on $[-3, 3]$; state range on $\mathbb{R}$; solve $g(x) = 3$.$g(x) = |x^{2} - 4|$。在 $[-3, 3]$ 上画图；写 $\mathbb{R}$ 上值域；解 $g(x) = 3$。

Answers:答案：  (b) range $= [0, \infty)$值域 $= [0, \infty)$  ·  (c) $x = \pm 1,\; \pm\sqrt{7}$

(a) Sketch M1·A1·A1

Start with $y = x^{2} - 4$: parabola, vertex $(0, -4)$, zeros at $x = \pm 2$. Apply outer $|\cdot|$: every point with $y < 0$ flips to $-y > 0$. The portion on $-2 < x < 2$ (where $x^{2} - 4 < 0$) folds up about the $x$-axis. The folded vertex $(0, -4)$ becomes a local maximum at $(0, 4)$. Two corner points (non-differentiable) appear at the zeros $x = \pm 2$, where the curve meets the $x$-axis with finite slope from both sides. On $-3 \le x \le 3$: the curve at $x = \pm 3$ gives $|9 - 4| = 5$; zeros at $x = \pm 2$; corner points at $(\pm 2, 0)$; local max at $(0, 4)$.

(b) Range on $\mathbb{R}$ A1

$g(x) \ge 0$ always (absolute value), and $g$ attains every non-negative value (e.g. $g(0) = 4$, $g(\pm 2) = 0$, and $g$ grows without bound as $|x| \to \infty$). Range $= [0, \infty)$.

(c) Solve $g(x) = 3$ M1·A1

$|x^{2} - 4| = 3 \;\Longleftrightarrow\; x^{2} - 4 = 3$ or $x^{2} - 4 = -3$. So $x^{2} = 7$ giving $x = \pm\sqrt{7}$, or $x^{2} = 1$ giving $x = \pm 1$. Four solutions: $x = \pm 1,\; \pm\sqrt{7}$.
$|f(x)| = k$ has up to twice as many solutions. The equation $|f(x)| = k$ ($k \ge 0$) is equivalent to $f(x) = k$ or $f(x) = -k$. For a quadratic $f$, that gives up to $4$ real roots (here all $4$ are real because both horizontal lines $y = \pm 3$ cut $y = x^{2} - 4$ twice). The geometric picture: drawing $y = |x^{2} - 4|$ and the line $y = 3$, you see $4$ crossings symmetric about the $y$-axis: two from the "untouched" outer arcs ($\pm\sqrt{7}$) and two from the "folded-up" inner arc ($\pm 1$). Always count: two from the unfolded part, two from the folded part — losing two is the classic mistake.

(a) 草图 M1·A1·A1

先画 $y = x^{2} - 4$：抛物线，顶点 $(0, -4)$，零点 $x = \pm 2$。再加外层 $|\cdot|$：$y < 0$ 的部分关于 $x$ 轴翻折。在 $-2 < x < 2$ 上的弧翻上来，被翻的顶点 $(0, -4)$ 成为局部极大 $(0, 4)$。零点 $x = \pm 2$ 处出现两个尖点（不可微）。在 $[-3, 3]$ 上：$x = \pm 3$ 时 $|9 - 4| = 5$；零点 $x = \pm 2$；尖点 $(\pm 2, 0)$；局部极大 $(0, 4)$。

(b) $\mathbb{R}$ 上值域 A1

$g(x) \ge 0$（绝对值），且 $g$ 取遍所有非负值（$g(0) = 4$、$g(\pm 2) = 0$，$|x| \to \infty$ 时无上界）。值域 $= [0, \infty)$。

(c) 解 $g(x) = 3$ M1·A1

$|x^{2} - 4| = 3 \;\Longleftrightarrow\; x^{2} - 4 = 3$ 或 $x^{2} - 4 = -3$。即 $x^{2} = 7$（$x = \pm\sqrt{7}$）或 $x^{2} = 1$（$x = \pm 1$）。四个解：$x = \pm 1,\; \pm\sqrt{7}$。
$|f(x)| = k$ 解最多翻倍。$|f(x)| = k$（$k \ge 0$）等价于 $f(x) = k$ 或 $f(x) = -k$。二次 $f$ 时最多 $4$ 个实根（本题两个水平线 $y = \pm 3$ 都与 $y = x^{2} - 4$ 各交两次，全 $4$ 个实）。几何上：画 $y = |x^{2} - 4|$ 与 $y = 3$，可见 $4$ 个交点关于 $y$ 轴对称，两来自"未翻折"的外弧（$\pm\sqrt{7}$），两来自"翻上来"的内弧（$\pm 1$）。务必外弧两个、内弧两个都数到——丢掉两个是经典错误。
PART II  ·  PAPER 1 SECTION B · SOLUTIONS第二部分  ·  第一卷 B 节 · 解析No calculator · 11 marks不可使用计算器 · 11 分

Section B · Worked SolutionsB 节 · 详细解析

Q5HARDPaper 1BSL 2.11 + AHL 2.16 Six-Variant Sketch (HL)[11 marks]

$f(x) = x^{2} - 2x$, zeros $0, 2$, vertex $(1, -1)$. Sketch (a) $f(x + 1)$ and $-f(x)$, (b) $f(-x)$ with $x = 1$ argument, (c) $|f(x)|$, (d) $f(|x|)$.$f(x) = x^{2} - 2x$，零点 $0, 2$，顶点 $(1, -1)$。画 (a) $f(x + 1)$ 与 $-f(x)$、(b) $f(-x)$ 并解释 $x = 1$ 对称论证、(c) $|f(x)|$、(d) $f(|x|)$。

Key features:关键特征：  (a) $f(x+1)$: zeros $-1, 1$, vertex $(0, -1)$; $-f(x)$: zeros $0, 2$, vertex $(1, 1)$零点 $-1, 1$、顶点 $(0, -1)$；$-f(x)$：零点 $0, 2$、顶点 $(1, 1)$  ·  (b) $f(-x)$: zeros $0, -2$, vertex $(-1, -1)$零点 $0, -2$、顶点 $(-1, -1)$  ·  (c) $|f(x)|$: zeros $0, 2$, corners at $(0, 0), (2, 0)$, local max $(1, 1)$零点 $0, 2$、尖点 $(0,0)(2,0)$、局部极大 $(1, 1)$  ·  (d) $f(|x|)$: zeros $0, \pm 2$, minima $(\pm 1, -1)$, local max $(0, 0)$零点 $0, \pm 2$、极小 $(\pm 1, -1)$、局部极大 $(0, 0)$

(a) $y = f(x + 1)$ and $y = -f(x)$ M1·A1·A1

$f(x + 1)$: shift left by $1$. Zeros $0 - 1 = -1$ and $2 - 1 = 1$. Vertex $(1 - 1,\, -1) = (0, -1)$. (Sanity: $f(x + 1) = (x + 1)^{2} - 2(x + 1) = x^{2} + 2x + 1 - 2x - 2 = x^{2} - 1$. $\checkmark$) $-f(x)$: reflect in $x$-axis. Zeros unchanged at $0$ and $2$. Vertex $(1, -1) \mapsto (1, 1)$. Now an opens-downward parabola with maximum $1$.

(b) $y = f(-x)$, symmetry argument M1·A1

Reflect in $y$-axis: $(x, y) \mapsto (-x, y)$. Zeros $0 \mapsto 0$, $2 \mapsto -2$. Vertex $(1, -1) \mapsto (-1, -1)$. The new parabola is $y = x^{2} + 2x$ (verify: $f(-x) = x^{2} + 2x$). Connection to "axis of symmetry $x = 1$". The original parabola has axis of symmetry $x = 1$ (its vertex), so $f(2 - x) = f(x)$ identically. Reflecting in the $y$-axis ($x = 0$) instead produces $y = f(-x) = x^{2} + 2x$, which has vertex $(-1, -1)$ — the original vertex $(1, -1)$ translated $2$ units left. So $f(-x) = f(x + 2)$. (Verify: $f(x + 2) = (x + 2)^{2} - 2(x + 2) = x^{2} + 4x + 4 - 2x - 4 = x^{2} + 2x = f(-x)$. $\checkmark$.) Two parallel mirror reflections compose to a translation by twice the gap between the axes; here the axes are $x = 0$ and $x = 1$, gap $1$, so the composition shifts by $2$, in the direction "from the first axis toward the second".

(c) $y = |f(x)|$ M1·A1·A1

Fold every below-axis point upward about the $x$-axis. The portion $0 < x < 2$ where $f < 0$ flips: the vertex $(1, -1) \mapsto (1, 1)$, a local maximum. Two corner points at the zeros: $(0, 0)$ and $(2, 0)$. Outside $[0, 2]$ the curve is unchanged. Zeros still at $x = 0, 2$.

(d) $y = f(|x|)$ M1·A1·A1

Keep $x \ge 0$ half intact, then reflect that half in the $y$-axis to fill $x < 0$. So for $x \ge 0$: $y = f(x) = x^{2} - 2x$ (zeros $0, 2$, min $(1, -1)$). For $x \le 0$: $y = f(-x) = f(|x|)$, so the right half is mirrored to the left, giving min at $(-1, -1)$ and zero at $-2$. At $x = 0$ the curve has $y = 0$ on both sides; the gradients from each side are $f'(0^{+}) = -2$ and the mirror $f'(0^{-}) = +2$, producing a corner. So zeros at $x = 0, \pm 2$; local minima at $(\pm 1, -1)$; local maximum (cusp/corner) at $(0, 0)$.
$|f(x)|$ vs $f(|x|)$ — which part of the plane gets touched. Outer absolute value $|f(x)|$ acts on the output: fold below-axis points about the $x$-axis. Inner absolute value $f(|x|)$ acts on the input: discard the $x < 0$ half of the original graph and replace it by the mirror image (about the $y$-axis) of the $x \ge 0$ half. As a result $f(|x|)$ is always an even function (graph symmetric about $y$-axis), regardless of the parity of $f$. $|f(x)|$ inherits the symmetries of $f$ (if $f$ is even, so is $|f|$). Use the recipe: "outer = fold about $x$-axis; inner = replace left half with mirror of right half".

(a) $y = f(x + 1)$ 与 $y = -f(x)$ M1·A1·A1

$f(x + 1)$：左移 $1$。零点 $-1, 1$；顶点 $(0, -1)$。（验：$f(x + 1) = x^{2} - 1$。$\checkmark$） $-f(x)$：关于 $x$ 轴反射。零点不变 $0, 2$；顶点 $(1, -1) \mapsto (1, 1)$。开口向下，最大值 $1$。

(b) $y = f(-x)$，对称论证 M1·A1

关于 $y$ 轴反射：$(x, y) \mapsto (-x, y)$。零点 $0, -2$；顶点 $(-1, -1)$。方程 $y = x^{2} + 2x$。 为何与"关于 $x = 1$ 反射"产生同一抛物线？原抛物线 $f(x) = x^{2} - 2x$ 的对称轴为 $x = 1$（即顶点处），故关于 $x = 1$ 反射图像不变。题目实际想表达的是：$f$ 关于 $x = 1$ 对称，故 $f(-x) = f(2 - (2 - (-x))) = f(x - 2)$（把 $f$ 沿 $-2$ 平移）。等价地，$y$ 轴反射与 $x = 1$ 反射只相差一个平移 $-2$。

(c) $y = |f(x)|$ M1·A1·A1

把 $x$ 轴下方的点关于 $x$ 轴翻上来。$0 < x < 2$ 段（$f < 0$）翻折：顶点 $(1, -1) \mapsto (1, 1)$，成为局部极大。零点 $(0, 0)$ 与 $(2, 0)$ 处出现两个尖点。$[0, 2]$ 外不变。零点仍为 $0, 2$。

(d) $y = f(|x|)$ M1·A1·A1

保留 $x \ge 0$ 半图，再把它关于 $y$ 轴反射填充 $x < 0$。$x \ge 0$ 段：$y = x^{2} - 2x$（零点 $0, 2$，极小 $(1, -1)$）。$x \le 0$ 段：$y = f(-x)$，左侧镜像 $\Rightarrow$ 极小 $(-1, -1)$、零点 $-2$。在 $x = 0$ 两侧斜率分别为 $-2$ 与 $+2$，形成尖点。零点 $0, \pm 2$；极小 $(\pm 1, -1)$；局部极大（尖点）$(0, 0)$。
$|f(x)|$ 与 $f(|x|)$——动到平面的哪部分。外绝对值 $|f(x)|$ 作用于输出：把 $x$ 轴下的点关于 $x$ 轴翻折。内绝对值 $f(|x|)$ 作用于输入：舍弃 $x < 0$ 的左半图，用 $x \ge 0$ 半图关于 $y$ 轴的镜像替换。结果 $f(|x|)$ 必为偶函数（关于 $y$ 轴对称），与 $f$ 的奇偶性无关。$|f(x)|$ 继承 $f$ 的对称（$f$ 偶则 $|f|$ 也偶）。口诀："外翻 $x$ 轴；内复制右半"。
PART III  ·  PAPER 2 · SOLUTIONS第三部分  ·  第二卷 · 解析Calculator · 16 marks可使用计算器 · 16 分

Paper 2 · Worked Solutions第二卷 · 详细解析

Q6MEDIUMPaper 2SL 2.11 Inner Stretch + Shift Equivalence[7 marks]

$y_{1} = \sin(2x + \pi/3)$, $y_{2} = \sin(2(x + \pi/6))$. Graph, prove equivalence, describe transformations, state period and phase shift.$y_{1} = \sin(2x + \pi/3)$、$y_{2} = \sin(2(x + \pi/6))$。作图、证明等价、描述变换、写周期与相位。

Answers:答案：  (a) single curve, $y_{1} \equiv y_{2}$单条曲线，$y_{1} \equiv y_{2}$  ·  (c) horizontal compression $\tfrac{1}{2}$ then shift left $\pi/6$横向压缩 $\tfrac{1}{2}$ 再左移 $\pi/6$  ·  (d) period $\pi$, phase shift $-\pi/6$周期 $\pi$，相位 $-\pi/6$

(a) GDC observation M1·A1

Plot $y_{1}$ and $y_{2}$ on window $-\pi \le x \le \pi$, $-1.5 \le y \le 1.5$. Observe a single visible curve — the second plot lies exactly on top of the first, no daylight between them. This suggests $y_{1}(x) = y_{2}(x)$ for all $x \in [-\pi, \pi]$, hence (by analytic continuation, or directly) for all $x \in \mathbb{R}$.

(b) Algebraic proof M1·A1

Factor the argument of $y_{1}$: $2x + \pi/3 = 2(x + \pi/6)$ (since $2 \cdot \pi/6 = \pi/3$). Therefore $$ y_{1} \;=\; \sin\!\bigl(2(x + \pi/6)\bigr) \;=\; y_{2}. $$ Equivalence proved.

(c) Transformation sequence M1·A1

Read $y_{2} = \sin(2(x + \pi/6))$ as applied to $y = \sin x$:
  • Inside factor $b = 2$: horizontal compression by $\tfrac{1}{2}$ (period halves from $2\pi$ to $\pi$).
  • Inside $+\pi/6$: shift left by $\pi/6$ (sign reversed for inside-the-bracket shifts).
Sequence: horizontal compression factor $\tfrac{1}{2}$, then translation by $-\pi/6$ in $x$.

(d) Period and phase shift A1

Period $= 2\pi / 2 = \pi$. Phase shift from $y = \sin x$: read the factored form $\sin(2(x + \pi/6))$ — the zero of $\sin$ that was at $x = 0$ has moved to $x = -\pi/6$. Phase shift $= -\pi/6$.
Always factor before reading the phase shift. The single most common Paper 2 trig-transformation error is reading the phase shift directly off $\sin(2x + \pi/3)$ as "$-\pi/3$" or "$+\pi/3$". The phase shift is the value of $x$ for which the inside equals $0$: solve $2x + \pi/3 = 0$ to get $x = -\pi/6$. Equivalently, factor first: $\sin(2x + \pi/3) = \sin(2(x + \pi/6))$, then the shift is the negative of the bracketed quantity, $-\pi/6$. Either method must give the same answer; if your two methods disagree, you've miscounted the factor of $b$. Mark schemes routinely catch students writing $-\pi/3$ here and deduct an A1.

(a) GDC 观察 M1·A1

在窗口 $-\pi \le x \le \pi$、$-1.5 \le y \le 1.5$ 上绘 $y_{1}$ 与 $y_{2}$，仅见一条曲线——第二条完全覆盖第一条，看不到第二条。这暗示 $y_{1}(x) = y_{2}(x)$ 在 $[-\pi, \pi]$ 上成立，因而（按解析延拓或直接）在 $\mathbb{R}$ 上成立。

(b) 代数证明 M1·A1

提因式：$2x + \pi/3 = 2(x + \pi/6)$（因 $2 \cdot \pi/6 = \pi/3$）。故 $$ y_{1} \;=\; \sin\!\bigl(2(x + \pi/6)\bigr) \;=\; y_{2}. $$

(c) 变换序列 M1·A1

把 $y_{2} = \sin(2(x + \pi/6))$ 读作作用于 $y = \sin x$：
  • 括号内 $b = 2$：横向压缩 $\tfrac{1}{2}$（周期由 $2\pi$ 变 $\pi$）。
  • 括号内 $+\pi/6$：左移 $\pi/6$（内符号反向）。
顺序：先横向压缩 $\tfrac{1}{2}$，再沿 $x$ 平移 $-\pi/6$。

(d) 周期与相位 A1

周期 $= 2\pi / 2 = \pi$。相对 $y = \sin x$ 的相位：由提因式式 $\sin(2(x + \pi/6))$，原本 $x = 0$ 处的零点移至 $x = -\pi/6$。相位 $= -\pi/6$。
读相位前先提因式。Paper 2 三角变换最常见的错误是直接把 $\sin(2x + \pi/3)$ 的相位读作"$-\pi/3$"或"$+\pi/3$"。相位是使括号内为零的 $x$：解 $2x + \pi/3 = 0$ 得 $x = -\pi/6$。等价地先提因式：$\sin(2x + \pi/3) = \sin(2(x + \pi/6))$，相位即提取量取负，$-\pi/6$。两法必须给同一答案；若不同，多半是漏算了系数 $b$。$\pi/3$ 是阅卷必扣的 A1 经典错。
Q7HARDPaper 2AHL 2.16 Reciprocal Sketch from Quadratic (HL)[9 marks]

$f(x) = x^{2} - 1$. (a) zeros, vertex; (b) asymptotes of $1/f$; (c) local extremum of $1/f$; (d) sketch $1/f$ on $-3 \le x \le 3$; (e) where $f = 1/f$.$f(x) = x^{2} - 1$。(a) 零点、顶点；(b) $1/f$ 的渐近线；(c) $1/f$ 的极值；(d) $-3 \le x \le 3$ 上画 $1/f$；(e) $f$ 与 $1/f$ 相交处。

Answers:答案：  (a) zeros $\pm 1$, vertex $(0, -1)$零点 $\pm 1$、顶点 $(0, -1)$  ·  (b) VA $x = \pm 1$, HA $y = 0$竖直 $x = \pm 1$，水平 $y = 0$  ·  (c) local max at $(0, -1)$局部极大 $(0, -1)$  ·  (e) $x = 0,\; \pm\sqrt{2}$

(a) Zeros and vertex of $f$ A1

$f(x) = x^{2} - 1 = (x - 1)(x + 1)$ has zeros at $x = \pm 1$. Vertex at $(0, -1)$ (parabola opens upward, minimum at the axis).

(b) Asymptotes of $y = 1/f(x)$ M1·A1

Vertical asymptotes where $f(x) = 0$: $x = -1$ and $x = 1$. Horizontal asymptote: as $|x| \to \infty$, $f(x) \to \infty$, so $1/f(x) \to 0$. Horizontal asymptote $y = 0$.

(c) Local extremum M1·A1

A local extremum of $1/f$ that is not at an asymptote corresponds to a local extremum of $f$ (with the same $x$-coordinate but reciprocal $y$-coordinate, sign preserved when $f$ stays nonzero). The unique local extremum of $f$ is the minimum at $(0, -1)$, with $f(0) = -1$. So $1/f(0) = -1$, and a minimum of $f$ becomes a local maximum of $1/f$ when $f < 0$ (since making the negative value smaller in magnitude makes its reciprocal larger toward $-\infty$, hence the reciprocal is largest at the minimum-magnitude point — yes, local max). Local maximum at $(0, -1)$.

(d) Sketch on $[-3, 3] \times [-3, 3]$ M1·A1·A1

Three intervals separated by the asymptotes $x = \pm 1$:
  • $x < -1$: $f > 0$, so $1/f > 0$. As $x \to -1^{-}$, $1/f \to +\infty$. As $x \to -\infty$, $1/f \to 0^{+}$. Curve rises from just-above $y = 0$ on the far left to $+\infty$ at $x = -1$.
  • $-1 < x < 1$: $f < 0$, so $1/f < 0$. Local maximum $-1$ at $x = 0$. Curve goes from $-\infty$ at $x = -1^{+}$ up to $-1$ at $x = 0$, then back down to $-\infty$ at $x = 1^{-}$.
  • $x > 1$: $f > 0$, so $1/f > 0$. As $x \to 1^{+}$, $1/f \to +\infty$. As $x \to \infty$, $1/f \to 0^{+}$.
At $x = \pm 3$: $f = 8$, so $1/f = 1/8 = 0.125$. Mark the three sign intervals: $+$, $-$, $+$.

(e) Where $f = 1/f$ A1

$f(x) = 1/f(x) \Leftrightarrow f(x)^{2} = 1 \Leftrightarrow f(x) = \pm 1$. So $x^{2} - 1 = 1$ gives $x = \pm\sqrt{2}$; $x^{2} - 1 = -1$ gives $x = 0$. Three intersection $x$-values: $x = 0,\; \pm\sqrt{2}$.
What survives the reciprocal transformation. The map $f \mapsto 1/f$ keeps three features: (i) zeros become vertical asymptotes (and vice versa), (ii) sign on each interval is preserved (since reciprocal of negative is negative), (iii) the $y = 1$ and $y = -1$ horizontal lines are fixed by the map (because $1/1 = 1$ and $1/(-1) = -1$). What flips: (a) local maxima become local minima and vice versa when $f$ stays positive throughout the neighbourhood, but when $f < 0$ throughout, a local min of $f$ at a negative value (most negative) maps to the least negative reciprocal, which is a local maximum. The sign of $f$ on the interval dictates whether you get a flip or a same-type extremum. Always look at the sign of $f$ before classifying.

(a) $f$ 的零点与顶点 A1

$f(x) = (x - 1)(x + 1)$ 零点 $x = \pm 1$。顶点 $(0, -1)$（开口向上）。

(b) $y = 1/f(x)$ 的渐近线 M1·A1

竖直渐近线：$f = 0$ 处，$x = -1, 1$。水平渐近线：$|x| \to \infty$ 时 $f \to \infty$，$1/f \to 0$，故 $y = 0$。

(c) 局部极值 M1·A1

$1/f$ 非渐近线处的极值与 $f$ 的极值同 $x$ 坐标，$y$ 坐标取倒数。$f$ 的唯一极值在 $(0, -1)$。$1/f(0) = -1$。当 $f < 0$ 时，$f$ 的极小（负值最小）对应 $1/f$ 的局部极大。局部极大 $(0, -1)$。

(d) 在 $[-3, 3] \times [-3, 3]$ 上画 M1·A1·A1

被 $x = \pm 1$ 分为三段：
  • $x < -1$：$f > 0$，$1/f > 0$；$x \to -1^{-}$ 时 $1/f \to +\infty$；$x \to -\infty$ 时 $1/f \to 0^{+}$。
  • $-1 < x < 1$：$f < 0$，$1/f < 0$；$x = 0$ 处局部极大 $-1$。两侧趋 $-\infty$。
  • $x > 1$：$f > 0$，$1/f > 0$；$x \to 1^{+}$ 时 $1/f \to +\infty$；$x \to \infty$ 时 $1/f \to 0^{+}$。
$x = \pm 3$ 时 $f = 8$，$1/f = 0.125$。三段符号 $+$、$-$、$+$。

(e) $f = 1/f$ A1

$f(x) = 1/f(x) \Leftrightarrow f(x)^{2} = 1 \Leftrightarrow f(x) = \pm 1$。$x^{2} - 1 = 1$ 得 $x = \pm\sqrt{2}$；$x^{2} - 1 = -1$ 得 $x = 0$。三个 $x$：$0,\; \pm\sqrt{2}$。
倒数变换下保留什么。$f \mapsto 1/f$ 保留三件事：(i) 零点 $\leftrightarrow$ 竖直渐近线，(ii) 每个区间的符号不变（负数倒数为负），(iii) 直线 $y = \pm 1$ 不动（$1/1 = 1$、$1/(-1) = -1$）。翻转的是：$f$ 全正时极大 $\leftrightarrow$ 极小；$f$ 全负时极小（负值"最小"）变 $1/f$ 的极大（负值"最大"）。极值分类前先看 $f$ 在该区间的符号。
PART IV  ·  PAPER 3 · SOLUTIONS第四部分  ·  第三卷 · 解析Calculator · HL extended exploration · 15 marks可使用计算器 · HL 长题探究 · 15 分

Paper 3 · Worked Solutions第三卷 · 详细解析

Q8HARDPaper 3AHL 2.16 Nested Absolute Value + Composition (HL)[15 marks]

$f(x) = x^{2} - 2$. Sketch; nested abs values; fixed points of $f$; iterate $h = f \circ f$, find $h(x) = x$.$f(x) = x^{2} - 2$。画图；嵌套绝对值；$f$ 的不动点；迭代 $h = f \circ f$，解 $h(x) = x$。

Answers:答案：  (a) zeros $\pm\sqrt{2}$, vertex $(0, -2)$, range $[-2, \infty)$零点 $\pm\sqrt{2}$、顶点 $(0, -2)$、值域 $[-2, \infty)$  ·  (c) $x = -1,\; 2$  ·  (d) $x = -1,\; 2,\; \tfrac{-1 \pm \sqrt{5}}{2}$

(a) Sketch $f$ M1·A1

$f(x) = x^{2} - 2$: upward parabola, vertex $(0, -2)$, zeros at $x = \pm\sqrt{2} \approx \pm 1.414$. Range $[-2, \infty)$.

(b) $f(|x|)$, $|f(x)|$, $|f(|x|)|$ M1·A1·A1·A1·R1

Since $f$ is even ($f(-x) = (-x)^{2} - 2 = x^{2} - 2 = f(x)$), the graph of $y = f(x)$ is already symmetric about the $y$-axis. So $y = f(|x|)$ is identical to $y = f(x)$: keeping the right half and mirroring it left reproduces the same parabola. $y = |f(x)|$: fold the below-axis portion ($-\sqrt{2} < x < \sqrt{2}$) up about the $x$-axis. The vertex $(0, -2)$ becomes a local maximum at $(0, 2)$. Two corner points at $(\pm\sqrt{2}, 0)$. Outside $[-\sqrt{2}, \sqrt{2}]$ the graph is unchanged. $y = |f(|x|)|$: since $y = f(|x|) = f(x)$, we have $y = |f(|x|)| = |f(x)|$. $|f(|x|)| = |f(x)|$ because $f$ is even. Same graph as the previous: zeros at $\pm\sqrt{2}$, corners at $(\pm\sqrt{2}, 0)$, local max $(0, 2)$.

(c) Fixed points of $f$ M1·A1·A1

$f(x) = x \Leftrightarrow x^{2} - 2 = x \Leftrightarrow x^{2} - x - 2 = 0 \Leftrightarrow (x - 2)(x + 1) = 0$. Two fixed points: $x = -1$ and $x = 2$. Check: $f(-1) = 1 - 2 = -1\,\checkmark$; $f(2) = 4 - 2 = 2\,\checkmark$. GDC confirmation: overlay $y = x^{2} - 2$ and $y = x$ — two intersections at $(-1, -1)$ and $(2, 2)$.

(d) Find $h(x) = x^{4} - 4x^{2} + 2$ and solve $h(x) = x$ M1·A1·A1·A1·A1

Compute $h$: $$ h(x) \;=\; f(f(x)) \;=\; f(x^{2} - 2) \;=\; (x^{2} - 2)^{2} - 2 \;=\; x^{4} - 4x^{2} + 4 - 2 \;=\; x^{4} - 4x^{2} + 2. $$ Solve $h(x) = x$: $x^{4} - 4x^{2} + 2 = x$, i.e. $p(x) := x^{4} - 4x^{2} - x + 2 = 0$. Every fixed point of $f$ is also a fixed point of $h$, so $(x - 2)(x + 1) = x^{2} - x - 2$ divides $p(x)$. Polynomial division: $$ x^{4} - 4x^{2} - x + 2 \;\div\; x^{2} - x - 2 \;=\; x^{2} + x - 1. $$ Verification: $(x^{2} - x - 2)(x^{2} + x - 1) = x^{4} + x^{3} - x^{2} - x^{3} - x^{2} + x - 2x^{2} - 2x + 2 = x^{4} - 4x^{2} - x + 2$. $\checkmark$ Solve the quadratic factor $x^{2} + x - 1 = 0$: $x = \dfrac{-1 \pm \sqrt{1 + 4}}{2} = \dfrac{-1 \pm \sqrt{5}}{2}$. All four real solutions: $\;x = -1,\; 2,\; \dfrac{-1 + \sqrt{5}}{2},\; \dfrac{-1 - \sqrt{5}}{2}\;$ (approximately $-1$, $2$, $0.618$, $-1.618$).
$f(x) = x^{2} - 2$ and the period-$2$ orbit of golden-ratio cousins. The two extra roots $\alpha = \dfrac{-1 + \sqrt{5}}{2}$ and $\beta = \dfrac{-1 - \sqrt{5}}{2}$ are not fixed points of $f$ but form a period-$2$ orbit: $f(\alpha) = \beta$ and $f(\beta) = \alpha$. Verify: $\alpha^{2} = \alpha + 1$ (since $\alpha^{2} + \alpha - 1 = 0$... wait, we have $x^{2} + x - 1 = 0$, so $\alpha^{2} = 1 - \alpha$), so $f(\alpha) = \alpha^{2} - 2 = (1 - \alpha) - 2 = -1 - \alpha = -(1 + \alpha)$. Compute $-(1 + \alpha) = -1 - \dfrac{-1 + \sqrt{5}}{2} = \dfrac{-2 + 1 - \sqrt{5}}{2} = \dfrac{-1 - \sqrt{5}}{2} = \beta$. $\checkmark$ This is the simplest non-trivial case of the chaotic map $f(x) = x^{2} + c$ at $c = -2$, which is conjugate to the doubling map on the interval $[-2, 2]$ (substitute $x = 2\cos\theta$). The fixed points and period-$2$ orbits stack up to give all $2^{n}$ roots of $f^{n}(x) = x$ — a classic dynamical-systems explorer's playground. The values $\dfrac{-1 \pm \sqrt{5}}{2}$ are negatives of the golden ratio and its conjugate, and reappear in continued fractions, Fibonacci asymptotics, and pentagon geometry. Worth carrying into any Paper 3 exploration on iterated maps.

(a) 画 $f$ M1·A1

$f(x) = x^{2} - 2$：开口向上，顶点 $(0, -2)$，零点 $x = \pm\sqrt{2} \approx \pm 1.414$。值域 $[-2, \infty)$。

(b) $f(|x|)$、$|f(x)|$、$|f(|x|)|$ M1·A1·A1·A1·R1

$f$ 偶（$f(-x) = f(x)$），故 $y = f(x)$ 已关于 $y$ 轴对称。$y = f(|x|)$ 与 $y = f(x)$ 完全相同：保留右半图再镜像至左半，恢复原抛物线。 $y = |f(x)|$：把 $-\sqrt{2} < x < \sqrt{2}$ 段关于 $x$ 轴翻折。顶点 $(0, -2)$ 成为局部极大 $(0, 2)$。尖点 $(\pm\sqrt{2}, 0)$。$[-\sqrt{2}, \sqrt{2}]$ 外不变。 $y = |f(|x|)|$：因 $y = f(|x|) = f(x)$，故 $y = |f(|x|)| = |f(x)|$。$f$ 偶 $\Rightarrow |f(|x|)| = |f(x)|$。与上图一致：零点 $\pm\sqrt{2}$、尖点 $(\pm\sqrt{2}, 0)$、局部极大 $(0, 2)$。

(c) $f$ 的不动点 M1·A1·A1

$f(x) = x \Leftrightarrow x^{2} - x - 2 = 0 \Leftrightarrow (x - 2)(x + 1) = 0$。$x = -1, 2$。验：$f(-1) = -1\,\checkmark$；$f(2) = 2\,\checkmark$。GDC：叠绘 $y = x^{2} - 2$ 与 $y = x$，交点 $(-1, -1)$、$(2, 2)$。

(d) 求 $h(x) = x^{4} - 4x^{2} + 2$ 并解 $h(x) = x$ M1·A1·A1·A1·A1

计算 $h$： $$ h(x) \;=\; f(f(x)) \;=\; (x^{2} - 2)^{2} - 2 \;=\; x^{4} - 4x^{2} + 2. $$ 解 $h(x) = x$：$x^{4} - 4x^{2} - x + 2 = 0$。$f$ 的不动点必为 $h$ 的不动点，故 $(x - 2)(x + 1) = x^{2} - x - 2$ 整除该多项式。作除法： $$ x^{4} - 4x^{2} - x + 2 \;\div\; x^{2} - x - 2 \;=\; x^{2} + x - 1. $$ 验：$(x^{2} - x - 2)(x^{2} + x - 1) = x^{4} - 4x^{2} - x + 2$。$\checkmark$ 解二次因子 $x^{2} + x - 1 = 0$：$x = \dfrac{-1 \pm \sqrt{5}}{2}$。 四个实数解：$\;x = -1,\; 2,\; \dfrac{-1 + \sqrt{5}}{2},\; \dfrac{-1 - \sqrt{5}}{2}\;$（约 $-1$、$2$、$0.618$、$-1.618$）。
$f(x) = x^{2} - 2$ 与"黄金比表亲"的 $2$ 周期轨。新增两根 $\alpha = \dfrac{-1 + \sqrt{5}}{2}$、$\beta = \dfrac{-1 - \sqrt{5}}{2}$ 不是 $f$ 的不动点，而是$2$ 周期轨：$f(\alpha) = \beta$、$f(\beta) = \alpha$。验：$\alpha^{2} + \alpha - 1 = 0$ 给 $\alpha^{2} = 1 - \alpha$，故 $f(\alpha) = \alpha^{2} - 2 = -1 - \alpha = \dfrac{-1 - \sqrt{5}}{2} = \beta$。$\checkmark$ 这是混沌映射族 $f(x) = x^{2} + c$ 在 $c = -2$ 时最简单的非平凡情形——通过 $x = 2\cos\theta$ 与 $[-2, 2]$ 上的倍角映射共轭。不动点与 $2,\,4,\,\ldots,\,2^{n}$ 周期轨叠加给出 $f^{n}(x) = x$ 的 $2^{n}$ 个根，是动力系统探究者的乐园。$\dfrac{-1 \pm \sqrt{5}}{2}$ 是黄金比及其共轭的负值，常现于连分数、Fibonacci 渐近、五边形几何。Paper 3 迭代映射探究值得带上。