📚 Support Pack 2: Contents (continued) | 支持包2：内容续篇
This article continues the detailed breakdown of Support Pack 2 for AQA A-Level Mathematics, focusing on the remaining topics that build upon the foundational skills covered in the first half. We will explore advanced algebraic techniques, coordinate geometry, and the early stages of calculus, all essential for your Paper 1 and Paper 2 assessments.
本文继续深入解析 AQA A-Level 数学 Support Pack 2 的后半部分内容，重点关注建立在基础技能之上的进阶主题。我们将探讨高级代数技巧、坐标几何以及微积分的入门知识，这些内容对于 Paper 1 和 Paper 2 的考试都至关重要。
1. Quadratic Inequalities | 二次不等式
Solving quadratic inequalities requires a clear understanding of the graph of a quadratic function. The key steps involve finding the critical roots and then determining the regions of x that satisfy the inequality, depending on whether the coefficient of x² is positive or negative.
求解二次不等式需要清晰理解二次函数的图像。关键步骤包括找出临界根，然后根据 x² 系数的正负，确定满足不等式的 x 区域。
For a quadratic inequality in the form ax² + bx + c > 0, where a > 0, the solution is found by first solving the corresponding equation ax² + bx + c = 0. If the roots are α and β (where α < β), the inequality > 0 is satisfied when x < α or x > β, while the inequality < 0 is satisfied when α < x < β. This is because the parabola opens upwards, forming a U-shape.
对于形式为 ax² + bx + c > 0（其中 a > 0）的二次不等式，求解时先解对应的方程 ax² + bx + c = 0。若根为 α 和 β（且 α < β），则不等式 > 0 在 x < α 或 x > β 时成立，而不等式 < 0 在 α < x < β 时成立。这是因为抛物线开口向上，呈 U 形。
Special cases arise when the discriminant b² − 4ac is negative. In such instances, the quadratic never crosses the x-axis, and the inequality is either always true or always false for all real x, depending on the sign of a and the direction of the inequality.
当判别式 b² − 4ac 为负时会出现特殊情况。此时二次函数不穿过 x 轴，不等式要么对所有实数 x 恒成立，要么恒不成立，具体取决于 a 的符号和不等号的方向。
When solving more complex inequalities that involve fractions or rational expressions, you must be careful: multiplying through by a denominator that may be negative flips the inequality sign. In AQA examinations, it is often safer to bring all terms to one side and form a single rational expression, then analyse the sign using a sign diagram or graphical reasoning.
在解涉及分数或有理表达式的复杂不等式时，必须小心：乘以可能为负的分母会改变不等号方向。在 AQA 考试中，更稳妥的做法是将所有项移到一侧并合并为单一有理表达式，然后通过符号图或图形推理分析符号。
2. The Factor Theorem and Polynomial Division | 因式定理与多项式除法
The Factor Theorem is a direct consequence of the Remainder Theorem. It states that if f(x) is a polynomial and f(a) = 0, then (x − a) is a factor of f(x). This theorem is the primary tool for factorising cubic and higher-order polynomials.
因式定理是余数定理的直接推论。它表明：若 f(x) 是一个多项式且 f(a) = 0，则 (x − a) 是 f(x) 的一个因式。该定理是分解三次及更高次多项式的核心工具。
To apply the theorem, you first test potential integer roots using the factors of the constant term. Once you locate a root, perform polynomial division to reduce the cubic to a quadratic, then factorise the quadratic normally. For example, for f(x) = x³ − 6x² + 11x − 6, testing x = 1 gives f(1) = 0, so (x − 1) is a factor. Dividing yields x² − 5x + 6, which factorises further into (x − 2)(x − 3).
应用该定理时，先利用常数项的因数测试可能的整数根。找到根后，进行多项式除法将三次式降为二次式，然后正常分解二次式。例如，对于 f(x) = x³ − 6x² + 11x − 6，测试 x = 1 得 f(1) = 0，所以 (x − 1) 是因式。相除得到 x² − 5x + 6，进一步分解为 (x − 2)(x − 3)。
Polynomial division can also be performed using the grid method or synthetic division. In AQA mark schemes, full marks require clear working, so choose the method you feel most confident with, as long as every step is visible to the examiner.
多项式除法也可以使用网格法或综合除法进行。在 AQA 评分标准中，满分需要清晰的解题过程，因此请选择你最有把握的方法，但每个步骤都应明确展示给阅卷人。
When a cubic has a repeated root, such as f(x) = (x − 2)²(x + 1), the Factor Theorem will identify x = 2 as a root once, but you must check whether it divides again. Differentiation can also be used: if f(a) = 0 and f'(a) = 0, then x = a is a repeated root of order at least 2.
当三次式有重根时，例如 f(x) = (x − 2)²(x + 1)，因式定理只会识别出 x = 2 是一次根，但你需要检查它是否能再次整除。也可以使用微分：若 f(a) = 0 且 f'(a) = 0，则 x = a 是至少二阶的重根。
3. The Remainder Theorem | 余数定理
The Remainder Theorem states that when a polynomial f(x) is divided by a linear divisor of the form (x − a), the remainder is f(a). This provides a rapid method for evaluating polynomials at a given point without performing full division.
余数定理指出：当多项式 f(x) 除以形如 (x − a) 的一次因式时，余数为 f(a)。这提供了一种快速求多项式在给定点值的方法，无需进行完整除法。
For example, dividing f(x) = 2x³ − 3x² + 5x − 7 by (x − 2) gives a remainder equal to f(2) = 2(8) − 3(4) + 10 − 7 = 16 − 12 + 10 − 7 = 7. Hence, the remainder is 7.
例如，用 (x − 2) 除 f(x) = 2x³ − 3x² + 5x − 7，得到的余数等于 f(2) = 2(8) − 3(4) + 10 − 7 = 16 − 12 + 10 − 7 = 7。因此，余数为 7。
Note that the divisor may not always be of the form (x − a). If the divisor is (2x − 1), rewrite it as 2(x − ½). The remainder is f(½), not f(2). This subtle distinction is a common source of errors in examinations, so always normalise the divisor to the form (x − a) before substitution.
注意除数并不总是 (x − a) 的形式。若除数为 (2x − 1)，将其改写为 2(x − ½) 后，余数是 f(½) 而非 f(2)。这个细微差别是考试中常见的错误来源，因此代入前务必将除数规范化为 (x − a) 的形式。
4. Coordinate Geometry: The Equation of a Circle | 坐标几何：圆的方程
The equation of a circle with centre (a, b) and radius r is given by (x − a)² + (y − b)² = r². This standard form is the foundation for all circle geometry problems in AQA A-Level Mathematics.
圆心为 (a, b)、半径为 r 的圆的方程为 (x − a)² + (y − b)² = r²。这一标准形式是 AQA A-Level 数学中所有圆几何问题的基础。
When given the general form x² + y² + 2gx + 2fy + c = 0, you can complete the square to find the centre as (−g, −f) and the radius as √(g² + f² − c). For a real circle to exist, the expression g² + f² − c must be positive.
当给出一般式 x² + y² + 2gx + 2fy + c = 0 时，可以通过配方法得到圆心为 (−g, −f)，半径为 √(g² + f² − c)。圆存在的条件是 g² + f² − c 必须为正。
To find the equation of a circle from three points on its circumference, set up a system of three simultaneous equations by substituting each point into the general form. Solving for g, f, and c yields the equation of the circle.
要从圆上的三个点求圆的方程，可将每个点代入一般式，建立三个联立方程。解出 g、f 和 c 即可得到圆的方程。
In many AQA problems, you will also need to find the equation of a tangent to a circle at a given point. The tangent is perpendicular to the radius at the point of contact. If the circle has centre C and the point of tangency is P, then the gradient of the tangent is the negative reciprocal of the gradient of CP.
在众多 AQA 问题中，你还需要求圆在某一点处的切线方程。切线在切点处垂直于半径。若圆心为 C，切点为 P，则切线的斜率是 CP 斜率的负倒数。
5. Circle Properties: Chords and Tangents | 圆的性质：弦与切线
A chord is a line segment connecting two points on a circle. The perpendicular bisector of any chord always passes through the centre of the circle. This property allows you to find the centre when given two chords, and is frequently tested in exam questions.
弦是连接圆上两点的线段。任何弦的垂直平分线必过圆心。这一性质使你能够在已知两条弦时求出圆心，是考试中的常见考点。
The tangent properties are equally important: a tangent is perpendicular to the radius at the point of contact; from an external point, two tangents of equal length can be drawn to a circle; and the angle between a tangent and a chord through the point of contact equals the angle in the alternate segment (the Alternate Segment Theorem).
切线性质同样重要：切线在切点处垂直于半径；从圆外一点可作两条等长的切线；切线与过切点的弦所夹的角等于弦切角定理中的相应圆周角（即弦切角定理）。
To determine whether a line intersects a circle at two points, is tangent, or does not intersect, substitute the equation of the line into the equation of the circle and examine the discriminant of the resulting quadratic. If the discriminant is positive, there are two distinct intersections; if it is zero, the line is tangent; and if negative, there is no intersection.
判断直线与圆是相交于两点、相切还是不相交，可将直线方程代入圆方程，然后检查所得二次方程的判别式。若判别式为正，则有两个不同交点；若为零，则直线为切线；若为负，则无交点。
6. The Gradient Function and Differentiation | 梯度函数与微分
Differentiation is the process of finding the gradient function, denoted f'(x) or dy/dx, which gives the slope of the tangent to a curve at any point. For a power function y = xⁿ, the derivative is dy/dx = nxⁿ⁻¹.
微分是求梯度函数的过程，记为 f'(x) 或 dy/dx，它给出曲线上任意一点处切线的斜率。对于幂函数 y = xⁿ，其导数为 dy/dx = nxⁿ⁻¹。
This rule extends to sums and differences: if y = f(x) ± g(x), then dy/dx = f'(x) ± g'(x). Constant multiples are preserved: if y = kxⁿ, then dy/dx = knxⁿ⁻¹. In AQA examinations, you are expected to differentiate polynomials confidently without the use of a calculator for basic examples.
该法则可推广到和与差：若 y = f(x) ± g(x)，则 dy/dx = f'(x) ± g'(x)。常数倍数保持不变：若 y = kxⁿ，则 dy/dx = knxⁿ⁻¹。在 AQA 考试中，你需要能够不借助计算器熟练地对基本多项式进行求导。
The gradient at a specific point, say x = 2, is obtained by substituting x = 2 into the gradient function. This value represents the slope of the tangent at that precise location. The normal to the curve at that point has a gradient equal to the negative reciprocal of the tangent’s gradient.
在某特定点（如 x = 2）处的梯度，通过将该值代入梯度函数得到。该值表示该精确位置处切线的斜率。曲线在该点的法线的斜率等于切线斜率的负倒数。
7. Increasing and Decreasing Functions | 递增与递减函数
A function f(x) is said to be increasing on an interval if f'(x) > 0 for all x in that interval. It is decreasing if f'(x) < 0 for all x in the interval. These conditions provide a straightforward method for analysing the monotonic behaviour of a function.
函数 f(x) 在某个区间上被称为递增函数，若该区间内所有 x 都满足 f'(x) > 0。若区间内所有 x 都满足 f'(x) < 0，则为递减函数。这些条件为分析函数的单调性提供了一种直接方法。
For example, consider f(x) = x³ − 3x² + 2. Its derivative is f'(x) = 3x² − 6x = 3x(x − 2). Setting f'(x) = 0 gives the critical points x = 0 and x = 2. Analysing the sign of f'(x) on the intervals (−∞, 0), (0, 2), and (2, ∞), we find that f is increasing on (−∞, 0) ∪ (2, ∞) and decreasing on (0, 2).
例如，考虑 f(x) = x³ − 3x² + 2。其导数为 f'(x) = 3x² − 6x = 3x(x − 2)。令 f'(x) = 0 得临界点 x = 0 和 x = 2。分析 f'(x) 在区间 (−∞, 0)、(0, 2) 和 (2, ∞) 上的符号，我们发现 f 在 (−∞, 0) ∪ (2, ∞) 上递增，在 (0, 2) 上递减。
Where the first derivative changes sign from positive to negative, there is a local maximum. Where it changes from negative to positive, there is a local minimum. This forms the basis of the first derivative test, a standard requirement in AQA.
当一阶导数由正变负时，存在局部极大值。当由负变正时，存在局部极小值。这是一阶导数检验法的基础，也是 AQA 的标准要求。
8. Stationary Points and Their Nature | 驻点及其性质
Stationary points occur where the gradient function equals zero, i.e., f'(x) = 0. These points can be classified as local maxima, local minima, or points of inflection with zero gradient. The second derivative test is the most efficient classification method.
驻点出现在梯度函数为零的位置，即 f'(x) = 0。这些点可以分为局部极大值、局部极小值或梯度为零的拐点。二阶导数检验法是最有效的分类方法。
At a stationary point, evaluate the second derivative f”(x). If f”(x) > 0, the point is a local minimum. If f”(x) < 0, the point is a local maximum. If f''(x) = 0, the test is inconclusive, and you should examine the sign of the first derivative around the point or consider higher-order derivatives.
在驻点处，计算二阶导数 f”(x)。若 f”(x) > 0，则该点为局部极小值。若 f”(x) < 0，则为局部极大值。若 f''(x) = 0，检验法无法判定，此时应检查一阶导数在该点附近的符号或考虑更高阶导数。
Consider the function y = x³. At x = 0, both f'(0) = 0 and f”(0) = 0. However, x = 0 is neither a maximum nor a minimum; it is a point of inflection with a horizontal tangent. The sign of f'(x) = 3x² does not change as x passes through 0, confirming this classification.
考虑函数 y = x³。在 x = 0 处，f'(0) = 0 且 f”(0) = 0。然而，x = 0 既不是极大值也不是极小值，而是一个具有水平切线的拐点。f'(x) = 3x² 的符号在 x 经过 0 时不发生变化，证实了该分类。
A typical AQA optimisation problem will require you to set up an expression for a quantity, differentiate it, set the derivative to zero, solve for the variable, and then use the second derivative to confirm the nature of the stationary point before calculating the final answer.
典型的 AQA 优化问题要求你建立某个量的表达式，对其求导，令导数为零，解出变量，然后使用二阶导数确认驻点的性质，最后计算最终答案。
9. Sketching Graphs of Functions | 函数图形绘制
Sketching a graph requires the identification of key features: intercepts with the coordinate axes, stationary points, the behaviour as x → ±∞, and any points of inflection. The gradient function provides information about the slope, allowing you to draw a smooth and accurate sketch.
绘制函数图形需要识别关键特征：坐标轴截距、驻点、x → ±∞ 时的行为以及任何拐点。导数函数提供斜率信息，使你能够绘制平滑且准确的草图。
For a cubic y = ax³ + bx² + cx + d, begin by finding the y-intercept (set x = 0) and the x-intercepts (set y = 0 and factorise). Then differentiate to find the coordinates of any stationary points. Finally, check the leading coefficient a: if a > 0, the graph rises to the right and falls to the left; if a < 0, the opposite is true.
对于三次函数 y = ax³ + bx² + cx + d，首先求 y 截距（令 x = 0）和 x 截距（令 y = 0 并分解因式）。然后求导以找出驻点坐标。最后检查首项系数 a：若 a > 0，图形向右上升、向左下降；若 a < 0，则相反。
Quadratics are simpler to sketch: identify whether they open upwards or downwards, find the vertex by completing the square or by differentiating, and locate the intercepts. Remember that the vertex is a stationary point, so setting the derivative to zero yields its x-coordinate directly.
二次函数的图形绘制更为简单：判断其开口方向，通过配方法或求导找顶点，并确定截距。记住顶点是一个驻点，因此令导数为零可直接得到其 x 坐标。
For rational functions, you must also check for vertical and horizontal asymptotes. Vertical asymptotes occur where the denominator is zero, and horizontal asymptotes are determined by comparing the degrees of the numerator and denominator.
对于有理函数，还必须检查垂直和水平渐近线。垂直渐近线出现在分母为零处，水平渐近线通过比较分子和分母的次数来确定。
10. Practical Optimisation Problems | 实际优化问题
Optimisation problems ask you to find the maximum or minimum value of a quantity subject to given constraints. The procedure is systematic: define the quantity to be optimised, express it as a function of a single variable, differentiate, find stationary points, and verify the nature of each point.
优化问题要求你在给定约束条件下找到某个量的最大值或最小值。解题步骤是系统化的：定义要优化的量，将其表示为单变量函数，求导，找到驻点，并验证各点的性质。
A common example involves finding the maximum area of a rectangle with a fixed perimeter. If the perimeter is 100 m, let the length be x and the width be 50 − x. The area is A = x(50 − x) = 50x − x². Differentiating gives dA/dx = 50 − 2x. Setting this to zero yields x = 25, and the second derivative d²A/dx² = −2 < 0 confirms a maximum. The maximum area is 625 m².
一个常见的例子是求固定周长下矩形的最大面积。若周长为 100 米，设长为 x，宽为 50 − x。面积为 A = x(50 − x) = 50x − x²。求导得 dA/dx = 50 − 2x。令其为零得 x = 25，二阶导数 d²A/dx² = −2 < 0 确认是最大值。最大面积为 625 平方米。
In more advanced problems, the quantity is a more complex algebraic expression, and you may need to apply the chain rule or product rule before setting the derivative to zero. Grid and geometric constraints from the problem statement must be integrated into your expression at the modelling stage.
在更高级的问题中，目标量是更复杂的代数表达式，你可能需要先应用链式法则或乘积法则，再令导数为零。问题陈述中的几何约束必须在建模阶段就被整合进你的表达式中。
AQA examination questions typically present the diagram and ask you to verify that a given expression for the quantity is correct, before proceeding to differentiation. Even if you cannot derive the expression, you should still differentiate the given one and complete the optimisation steps for partial credit.
AQA 考试题通常给出图形，要求你先验证给定量的表达式是否正确，然后再进行求导。即使你无法推导出表达式，也应对给定的式子进行求导并完成优化步骤，以获得部分分数。
11. Kinematics: Motion in a Straight Line | 运动学：直线运动
Kinematics connects the calculus of differentiation with the physical concepts of displacement, velocity, and acceleration. If displacement is given by s(t), then velocity is v(t) = ds/dt, and acceleration is a(t) = dv/dt = d²s/dt².
运动学将微分与位移、速度和加速度的物理概念联系起来。若位移由 s(t) 给出，则速度 v(t) = ds/dt，加速度 a(t) = dv/dt = d²s/dt²。
When interpreting these quantities, pay close attention to signs. A positive velocity indicates motion in the positive direction, while a negative velocity indicates motion in the opposite direction. An object is at rest when v(t) = 0, and it changes direction when the velocity changes sign.
在解读这些量时，要特别注意符号。正速度表示向正方向运动，负速度表示向反方向运动。物体静止时 v(t) = 0，速度方向改变时物体改变运动方向。
To find the total distance travelled, you must integrate the absolute value of velocity over the time interval, rather than simply evaluating the change in displacement. This is because displacement can be negative, whereas distance is always positive.
求总路程时，必须对速度的绝对值在时间区间内进行积分，而不是简单计算位移的变化量。这是因为位移可能为负，而路程始终为正。
Common AQA problems ask you to find the displacement after a given time, the times at which the particle is at a specific position, or the maximum height reached by a projectile. For the latter, set v(t) = 0 and solve for t, then substitute back into s(t) to find the maximum displacement.
AQA 常见题型要求求给定时间后的位移、粒子经过某一位置的时刻，或抛体达到的最大高度。对于后者，令 v(t) = 0 解出 t，再代回 s(t) 求得最大位移。
12. Connected Rates of Change | 关联变化率
Rates of change problems involve finding how quickly one quantity changes with respect to time when given the rate of change of another quantity. The chain rule is the fundamental tool: if y depends on x and x depends on t, then dy/dt = dy/dx × dx/dt.
变化率问题涉及在已知一个量关于时间的变化率时，求另一个量的变化速率。链式法则是基本工具：若 y 依赖于 x 且 x 依赖于 t，则 dy/dt = dy/dx × dx/dt。
A classic example is the expansion of a circular ripple on a pond. If the radius r increases at a constant rate of 0.5 m/s, the area A = πr² changes at a rate dA/dt = dA/dr × dr/dt = 2πr × 0.5 = πr m²/s. At r = 3 m, the area is increasing at 3π m²/s.
一个经典例子是池塘中圆形波纹的扩散。若半径 r 以 0.5 米/秒的恒定速率增大，则面积 A = πr² 的变化率为 dA/dt = dA/dr × dr/dt = 2πr × 0.5 = πr 平方米/秒。当 r = 3 米时，面积以 3π 平方米/秒的速率增加。
In these problems, it is essential to write down the relationship between the variables before differentiating with respect to time. Be careful to substitute known values only after performing the differentiation, as using numeric values too early often leads to incorrect derivatives.
在这类问题中，必须先写出变量之间的关系，再进行对时间的微分。注意只能在进行完微分之后再代入已知数值，过早使用数值常会导致错误的导数结果。
When a diagram involves similar triangles, such as a ladder sliding down a wall, the relationships may require implicit differentiation. Recognising which quantities are constant and which vary with time is the key to a correct solution.
当图形涉及相似三角形时，例如梯子沿墙下滑的问题，关系式可能需要使用隐函数求导。识别哪些量是常量、哪些量随时间变化是正确解题的关键。
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