Miscellaneous Exercises 3: Mixed AQA A-Level Maths Practice | 综合练习三：AQA A-Level 数学混合训练

📚 Miscellaneous Exercises 3: Mixed AQA A-Level Maths Practice | 综合练习三：AQA A-Level 数学混合训练

This set of miscellaneous exercises brings together key topics from the AQA A-Level Mathematics specification, including pure mathematics, statistics and mechanics. Use it to test your fluency, identify weak areas and practise writing clear, structured solutions.

本套综合练习汇集了 AQA A-Level 数学大纲中的核心主题，涵盖纯数、统计和力学。用它来检验熟练度、找出薄弱环节并练习规范书写解答。

1. Algebraic Simplification and Surds | 代数化简与根式

Simplify (√75 + √12)² and rationalise the denominator of 5/(3 − √2).

化简 (√75 + √12)²，并将 5/(3 − √2) 的分母有理化。

Write √75 = √25×3 = 5√3 and √12 = √4×3 = 2√3, so √75 + √12 = 7√3. Squaring gives (7√3)² = 49×3 = 147.

写出 √75 = √25×3 = 5√3，√12 = √4×3 = 2√3，因此 √75 + √12 = 7√3。平方得到 (7√3)² = 49×3 = 147。

For 5/(3 − √2), multiply numerator and denominator by the conjugate 3 + √2: 5(3 + √2)/((3 − √2)(3 + √2)) = 5(3 + √2)/(9 − 2) = 5(3 + √2)/7.

对于 5/(3 − √2)，将分子和分母同乘共轭根式 3 + √2：5(3 + √2)/((3 − √2)(3 + √2)) = 5(3 + √2)/(9 − 2) = 5(3 + √2)/7。

(√75 + √12)² = 147 and 5/(3 − √2) = 5(3 + √2)/7

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2. Quadratics and the Discriminant | 二次函数与判别式

f(x) = 2x² − 8x + k, has two distinct real roots. Find the range of possible values of k.

方程 f(x) = 0 中 f(x) = 2x² − 8x + k 有两个不相等的实根。求 k 的取值范围。

² − 4ac is greater than zero. Here a = 2, b = −8, c = k, so Δ = (−8)² − 4(2)(k) = 64 − 8k.

对于二次式 ax² + bx + c，当判别式 Δ = b² − 4ac 大于零时有两个不相等的实根。这里 a = 2，b = −8，c = k，因此 Δ = (−8)² − 4(2)(k) = 64 − 8k。

Set Δ > 0: 64 − 8k > 0, so −8k > −64, hence k < 8. The possible values are k < 8.

令 Δ > 0：64 − 8k > 0，即 −8k > −64，因此 k < 8。k 的取值范围是 k < 8。

k < 8

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3. Polynomial Division and the Factor Theorem | 多项式除法与因式定理

) is a factor, fully factorise f(x), and find the remainder when f(x) is divided by (x + 1).

设 f(x) = x³ − 4x² + x + 6。证明 (x − 2) 是一个因式，将 f(x) 完全因式分解，并求 f(x) 除以 (x + 1) 的余数。

actor, then f(2) = 0. Substitute x = 2: f(2) = 8 − 16 + 2 + 6 = 0, so (x − 2) is a factor.

利用因式定理，若 (x − 2) 是因式，则 f(2) = 0。代入 x = 2：f(2) = 8 − 16 + 2 + 6 = 0，所以 (x − 2) 是一个因式。

factor: f(x) = (x − 2)(x² − 2x − 3). Fully factorising gives f(x) = (x − 2)(x − 3)(x + 1).

将 f(x) 除以 (x − 2) 得到二次因式：f(x) = (x − 2)(x² − 2x − 3)。完全因式分解得到 f(x) = (x − 2)(x − 3)(x + 1)。

(x + 1) is f(−1) = −1 − 4 − 1 + 6 = 0, so (x + 1) is also a factor and the remainder is 0.

当 f(x) 除以 (x + 1) 时余数为 f(−1) = −1 − 4 − 1 + 6 = 0，所以 (x + 1) 也是因式，余数为 0。

f(x) = (x − 2)(x − 3)(x + 1), remainder = 0

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4. Exponentials and Logarithms | 指数与对数

Solve the equations: (a) 3e²ˣ = 21, and (b) log₂(x + 1) + log₂(x − 1) = 3.

解方程：(a) 3e²ˣ = 21，以及 (b) log₂(x + 1) + log₂(x − 1) = 3。

For (a), divide both sides by 3: e²ˣ = 7. Take the natural logarithm: 2x = ln 7, so x = (ln 7)/2.

对于 (a)，两边同时除以 3：e²ˣ = 7。取自然对数：2x = ln 7，因此 x = (ln 7)/2。

gives x = 3 or x = −3. Since the domain requires x > 1, the only valid solution is x = 3.

对于 (b)，合并对数：log₂((x + 1)(x − 1)) = 3。于是 (x + 1)(x − 1) = 2³ = 8，所以 x² − 1 = 8，x² = 9。得到 x = 3 或 x = −3。由于定义域要求 x > 1，唯一有效解是 x = 3。

(a) x = (ln 7)/2, (b) x = 3

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5. Trigonometric Equations | 三角方程

Solve 2 sin² θ − sin θ − 1 = 0 for 0° ≤ θ ≤ 360°.

在 0° ≤ θ ≤ 360° 范围内解方程 2 sin² θ − sin θ − 1 = 0。

becomes 2u² − u − 1 = 0, which factorises as (2u + 1)(u − 1) = 0. Hence u = 1 or u = −1/2.

令 u = sin θ。方程变为 2u² − u − 1 = 0，因式分解为 (2u + 1)(u − 1) = 0。因此 u = 1 或 u = −1/2。

sin θ = 1, then θ = 90°. If sin θ = −1/2, then θ = 210° or θ = 330° in the given interval.

若 sin θ = 1，则 θ = 90°。若 sin θ = −1/2，则在指定区间内 θ = 210° 或 θ = 330°。

θ = 90°, 210°, 330°

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6. Differentiation Techniques | 微分技巧

espect to x, and find the equation of the tangent to y = x³ − 3x at the point where x = 2.

对 y = (3x² − 2)⁴ 求关于 x 的导数，并求曲线 y = x³ − 3x 在 x = 2 处的切线方程。

Using the chain rule, dy/dx = 4(3x² − 2)³ × 6x = 24x(3x² − 2)³.

利用链式法则，dy/dx = 4(3x² − 2)³ × 6x = 24x(3x² − 2)³。

2 = 8 − 6 = 2. The derivative is dy/dx = 3x² − 3, so at x = 2 the gradient is 3×4 − 3 = 9.

对于 y = x³ − 3x 在 x = 2 处，y 坐标为 2³ − 3×2 = 8 − 6 = 2。导数为 dy/dx = 3x² − 3，因此在 x = 2 处斜率为 3×4 − 3 = 9。

sses through (2, 2), so its equation is y − 2 = 9(x − 2), which simplifies to y = 9x − 16.

切线斜率为 9 且经过点 (2, 2)，因此其方程为 y − 2 = 9(x − 2)，化简为 y = 9x − 16。

dy/dx = 24x(3x² − 2)³, tangent: y = 9x − 16

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7. Integration and Area Under Curves | 积分与曲线下面积

3) dx, evaluate ∫₁² (3x² − 2x) dx, and find the area under y = x² + 1 from x = 0 to x = 2.

求 ∫ (4x³ − 2/x + 3) dx，计算 ∫₁² (3x² − 2x) dx，并求 y = x² + 1 从 x = 0 到 x = 2 的曲线下面积。

Integrate term by term: ∫ 4x³ dx = x⁴, ∫ −2/x dx = −2 ln|x|, and ∫ 3 dx = 3x. Therefore the indefinite integral is x⁴ − 2 ln|x| + 3x + c.

逐项积分：∫ 4x³ dx = x⁴，∫ −2/x dx = −2 ln|x|，∫ 3 dx = 3x。因此不定积分为 x⁴ − 2 ln|x| + 3x + c。

ntiderivative of 3x² − 2x is x³ − x². Evaluate from 1 to 2: (8 − 4) − (1 − 1) = 4 − 0 = 4.

对于定积分，3x² − 2x 的原函数是 x³ − x²。从 1 到

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