📚 Miscellaneous Exercises 1 | 杂项练习 1
This article is tailored for AQA A-Level Mathematics students who are working through the first block of pure mathematics content. The “Miscellaneous Exercises 1” set is designed to test everything you have learned in the opening chapters — from indices and surds to quadratics, simultaneous equations, inequalities, straight lines and circles. We will work through each major skill, with fully worked examples and the exact notation you need under exam conditions.
本文专为 AQA A-Level 数学学生撰写，针对教材第一单元的“杂项练习 1”进行系统复习。该练习旨在检验开头几章所学的全部内容，包括指数与根式、二次方程、联立方程、不等式、直线与圆。我们将逐项梳理核心技能，给出完整解答过程，并展示考试中要求的规范书写格式。
1. Laws of Indices | 指数法则
All AQA Pure Mathematics papers expect confident manipulation of indices. The three core laws are aᵐ × aⁿ = aᵐ⁺ⁿ, aᵐ ÷ aⁿ = aᵐ⁻ⁿ, and (aᵐ)ⁿ = aᵐⁿ. You must also recall that a⁰ = 1, a⁻ⁿ = 1/aⁿ, and a^(m/n) = (ⁿ√a)ᵐ.
AQA 纯数试卷要求熟练掌握指数运算。三大基本法则为 aᵐ × aⁿ = aᵐ⁺ⁿ、aᵐ ÷ aⁿ = aᵐ⁻ⁿ、以及 (aᵐ)ⁿ = aᵐⁿ。你还需要记住 a⁰ = 1、a⁻ⁿ = 1/aⁿ 以及 a^(m/n) = (ⁿ√a)ᵐ。
Worked example: simplify (3x⁻²y³)² ÷ (6x⁻⁴y²), giving your answer with positive indices.
示例：化简 (3x⁻²y³)² ÷ (6x⁻⁴y²)，最终结果中的指数须为正数。
(3x⁻²y³)² ÷ (6x⁻⁴y²) = 9x⁻⁴y⁶ ÷ 6x⁻⁴y² = (3/2)y⁴
First square the bracket: multiply the index 2 by each inner index to get 9x⁻⁴y⁶. Then divide the coefficients (9 ÷ 6 = 3/2) and subtract the indices: x⁻⁴⁻⁽⁻⁴⁾ = x⁰ = 1, and y⁶⁻² = y⁴. The final answer contains no x term at all.
首先将括号平方：用指数 2 分别乘以括号内各指数，得到 9x⁻⁴y⁶。然后系数相除（9 ÷ 6 = 3/2），指数相减：x⁻⁴⁻⁽⁻⁴⁾ = x⁰ = 1，y⁶⁻² = y⁴。最终结果中不含有 x 项。
2. Surds and Rationalisation | 根式与有理化
Surds are exact roots such as √2, √5 or 2√3. In AQA exams you are often asked to simplify expressions like √18 = 3√2, or to rationalise a denominator. The key identity is (√a + √b)(√a − √b) = a − b.
根式是精确的无理根，如 √2、√5 或 2√3。AQA 考试常要求化简根式，例如 √18 = 3√2，或对分母进行有理化。核心恒等式为 (√a + √b)(√a − √b) = a − b。
Worked example: rationalise the denominator of 5/(2 − √3), leaving your answer in the form a + b√3.
示例：将 5/(2 − √3) 的分母有理化，结果表示为 a + b√3 的形式。
5/(2 − √3) × (2 + √3)/(2 + √3) = 5(2 + √3)/(4 − 3) = 10 + 5√3
Multiply top and bottom by the conjugate surd 2 + √3. The denominator becomes 4 − 3 = 1, so the surd simply disappears. The result is 10 + 5√3, so a = 10 and b = 5. Always check whether the final surd can be simplified further.
将分子分母同时乘以共轭根式 2 + √3。分母变为 4 − 3 = 1，根式随即消失。结果为 10 + 5√3，因此 a = 10，b = 5。完成后要检查根式是否还能继续化简。
If the denominator is of the form √c only, multiply by √c/√c. For example, 3/√6 = 3√6/6 = √6/2.
若分母只是形如 √c 的单一根式，则乘以 √c/√c。例如 3/√6 = 3√6/6 = √6/2。
3. Solving Quadratic Equations | 解二次方程
There are three standard methods for solving ax² + bx + c = 0 in AQA exams: factorisation, completing the square, and the quadratic formula. You should choose the most efficient method for each question.
在 AQA 考试中，解 ax² + bx + c = 0 的三种标准方法为：因式分解、配方法以及求根公式。你应当针对不同题目选择最高效的方法。
Worked example: solve 2x² − 7x + 3 = 0.
示例：解方程 2x² − 7x + 3 = 0。
2x² − 7x + 3 = (2x − 1)(x − 3) = 0, hence x = 1/2 or x = 3
Check by expanding: (2x − 1)(x − 3) = 2x² − 6x − x + 3 = 2x² − 7x + 3. The solutions are the values of x that make each bracket zero. In an AQA examination, always state both solutions clearly and substitute them back if time permits.
检验展开：(2x − 1)(x − 3) = 2x² − 6x − x + 3 = 2x² − 7x + 3。令每个括号为零即得方程的解。考试中应清晰地写出两个解，时间允许时最好代回原方程验算。
If the quadratic does not factorise neatly, use the formula x = (−b ± √(b² − 4ac))/(2a). Do not round at intermediate steps; store values in your calculator or keep exact surd forms.
若二次式无法直接分解，则使用求根公式 x = (−b ± √(b² − 4ac))/(2a)。不要在中间步骤进行四舍五入；应利用计算器存储值或保留精确根式形式。
4. Completing the Square | 配方法
Completing the square rewrites x² + bx + c as (x + p)² + q. This form reveals the turning point of a quadratic graph and is required for solving equations when factorisation fails. It also appears frequently in AQA questions on ranges and transformations.
配方法将 x² + bx + c 改写为 (x + p)² + q 的形式，能够直接显示抛物线顶点的坐标，并在因式分解失效时用于解方程。AQA 试题还常借助配方法考查函数值域与图像变换。
Worked example: express x² + 6x + 11 in the form (x + p)² + q.
示例：将 x² + 6x + 11 化为 (x + p)² + q 的形式。
x² + 6x + 11 = (x + 3)² − 9 + 11 = (x + 3)² + 2
Take half of 6 to get p = 3, write (x + 3)², then subtract 3² = 9 to compensate. The constant 11 becomes +11 − 9 = +2. The turning point of y = x² + 6x + 11 is therefore (−3, 2).
取 6 的一半得 p = 3，写成 (x + 3)²，再减去 3² = 9 以保持等价。常数项 11 变为 11 − 9 = +2。因此 y = x² + 6x + 11 的顶点坐标为 (−3, 2)。
For a coefficient of x² greater than 1, factor it out first. For example, 2x² − 8x + 5 = 2(x² − 4x) + 5 = 2[(x − 2)² − 4] + 5 = 2(x − 2)² − 3.
当 x² 的系数大于 1 时，应先将该系数提出。例如 2x² − 8x + 5 = 2(x² − 4x) + 5 = 2[(x − 2)² − 4] + 5 = 2(x − 2)² − 3。
5. The Discriminant | 判别式
For a quadratic ax² + bx + c = 0, the discriminant is Δ = b² − 4ac. The sign of Δ tells you the number of real roots: Δ > 0 gives two distinct real roots, Δ = 0 gives one repeated root, and Δ < 0 gives no real roots.
对于二次方程 ax² + bx + c = 0，判别式为 Δ = b² − 4ac。Δ 的符号决定实根的个数：Δ > 0 有两个不等实根，Δ = 0 有一个重根，Δ < 0 没有实根。
Worked example: find the set of values of k for which the equation x² + kx + 9 = 0 has real roots.
示例：求使方程 x² + kx + 9 = 0 有实数根的 k 的取值范围。
Δ = k² − 4 × 1 × 9 = k² − 36 ≥ 0, so k² ≥ 36, hence k ≤ −6 or k ≥ 6
Setting Δ ≥ 0 gives k² − 36 ≥ 0. Be careful with the final inequality: taking square roots gives k ≤ −6 or k ≥ 6, not −6 ≤ k ≤ 6. A sketch of the parabola y = k² − 36 helps avoid this classic sign error.
令 Δ ≥ 0 得 k² − 36 ≥ 0。注意最后的不等式：两边开平方得到 k ≤ −6 或 k ≥ 6，而不是 −6 ≤ k ≤ 6。画出抛物线 y = k² − 36 的草图有助于避免这个经典符号错误。
The discriminant also connects to tangents: a line that just touches a curve corresponds to Δ = 0, while a line that misses the curve entirely corresponds to Δ < 0.
判别式还与切线有关：直线与曲线相切对应 Δ = 0，直线与曲线相离对应 Δ < 0。
6. Simultaneous Equations | 联立方程
AQA miscellaneous exercise sets frequently mix a linear equation with a quadratic equation. The substitution method is standard: rearrange the linear equation for one variable, substitute into the quadratic, then solve the resulting quadratic.
AQA 杂项练习常将线性方程与二次方程联立。标准做法是代入法：先从线性方程中解出一个变量，代入二次方程，再解所得二次方程。
Worked example: solve y = x² − 3x + 2 and y = x − 1 simultaneously.
示例：联立求解 y = x² − 3x + 2 与 y = x − 1。
x² − 3x + 2 = x − 1 ⟹ x² − 4x + 3 = 0 ⟹ (x − 1)(x − 3) = 0
So x = 1 or x = 3. Substitute back into y = x − 1 to get y = 0 when x = 1, and y = 2 when x = 3. The intersection points are (1, 0) and (3, 2).
因此 x = 1 或 x = 3。代回 y = x − 1 得：当 x = 1 时 y = 0；当 x = 3 时 y = 2。两交点分别为 (1, 0) 与 (3, 2)。
After substituting, you may obtain a quadratic with no real roots, meaning the line and curve do not intersect. This is a valid final answer — do not force solutions that do not exist.
代入后若所得二次方程无实根，说明直线与曲线不相交。这本身就是正确答案，切勿强行构造不存在的解。
7. Quadratic Inequalities | 二次不等式
To solve a quadratic inequality, first rearrange it so that one side is zero, then sketch the corresponding parabola. The solution set is read directly from the sketch: above the x-axis for a > 0, below the x-axis for a < 0.
解二次不等式时，先将不等式整理为一边为零的形式，再画出对应抛物线草图。通过观察图像直接写出解集：在 x 轴上方对应 a > 0，在 x 轴下方对应 a < 0。
Worked example: solve x² − 5x + 6 < 0.
示例：解不等式 x² − 5x + 6 < 0。
x² − 5x + 6 = (x − 2)(x − 3) < 0 ⟹ 2 < x < 3
The parabola crosses the x-axis at x = 2 and x = 3. Because the coefficient of x² is positive, the curve is U-shaped, so the region below zero lies between the two roots. Write the answer as 2 < x < 3, not as two separate inequalities.
抛物线在 x = 2 与 x = 3 处与 x 轴相交。由于 x² 系数为正，曲线开口向上，位于零以下的部分在两根之间。答案应写成 2 < x < 3，而不要分成两个独立的不等式。
If the inequality sign is reversed, such as x² − 5x + 6 > 0, the solution is the union x < 2 or x > 3. Memorise these two patterns carefully.
若不等号方向相反，例如 x² − 5x + 6 > 0，则解集为两段区域 x < 2 或 x > 3。务必牢记这两种典型情形。
8. Graph Transformations | 图像变换
Graph transformations are a heavy topic in AQA Pure Year 1. The key rules are: y = f(x) + a translates upwards by a; y = f(x + a) translates left by a; y = −f(x) reflects in the x-axis; and y = f(−x) reflects in the y-axis.
图像变换是 AQA 纯数第一年的大考点。核心规则为：y = f(x) + a 表示向上平移 a；y = f(x + a) 表示向左平移 a；y = −f(x) 表示关于 x 轴反射；y = f(−x) 表示关于 y 轴反射。
Worked example: describe the transformation that maps y = f(x) onto y = f(x + 2) − 3.
示例：描述由 y = f(x) 到 y = f(x + 2) − 3 的图像变换。
Translation by the vector (−2, −3)
The +2 inside the bracket moves the graph left by 2 units, and the −3 outside moves it down by 3 units. In AQA mark schemes, you must give both the type of transformation and the precise displacement. State “translation” and include the vector.
括号内的 +2 使图像向左移动 2 个单位，括号外的 −3 使图像向下移动 3 个单位。AQA 评分标准要求同时说明变换类型与精确位移：写“平移 (translation)”并给出向量 (−2, −3)。
Stretches appear in later chapters, but the translation and reflection rules above are the ones tested in most miscellaneous exercise sets for Chapter 1.
拉伸变换会在后续章节出现，但以上平移与反射规则正是第一章杂项练习中最常考查的内容。
9. Equations of Straight Lines | 直线方程
The equation of a straight line can be written as y = mx + c, where m is the gradient and c is the y-intercept. Alternatively, given a point (x₁, y₁) and gradient m, use y − y₁ = m(x − x₁).
直线方程可写成 y = mx + c，其中 m 为斜率，c 为 y 轴截距。若已知点 (x₁, y₁) 与斜率 m，可使用点斜式 y − y₁ = m(x − x₁)。
Worked example: find the equation of the line through (2, 3) that is perpendicular to y = 2x + 1.
示例：求过点 (2, 3) 且垂直于 y = 2x + 1 的直线方程。
m₁ × m₂ = −1 ⟹ 2 × m₂ = −1 ⟹ m₂ = −1/2
y − 3 = −1/2 (x − 2) ⟹ y = −x/2 + 4
Perpendicular gradients multiply to −1, so the new gradient is −1/2. Substituting the point (2, 3) gives y − 3 = −1/2(x − 2), which simplifies to y = −x/2 + 4. Always state the final equation in the form requested — often y = mx + c.
垂直直线的斜率之积为 −1，故新斜率为 −1/2。代入点 (2, 3) 得 y − 3 = −1/2(x − 2)，化简为 y = −x/2 + 4。最终答案应按照题目要求的形式书写，通常为 y = mx + c。
Parallel lines have equal gradients, and the distance between two points is d = √((x₂ − x₁)² + (y₂ − y₁)²). Keep exact forms where surds appear.
平行直线的斜率相等。两点间距离公式为 d = √((x₂ − x₁)² + (y₂ − y₁)²)。出现根式时请保留精确值。
10. The Equation of a Circle | 圆的方程
The equation of a circle with centre (a, b) and radius r is (x − a)² + (y − b)² = r². To identify the centre and radius from a general equation, you must complete the square in both x and y.
圆心为 (a, b)、半径为 r 的圆方程为 (x − a)² + (y − b)² = r²。若给出一般式，则需分别在 x 和 y 上使用配方法来确定圆心与半径。
Worked example: write down the equation of the circle with centre (3, −2) and radius 4.
示例：写出圆心为 (3, −2)、半径为 4 的圆的方程。
(x − 3)² + (y + 2)² = 16
Replace a = 3 and b = −2 directly into the standard form. Note that y − (−2) becomes y + 2, and the radius 4 is squared to give 16. This is the most straightforward circle question you will meet.
直接将 a = 3、b = −2 代入标准形式。注意 y − (−2) 变为 y + 2，半径 4 要平方为 16。这是最直接的圆的方程题目。
To check whether a point lies on a circle, substitute its coordinates into the left-hand side and confirm it equals r². If the value is less than r², the point lies inside; if greater, it lies outside.
要判断点是否在圆上，可将该点坐标代入方程左边，检查是否等于 r²。若小于 r²，则点在
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