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 algebraic step. For "from first principles", you must start from $\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ and let $h \to 0$ explicitly. State tangent / normal lines in the form $y = mx + c$. Classify stationary points using the first or second derivative test and name the test you used. No calculator permitted.写出每一步代数。题目要求"由定义（first principles）"时，必须从 $\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ 出发，并显式取 $h \to 0$。切线 / 法线写成 $y = mx + c$ 的形式。稳定点用一阶或二阶导数判别法分类，并写明所用判别法。不可使用计算器。
(a)Explain why direct substitution gives the indeterminate form $\tfrac{0}{0}$.说明为何直接代入会得到不定式 $\tfrac{0}{0}$。[1]
(b)Factor the numerator and simplify.分解分子并化简。[2]
(c)Take the limit as $x \to 2$ and state the value.取 $x \to 2$ 的极限并写出结果。[1]
Q2MEDIUMPaper 1AE1.2 First Principles[5 marks]
Let $f(x) = 3x^{2} - x$. Using the definition of the derivative from first principles, find $f'(x)$.设 $f(x) = 3x^{2} - x$。由导数定义（first principles），求 $f'(x)$。
(a)Write down the limit definition $\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ and substitute $f$.写出定义 $\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$，并代入 $f$。[1]
(b)Expand $f(x + h)$ and simplify the numerator. The $h$-free terms must cancel.展开 $f(x + h)$ 并化简分子。不含 $h$ 的项必须消去。[2]
(c)Divide numerator and denominator by $h$, then take the limit $h \to 0$ to state $f'(x)$.分子分母同除 $h$，再取 $h \to 0$ 写出 $f'(x)$。[2]
Q3MEDIUMPaper 1AE1.3 Tangent & Normal[6 marks]
The curve $C$ has equation $y = x^{2} - 4x + 6$. Find the equations of the tangent and the normal to $C$ at the point $P(3, 3)$.曲线 $C$ 的方程为 $y = x^{2} - 4x + 6$。求 $C$ 在点 $P(3, 3)$ 处的切线与法线方程。
(a)Verify that $P(3, 3)$ lies on $C$.验证 $P(3, 3)$ 在 $C$ 上。[1]
(b)Find $\dfrac{dy}{dx}$ and evaluate the slope of the tangent at $P$.求 $\dfrac{dy}{dx}$ 并在 $P$ 处求切线斜率。[2]
(c)State the equation of the tangent to $C$ at $P$ in the form $y = mx + c$.以 $y = mx + c$ 形式写出 $C$ 在 $P$ 处的切线方程。[1]
(d)State the equation of the normal to $C$ at $P$ in the form $y = mx + c$. (The normal is perpendicular to the tangent.)以 $y = mx + c$ 形式写出 $C$ 在 $P$ 处的法线方程。（法线垂直于切线。）[2]
(b)Use the second derivative test to classify each stationary point as a local maximum or local minimum.用二阶导数判别法将每个稳定点分类为局部极大或局部极小。[2]
(c)State the $y$-coordinate of each stationary point.写出每个稳定点的 $y$ 坐标。[1]
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
Show every algebraic step. Sketches must label all stationary points, points of inflection, intercepts, and the long-run behaviour at the ends. HL: a sign change of $f''$ at a candidate inflection is required; checking $f''(x_{0}) = 0$ alone is not sufficient.写出每一步代数。函数草图须标出所有稳定点、拐点、截距以及两端的长程行为。HL 要求：在候选拐点处必须验证 $f''$ 的符号变化；仅 $f''(x_{0}) = 0$ 不足以证明该点是拐点。
(a)Find the stationary points of $f$ and classify each (state the test used).求 $f$ 的稳定点，并对每个进行分类（写明所用判别法）。[4]
(b)Find the point of inflection of $f$. You must justify it is an inflection point by checking the sign of $f''$ on each side.求 $f$ 的拐点。必须通过检验 $f''$ 在两侧的符号来证明该点为拐点。[3]
(c)Sketch $y = f(x)$. On your sketch label: both stationary points (coordinates), the point of inflection (coordinates), the three $x$-intercepts, and the $y$-intercept.画出 $y = f(x)$ 的草图。在草图上标出：两个稳定点（坐标）、拐点（坐标）、三个 $x$ 截距以及 $y$ 截距。[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. You may compute derivatives, zeros, and intersections numerically on the GDC, but you must (i) state the function entered, (ii) record the GDC output to at least four significant figures, and (iii) interpret the result in words. Give exact answers where reasonable.需要图形计算器（GDC）。可在 GDC 上数值求导数、零点与交点，但必须（i）写出所输入的函数，（ii）记录至少四位有效数字的 GDC 输出，（iii）用文字解释结果。能精确求解的题目就给精确答案。
Q6MEDIUMPaper 2E1.3 GDC Tangent Line[7 marks]
Let $g(x) = x^{3} - 2x^{2} + x + 1$. Use a GDC to investigate the tangent to $y = g(x)$ at $x = 2$.设 $g(x) = x^{3} - 2x^{2} + x + 1$。使用 GDC 探究 $y = g(x)$ 在 $x = 2$ 处的切线。
(a)Compute $g(2)$ exactly. Then use the GDC's numerical derivative to find $g'(2)$.精确计算 $g(2)$。再用 GDC 的数值求导功能求 $g'(2)$。[3]
(b)Confirm the GDC value of $g'(2)$ by computing the derivative algebraically.通过解析求导验证 GDC 给出的 $g'(2)$。[2]
(c)Hence write the equation of the tangent to $y = g(x)$ at $x = 2$ in the form $y = mx + c$.由此以 $y = mx + c$ 形式写出 $y = g(x)$ 在 $x = 2$ 处的切线方程。[2]
Q7HARDPaper 2E1.5 Continuity vs Differentiability (HL)[10 marks]
(a)Sketch $y = f(x)$ on a GDC and reproduce the shape, identifying the two "corner" $x$-values.在 GDC 上画出 $y = f(x)$，复制其形状，并标出两个"折角"的 $x$ 值。[2]
(b)Write $f(x)$ as a piecewise linear function on the three intervals determined by the corners.将 $f(x)$ 写成分三个区间（由折角决定）的分段线性函数。[3]
(c)Show that $f$ is continuous at $x = 1$ and at $x = -2$ by checking that the one-sided limits agree with $f(1)$ and $f(-2)$ respectively.通过验证单侧极限分别等于 $f(1)$ 与 $f(-2)$，证明 $f$ 在 $x = 1$ 与 $x = -2$ 处连续。[2]
(d)Compute the left and right derivatives of $f$ at $x = 1$ and at $x = -2$. Hence show $f$ is not differentiable at either corner.分别计算 $f$ 在 $x = 1$ 与 $x = -2$ 的左、右导数。由此证明 $f$ 在两个折角处都不可导。[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. Each candidate inflection must be confirmed by a sign change of $f''$; each stationary point by the second derivative test (or a first-derivative sign chart). Tabulate the sign chart explicitly when more than two critical points appear.需要图形计算器（GDC）。方法分占主导。每个候选拐点必须由 $f''$ 的符号变化确认；每个稳定点必须由二阶导数判别法（或一阶导数符号表）分类。临界点超过两个时，须明确列出符号表。
(a)Find $f'(x)$ and $f''(x)$.求 $f'(x)$ 与 $f''(x)$。[2]
(b)Find all stationary points of $f$. Use the second derivative test to classify each as a local maximum or local minimum, and give the $y$-coordinate of each.求 $f$ 的所有稳定点。用二阶导数判别法将每个分类为局部极大或局部极小，并给出每个的 $y$ 坐标。[5]
(c)Find all candidate points of inflection by solving $f''(x) = 0$. For each candidate, verify the inflection by showing $f''$ changes sign across that $x$-value. State the coordinates of each inflection point.通过求解 $f''(x) = 0$ 找出所有候选拐点。对每个候选，通过证明 $f''$ 在该 $x$ 值两侧变号来确认拐点，并写出每个拐点的坐标。[5]
(d)State the intervals on which $f$ is concave up and concave down. Hence sketch $y = f(x)$, labelling stationary points, inflection points, and the $x$-intercepts.写出 $f$ 上凹与下凹的区间。由此画出 $y = f(x)$ 的草图，并标出稳定点、拐点与 $x$ 截距。[3]