📚 SQA Higher Maths: Glossary & Terminology Quick Memorisation Guide | SQA高等数学：词汇术语速记指南
Mastering the technical vocabulary of SQA Higher Mathematics is half the battle. This guide breaks down the essential terms into logical groups, giving you quick memory hooks and precise definitions in both English and Chinese. Use it alongside your revision to sharpen your exam responses and avoid common misunderstandings.
掌握 SQA 高等数学的专业术语是成功的一半。本指南将核心词汇按逻辑分组，为你提供中英双语的精准定义和速记窍门。结合复习使用，能帮你优化考试答题，避开常见误区。
1. Functions & Graph Terminology | 函数与图像术语
A function maps every input (x) to exactly one output (y). In SQA notation, you will see f(x), and you must state the domain (the set of allowed x‑values) and the range (the set of all possible y‑values). Quick memory: ‘domain is x, range is y’ – think ‘d for driver (the input), r for result (the output)’.
函数将每个输入 (x) 映射到唯一输出 (y)。SQA 中常出现 f(x)，并要求说明定义域（允许的 x 值集）和值域（所有可能的 y 值集）。速记：定义域针对 x，值域针对 y——可以联想‘d 如同方向盘（输入），r 如同结果（输出）’。
Composite functions such as f(g(x)) mean applying g first, then f. Remember: ‘inside first, then outside’. Always check the domain of the inner function before composing.
复合函数 f(g(x)) 指先进行 g 运算，再进行 f 运算。记法：‘先内后外’。复合前务必检查内层函数的定义域。
An inverse function f⁻¹(x) undoes the original mapping. You can find it by swapping x and y, then rearranging. Graphically, the inverse is the reflection of f(x) in the line y = x. The notation f⁻¹ does not mean 1/f(x).
反函数 f⁻¹(x) 会逆转原来的映射。可通过交换 x 与 y 后变形求得。图像上看，反函数是原函数关于直线 y = x 的对称图形。注意符号 f⁻¹ 不表示 1/f(x)。
Stationary points occur where f'(x) = 0. They can be a maximum turning point, a minimum turning point, or a horizontal point of inflection. Use a nature table or the second derivative to classify them.
驻点出现在 f'(x) = 0 处，可能是极大值点、极小值点或水平拐点。可用导数符号表或二阶导数来判断类型。
2. Differentiation Terms | 微分术语
The derivative f'(x) or dy/dx measures the rate of change of y with respect to x. The power rule states: if f(x) = xⁿ, then f'(x) = nxⁿ⁻¹. This is your most used tool.
导数 f'(x) 或 dy/dx 衡量 y 随 x 的变化率。幂函数法则：若 f(x) = xⁿ，则 f'(x) = nxⁿ⁻¹。这是最常用的工具。
d/dx [xⁿ] = n xⁿ⁻¹
The chain rule is used for composite functions: dy/dx = dy/du × du/dx. Trigger phrase: ‘differentiate the outside, keep the inside, then multiply by the derivative of the inside’. This is also called the ‘function of a function’ rule.
链式法则用于复合函数：dy/dx = dy/du × du/dx。口诀：‘先微外面，保留里面，再乘上里面的导数’。这也叫‘函数的函数’法则。
The product rule: if y = u v, then dy/dx = u’v + uv’. The quotient rule: if y = u/v, then dy/dx = (u’v – uv’)/v². In SQA, both can appear but the chain rule is tested most heavily.
乘法法则：若 y = u v，则 dy/dx = u’v + uv’。除法法则：若 y = u/v，则 dy/dx = (u’v – uv’)/v²。SQA 考试中两者都可能出现，但链式法则是重中之重。
The second derivative, written d²y/dx² or f”(x), tells you the rate of change of the gradient. It is used to determine concavity and to confirm the nature of stationary points.
二阶导数记作 d²y/dx² 或 f”(x)，表示斜率的变化率，用于判断凹凸性及确认驻点类型。
Differentiation from first principles uses the limit definition: f'(x) = limₕ→₀ (f(x+h) – f(x))/h. You must be able to do this for simple powers and understand its link to the gradient of a tangent.
从第一原理求导需用极限定义：f'(x) = limₕ→₀ (f(x+h) – f(x))/h。你必须能够对简单幂函数完成该推导，并理解它与切线斜率的关系。
3. Integration Terms | 积分术语
Integration reverses differentiation. The indefinite integral ∫ f(x) dx gives a family of antiderivatives plus a constant of integration ‘C’ . Always write ‘+ C’ unless the question says otherwise.
积分是微分的逆运算。不定积分 ∫ f(x) dx 得到一族原函数，并加上积分常数 C。除非题目另有说明，永远记得写‘+ C’。
∫ xⁿ dx = (xⁿ⁺¹)/(n+1) + C, n ≠ –1
A definite integral ∫ₐᵇ f(x) dx calculates the exact area between the curve and the x‑axis from x = a to x = b. Evaluate the antiderivative at the upper and lower limits, then subtract.
定积分 ∫ₐᵇ f(x) dx 计算从 x = a 到 x = b 曲线与 x 轴之间的准确面积。方法是将原函数在上、下限取值后相减。
When the curve dips below the x‑axis, the integral gives a negative value. To find the total physical area, split the integral at the x‑intercepts and take absolute values where needed.
当曲线落在 x 轴下方时，积分值为负。求实际总面积时，需在 x 截点分段，并对相应部分取绝对值。
Integration by substitution is a method for reverse chain rule. Let u equal the inner function, find du/dx, and rewrite the integral entirely in terms of u and du. Always switch limits for definite integrals.
换元积分法是链式法则的逆用。令 u 等于内层函数，求出 du/dx，并将整个积分用 u 与 du 表示。对定积分记得转换积分限。
The area between two curves is found by integrating the ‘top minus bottom’ function between their intersection points. Sketching the region is essential for SQA exams.
两曲线间的面积可通过在交点间对‘上函数减下函数’积分求得。SQA 考试中画图是必要步骤。
4. Exponentials & Logarithms | 指数与对数
The exponential function eˣ is its own derivative and integral. e is approximately 2.718. Its inverse is the natural logarithm, ln x, defined for x > 0. Key identity: ln(eˣ) = x and eˡⁿ ˣ = x.
指数函数 eˣ 的导数与积分都是它自身。e 约等于 2.718。其反函数是自然对数 ln x，定义域为 x > 0。核心恒等式：ln(eˣ) = x, eˡⁿ ˣ = x。
Log laws are vital: ln(ab) = ln a + ln b, ln(a/b) = ln a – ln b, and ln aᵏ = k ln a. These allow you to solve exponential equations by taking logs on both sides.
对数法则至关重要：ln(ab) = ln a + ln b，ln(a/b) = ln a – ln b，ln aᵏ = k ln a。通过两边取对数可解指数方程。
Exponential growth and decay models take the form A = A₀ eᵏᵗ. If k > 0, it is growth; if k < 0, it is decay. In SQA, you often combine this with logarithms to find the time to reach a certain value.
指数增长与衰减模型形如 A = A₀ eᵏᵗ。k > 0 为增长，k < 0 为衰减。SQA 常要求结合对数求出达到某值所需的时间。
To solve ln(2x+1)=3, rewrite as e³ = 2x+1. Translate every log equation into an exponential form to unlock the unknown.
解 ln(2x+1)=3 时，改写为 e³ = 2x+1。将对数方程转化为指数形式，即可解出未知数。
5. Trigonometry Core | 三角学核心
Radians are the standard angle measure in SQA Higher Maths: π rad = 180°. Exact values for sin, cos and tan of 0, π/6, π/4, π/3 and π/2 must be memorised. Draw the two standard triangles (45° and 60°‑30°) to recall them quickly.
弧度是 SQA 高等数学的标准角度单位：π rad = 180°。必须熟记 sin、cos、tan 在 0、π/6、π/4、π/3、π/2 的精确值。画出 45° 和 60°‑30° 两个标准三角形可快速回忆。
The CAST diagram shows which trigonometric functions are positive in each quadrant. Moving anti‑clockwise from the fourth quadrant: C (cos positive), A (all positive), S (sin positive), T (tan positive).
CAST 图表示各象限中的正函数。从第四象限逆时针：C（cos 正）、A（全正）、S（sin 正）、T（tan 正）。
Key identities: sin²θ + cos²θ ≡ 1, sin2θ ≡ 2 sinθ cosθ, cos2θ ≡ cos²θ – sin²θ ≡ 2cos²θ – 1 ≡ 1 – 2sin²θ. These are used extensively in solving trig equations.
关键恒等式：sin²θ + cos²θ ≡ 1，sin2θ ≡ 2 sinθ cosθ，cos2θ ≡ cos²θ – sin²θ ≡ 2cos²θ – 1 ≡ 1 – 2sin²θ。它们在解三角方程中大量使用。
For solving equations like sin x = 0.4, first find the reference angle using arcsin(0.4), then locate all solutions in the required range using the CAST diagram or sin graph symmetry.
解方程如 sin x = 0.4 时，先用 arcsin(0.4) 求参考角，再依据 CAST 图或正弦图像对称性找出给定范围内的所有解。
Amplitude, period and phase angle describe transformations of trig graphs. For y = a sin(bx + c), amplitude = |a|, period = 2π/b, and phase shift = –c/b. Memorable pattern: ‘A is height, B controls cycle, C shifts laterally’.
振幅、周期和相位角描述三角图像的变换。对 y = a sin(bx + c)，振幅 = |a|，周期 = 2π/b，相位位移 = –c/b。记忆模式：‘A 定高，B 控周期，C 平移’。
6. Vectors in 2D/3D | 二维/三维向量
A vector has both magnitude (length) and direction. In SQA, vectors are written in column form or using i, j, k unit vectors. The position vector of point A relative to the origin is usually denoted as a.
向量既有大小（长度）又有方向。SQA 中向量以列向量或 i, j, k 单位向量表示。点 A 相对于原点的位置向量常记作 a。
The magnitude of a vector a = (x, y) is |a| = √(x² + y²). This extends to 3D with the z component. A unit vector has magnitude 1 and is found by scaling the vector by the reciprocal of its magnitude.
向量 a = (x, y) 的模为 |a| = √(x² + y²)。三维情况加上 z 分量。单位向量的模为 1，由原向量除以模长得到。
The scalar (dot) product a·b = |a||b| cos θ = x₁x₂ + y₁y₂ (+ z₁z₂ in 3D). It is used to show perpendicularity (a·b = 0) and to find the angle between two vectors.
标量积（点乘）a·b = |a||b| cos θ = x₁x₂ + y₁y₂（三维加 z₁z₂）。可用于证明垂直（a·b = 0）及求两向量夹角。
Collinear points lie on the same straight line. You prove collinearity by showing that one vector is a scalar multiple of another and that a common point is shared.
共线点位于同一直线上。可通过证明一个向量是另一个向量的标量倍数，且有公共点，来证明共线。
The vector equation of a line is r = a + λ b, where a is a position vector on the line and b is the direction vector. Changing λ moves you along the line. Learn to convert between parametric and symmetric forms.
直线的向量方程为 r = a + λ b，其中 a 是线上一点的位置向量，b 是方向向量。改变 λ 使点沿直线移动。需学会参数式与对称式之间的转换。
7. Polynomials & Quadratic Functions | 多项式与二次函数
A polynomial is an expression like axⁿ + … + c. The degree is the highest power. Roots or zeros are the x‑values where the polynomial equals zero. Factorisation links roots to linear factors.
多项式是形如 axⁿ + … + c 的表达式，次数指最高次幂。根或零点是使多项式为零的 x 值。因式分解将根与一次因式联系起来。
For a quadratic ax² + bx + c = 0, the discriminant Δ = b² – 4ac tells you the nature of the roots: Δ > 0 → two distinct real roots; Δ = 0 → one repeated real root; Δ < 0 → no real roots (complex).
对二次方程 ax² + bx + c = 0，判别式 Δ = b² – 4ac 说明根的情况：Δ > 0 → 两不等实根；Δ = 0 → 一个重实根；Δ < 0 → 无实根。
Completing the square rewrites ax² + bx + c into a(x + p)² + q. This form instantly gives the vertex (–p, q) and helps with integration and solving equations. The process involves halving the coefficient of x.
配方法将 ax² + bx + c 写为 a(x + p)² + q。此形式直接给出顶点 (–p, q)，并有助于积分和解方程。配方的关键是取 x 系数的一半。
The quadratic formula x = [–b ± √(b² – 4ac)]/(2a) gives the roots of any quadratic. In SQA, you must be comfortable using it without a calculator for exact answers, leaving surds in simplest form.
二次公式 x = [–b ± √(b² – 4ac)]/(2a) 可求任何二次方程的根。SQA 要求能够不用计算器使用该公式，并以最简根式形式给出精确答案。
The Factor Theorem states: if f(p) = 0, then (x – p) is a factor of f(x). Use synthetic division (polynomial long division) to factorise cubics and quartics. The Remainder Theorem gives the remainder when dividing by (x – p).
因式定理：若 f(p) = 0，则 (x – p) 是 f(x) 的因式。用综合除法（多项式长除）对三次及四次式分解。余数定理给出除以 (x – p) 后的余数。
8. Recurrence Relations & Sequences | 递推关系与数列
A recurrence relation defines a term in a sequence using previous terms, e.g. uₙ₊₁ = a uₙ + b. The initial value u₀ or u₁ must be given. These relations model things like population growth, cooling and loans.
递推关系用前项定义数列中的项，如 uₙ₊₁ = a uₙ + b。必须给出初始值 u₀ 或 u₁。这类关系可用于模拟人口增长、冷却过程和贷款等。
A limit L exists if the sequence converges: as n → ∞, uₙ → L. For uₙ₊₁ = a uₙ + b, the limit satisfies L = a L + b, giving L = b/(1 – a), provided –1 < a < 1. Always check the condition before stating a limit.
若数列收敛，则存在极限 L：当 n → ∞ 时，uₙ → L。对 uₙ₊₁ = a uₙ + b，极限满足 L = a L + b，得 L = b/(1 – a)，条件是 –1 < a < 1。说明极限前务必检验该条件。
A linear recurrence relation can be solved to find a closed (nth‑term) formula. In SQA, you may just need to iterate and observe behaviour, or prove a limit. Practice proving by taking the limit on both sides.
线性递推关系可求解得到通项公式。SQA 可能只要求迭代并观察行为，或证明极限。要练习对递推式两边取极限的证明方法。
An arithmetic sequence has a common difference d; a geometric sequence has a common ratio r. Although mainly appearing in recurrence contexts, you must spot these patterns quickly.
等差数列有公差 d；等比数列有公比 r。虽然主要在递推环境中出现，但仍需快速识别这些模式。
9. Transformations of Graphs | 图像变换
Graph transformations move or stretch the curve of y = f(x). The four basic types are: translation, reflection, horizontal stretch and vertical stretch. You must know how each one changes the function’s equation.
图像变换可移动或拉伸 y = f(x) 的曲线。四种基本类型为：平移、对称、水平拉伸和垂直拉伸。必须掌握每种变换如何改变函数表达式。
f(x) + a shifts the graph vertically by a (up if a>0). f(x + a) shifts horizontally by –a (left if a>0). Memorable phrase: ‘outside moves up/down, inside moves opposite left/right’.
f(x) + a 使图像垂直移动 a（上移若 a>0）。f(x + a) 水平移动 –a（左移若 a>0）。口诀：‘外面管上下，里面管相反左右’。
–f(x) reflects in the x‑axis; f(–x) reflects in the y‑axis. When multiplying the function: af(x) is a vertical stretch by factor a; f(kx) is a horizontal compression by factor 1/k (often described as stretch by 1/k).
–f(x) 关于 x 轴对称；f(–x) 关于 y 轴对称。函数乘法：af(x) 是垂直拉伸 a 倍；f(kx) 是水平压缩 1/k 倍（常表述为拉伸 1/k）。
When multiple transformations are applied, apply them in the order ‘horizontal shift, horizontal stretch, reflection, vertical stretch, vertical shift’. SQA often asks to identify a sequence of transformations from one graph to another.
进行多次变换时，按‘水平平移、水平拉伸、对称、垂直拉伸、垂直平移’的顺序操作。SQA 常要求从一图像到另一图像识别变换序列。
The mapping notation (x, y) → ( … , … ) concisely describes what happens to every point. For example, y = 2 f(3x – 1) + 4 can be written as (x, y) → ( (x+1)/3 , 2y + 4 ). Practice converting between equation and mapping.
映射记法 (x, y) → ( … , … ) 简洁描述每个点的变化。例如 y = 2 f(3x – 1) + 4 可写为 (x, y) → ( (x+1)/3 , 2y + 4 )。需练习表达式与映射之间的互化。
10. Optimisation & Rates of Change | 优化与变化率
Optimisation problems ask you to find the maximum or minimum of a quantity. Steps: express the quantity as a function of one variable, differentiate, set f'(x)=0, solve, and prove max/min using a nature table or second derivative.
优化问题要求找出某量的最大值或最小值。步骤：将量表示为单变量函数、求导、令 f'(x)=0 求解，并用导数符号表或二阶导数证明极值类型。
Common SQA contexts include minimising surface area for a fixed volume, maximising area enclosed by a fence, and maximising revenue. Always isolate a variable using given constraints.
SQA 常见情景包括固定体积下最小化表面积，围栏圈地最大化面积，以及最大化收入。始终利用给定约束消去一个变量。
The rate of change dy/dt is found using the chain rule dy/dt = dy/dx × dx/dt. This links gradients to real‑world speeds, such as how fast a water level rises. Identify what is constant and what is changing.
变化率 dy/dt 通过链式法则 dy/dt = dy/dx × dx/dt 求取。这把斜率与真实世界速度联系起来，如水位上升速度。要区分常量与变量。
Stationary points in context: a turning point may represent the maximum profit or minimum cost. Always interpret your answers in the context of the problem and check the endpoints of the domain too.
实际问题中的驻点：极值点可能代表最大利润或最小成本。务必将答案放回问题情境中解释，并检查定义域端点。
Published by TutorHao | Maths Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导