Further Maths Unit Test Mock Paper Analysis | 进阶数学单元测试模拟卷解析

📚 Further Maths Unit Test Mock Paper Analysis | 进阶数学单元测试模拟卷解析

This article provides a detailed walkthrough of a typical Year 11 Cambridge Further Maths unit test mock paper. The paper covers functions, quadratics, exponentials, logarithms, coordinate geometry, trigonometry, polynomials, binomial expansion, vectors, differentiation and permutations – all core topics for Cambridge IGCSE Additional Mathematics (0606) or O Level Further Maths. Each solution is broken down step by step, with notes on common pitfalls and examiner expectations.

本文详细解析一套典型的 Year 11 剑桥进阶数学单元测试模拟卷。试卷涵盖函数、二次方程、指数、对数、坐标几何、三角学、多项式、二项式定理、向量、微分以及排列组合——均为剑桥 IGCSE 附加数学 (0606) 或 O Level 进阶数学的核心考点。每道题的解答都分步骤拆解，并指出常见错误与考官要求。

1. Functions and Inverse Functions | 函数与反函数

Question: Given f(x) = 2x + 3, find f⁻¹(x) and state its domain.

题目：已知 f(x) = 2x + 3，求 f⁻¹(x) 并说明其定义域。

tion as y = 2x + 3. To find the inverse, swap x and y to get x = 2y + 3, then solve for y.

将函数写成 y = 2x + 3。求反函数时，交换 x 与 y 得到 x = 2y + 3，然后解出 y。

Solving: x - 3 = 2y → y = (x - 3)/2. Therefore f⁻¹(x) = (x - 3)/2.

求解：x - 3 = 2y → y = (x - 3)/2。因此 f⁻¹(x) = (x - 3)/2。

The domain of f⁻¹ is the set of all real numbers, ℝ, because the original function f is a linear function with unrestricted domain and range. There are no denominators or square roots that would impose restrictions.

反函数的定义域为全体实数 ℝ，因为原函数 f 是一次函数，其定义域和值域均无限制。这里没有分母或平方根，不会形成限制。

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2. Quadratic Discriminant | 二次方程判别式

Question: Find the range of values of k for which the equation x² + kx + 4 = 0 has no real roots.

题目：求 k 的取值范围，使方程 x² + kx + 4 = 0 没有实数根。

eal roots, the discriminant Δ = b² - 4ac must be less than zero. Here a = 1, b = k, c = 4.

对于二次方程 ax² + bx + c = 0，要没有实数根，判别式 Δ = b² - 4ac 必须小于零。此处 a = 1, b = k, c = 4。

Calculate Δ = k² - 4(1)(4) = k² - 16. Set Δ < 0 → k² - 16 < 0 → k² < 16.

计算 Δ = k² - 4(1)(4) = k² - 16。令 Δ < 0 → k² - 16 < 0 → k² < 16。

Taking square roots yields -4 < k < 4. So the equation has no real roots when k is strictly between -4 and 4.

开平方得 -4 < k < 4。因此当 k 严格介于 -4 与 4 之间时，方程没有实数根。

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3. Exponential and Logarithmic Equations | 指数与对数方程

Question: Solve the equation 3²ˣ⁺¹ = 5 × 3ˣ - 2.

题目：解方程 3²ˣ⁺¹ = 5 × 3ˣ - 2。

Rewrite 3²ˣ⁺¹ as 3 × 3²ˣ = 3 × (3ˣ)². Let t = 3ˣ, where t > 0. The equation becomes 3t² = 5t - 2.

将 3²ˣ⁺¹ 写成 3 × 3²ˣ = 3 × (3ˣ)²。设 t = 3ˣ，其中 t > 0。方程化为 3t² = 5t - 2。

Rearrange to 3t² - 5t + 2 = 0. Factorise: (3t - 2)(t - 1) = 0, giving t = 2/3 or t = 1.

移项得 3t² - 5t + 2 = 0。因式分解：(3t - 2)(t - 1) = 0，解得 t = 2/3 或 t = 1。

Substitute back: 3ˣ = 1 → x = 0. For 3ˣ = 2/3 → take log base 3: x = log₃(2/3) = log₃2 - 1.

代回：3ˣ = 1 → x = 0。对于 3ˣ = 2/3 → 取以 3 为底的对数：x = log₃(2/3) = log₃2 - 1。

olutions are valid since t > 0 is satisfied. Exact answers are x = 0 and x = log₃2 - 1.

两个解均有效，满足 t > 0。精确答案为 x = 0 和 x = log₃2 - 1。

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4. Coordinate Geometry: Perpendicular Bisector | 坐标几何：垂直平分线

Question: Find the equation of the perpendicular bisector of the line segment joining A(1,2) and B(5,-4).

题目：求连接 A(1,2) 和 B(5,-4) 的线段之垂直平分线方程。

First find the midpoint M of AB: ((1+5)/2, (2+(-4))/2) = (3, -1).

先求 AB 的中点 M：((1+5)/2, (2+(-4))/2) = (3, -1)。

Gradient of AB = (y₂ - y₁)/(x₂ - x₁) = (-4 - 2)/(5 - 1) = -6/4 = -3/2.

AB 的斜率 = (y₂ - y₁)/(x₂ - x₁) = (-4 - 2)/(5 - 1) = -6/4 = -3/2。

The perpendicular gradient is the negative reciprocal: m = 2/3.

垂直线的斜率为其负倒数：m = 2/3。

Using point-slope form with M(3,-1): y - (-1) = (2/3)(x - 3) → y + 1 = (2/3)(x - 3).

利用点斜式代入 M(3,-1)：y - (-1) = (2/3)(x - 3) → y + 1 = (2/3)(x - 3)。

3: 3y + 3 = 2x - 6 → 2x - 3y - 9 = 0. This is the equation of the perpendicular bisector.

两边乘 3：3y + 3 = 2x - 6 → 2x - 3y - 9 = 0。此即垂直平分线方程。

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5. Trigonometric Equations | 三角方程

Question: Solve 2 sin² θ - cos θ = 1 for 0° ≤ θ ≤ 360°.

题目：解方程 2 sin² θ - cos θ = 1，其中 0° ≤ θ ≤ 360°。

Use the identity sin² θ = 1 - cos² θ. Substitute: 2(1 - cos² θ) - cos θ = 1.

利用恒等式 sin² θ = 1 - cos² θ 代入：2(1 - cos² θ) - cos θ = 1。

Expand: 2 - 2 cos² θ - cos θ = 1 → bring all terms to one side: -2 cos² θ - cos θ + 1 = 0.

展开：2 - 2 cos² θ - cos θ = 1 → 将所有项移到一边：-2 cos² θ - cos θ + 1 = 0。

Multiply by -1: 2 cos² θ + cos θ - 1 = 0. Factorise: (2 cos θ - 1)(cos θ + 1) = 0.

乘以 -1：2 cos² θ + cos θ - 1 = 0。因式分解：(2 cos θ - 1)(cos θ + 1) = 0。

Thus cos θ = 1/2 or cos θ = -1.

因此 cos θ = 1/2 或 cos θ = -1。

r cos θ = 1/2, θ = 60°, 300°. For cos θ = -1, θ = 180°. All are within the given interval.

对于 cos θ = 1/2，θ = 60°, 300°。对于 cos θ = -1，θ = 180°。所有解均在给定区间内。

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6. Polynomial Remainder Theorem | 多项式余数定理

Question: Given f(x) = 2x³ - 5x² + x + 2, find the remainder when f(x) is divided by (2x + 1).

题目：已知 f(x) = 2x³ - 5x² + x + 2，求 f(x) 除以 (2x + 1) 的余式。

By the Remainder Theorem, the remainder when dividing by (ax + b) is f(-b/a). Here divisor is 2x + 1, so evaluate f(-1/2).

根据余数定理，除以 (ax + b) 的余式为 f(-b/a)。此处除式为 2x + 1，因此计算 f(-1/2)。

f(-1/2) = 2(-1/2)³ - 5(-1/2)² + (-1/2) + 2 = 2(-1/8) - 5(1/4) - 1/2 + 2.

f(-1/2) = 2(-1/2)³ - 5(-1/2)² + (-1/2) + 2 = 2(-1/8) - 5(1/4) - 1/2 + 2。

-1/4 - 5/4 - 1/2 + 2 = (-1 -5)/4 - 1/2 + 2 = -6/4 - 1/2 + 2 = -3/2 - 1/2 + 2 = -2 + 2 = 0.

逐项计算：-1/4 - 5/4 - 1/2 + 2 = (-1 -5)/4 - 1/2 + 2 = -6/4 - 1/2 + 2 = -3/2 - 1/2 + 2 = -2 + 2 = 0。

The remainder is 0, meaning (2x + 1) is a factor of f(x).

余数为 0，说明 (2x + 1) 是 f(x) 的因式。

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7. Binomial Expansion Coefficient | 二项式展开系数

Question: Find the coefficient of x³ in the expansion of (2 + 3x)⁷.

题目：求 (2 + 3x)⁷ 的展开式中 x³ 的系数。

The general term in the binomial expansion is given by C(7, r) × 2^7-r × (3x)^r, where r = 0,1,...,7.

二项式展开的通项为 C(7, r) × 2^7-r × (3x)^r，其中 r = 0,1,...,7。

We need the term containing x³, so set r = 3. The coefficient is C(7,3) × 2^7-3 × 3³.

我们需要 x³ 项，故令 r = 3。系数为 C(7,3) × 2^7-3 × 3³。

C(7,3) = 35, 2⁴ = 16, 3³ = 27. Multiply: 35 × 16 = 560, then 560 × 27 = 15120.

C(7,3) = 35, 2⁴ = 16, 3³ = 27。连乘：35 × 16 = 560，560 × 27 = 15120。

Thus the coefficient of x³ is 15120. Note that only the numerical coefficient is required, not the full term.

因此 x³ 的系数为 15120。注意仅需数值系数，不需写出整个项。

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8. Vectors: Magnitude and Unit Vector | 向量：模与单位向量

Question: Given a = 3i - 2j and b = -i + 4j, find |a - 2b| and the unit vector in the direction of a + b.

题目：已知 a = 3i - 2j，b = -i + 4j，求 |a - 2b| 以及沿 a + b 方向的单位向量。

Compute a - 2b = (3i - 2j) - 2(-i + 4j) = 3i - 2j + 2i - 8j = 5i - 10j.

计算 a - 2b = (3i - 2j) - 2(-i + 4j) = 3i - 2j + 2i - 8j = 5i - 10j。

Magnitude: |a - 2b| = √(5² + (-10)²) = √25 + 100 = √125 = 5√5.

模：|a - 2b| = √(5² + (-10)²) = √25 + 100 = √125 = 5√5。

Next, a + b = (3i - 2j) + (-i + 4j) = 2i + 2j. Its magnitude is √2² + 2² = √8 = 2√2.

接着，a + b = (3i - 2j) + (-i + 4j) = 2i + 2j。其模为 √2² + 2² = √8 = 2√2。

The unit vector is (a + b) / |a + b| = (2i + 2j) / (2√2) = (i + j)/√2. Rationalising the denominator is optional.

单位向量为 (a + b) / |a + b| = (2i + 2j) / (2√2) = (i + j)/√2。分母有理化可视情况而定。

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9. Differentiation and Tangent | 微分与切线

Question: : Find the gradient of the curve y = 4x³ - 3/x + 2 at x = 2, and determine the equation of the tangent at that point.

题目：求曲线 y = 4x³ - 3/x + 2 在 x = 2 处的梯度，并求出该点处的切线方程。

Rewrite the function using negative exponents: y = 4x³ - 3x⁻¹ + 2.

用负指数重写函数：y = 4x³ - 3x⁻¹ + 2。

Differentiate: dy/dx = 12x² + 3x⁻². The derivative of -3x⁻¹ is -3(-1)x⁻² = +3x⁻².

求导：dy/dx = 12x² + 3x⁻²。-3x⁻¹ 的导数为 -3(-1)x⁻² = +3x⁻²。

At x = 2, gradient m = 12(2)² + 3(2)⁻² = 12(4) + 3/4 = 48 + 3/4 = 195/4 (= 48.75).

在 x = 2 处，梯度 m = 12(2)² + 3(2)⁻² = 12(4) + 3/4 = 48 + 3/4 = 195/4（或 48.75）。

Find the y-coordinate: y = 4(8) - 3/2 + 2 = 32 - 1.5 + 2 = 32.5 = 65/2.

求 y 坐标：y = 4(8) - 3/2 + 2 = 32 - 1.5 + 2 = 32.5 = 65/2。

o clear denominators: 4y - 130 = 195(x - 2) → 4y - 130 = 195x - 390 → 195x - 4y - 260 = 0.

利用点斜式求切线：y - 65/2 = (195/4)(x - 2)。乘以 4 去分母：4y - 130 = 195(x - 2) → 4y - 130 = 195x - 390 → 195x - 4y - 260 = 0。

This is the tangent equation. Always give the final answer in its simplest integer-coefficient form.

此即切线方程。最终答案一般化简为系数为整数的形式。

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10. Permutations with Restrictions | 限制排列

Question: In how many ways can 5 different books be arranged on a shelf if two particular books must not be placed next to each other?

题目：5 本不同的书排列在书架上，若其中两本特定的书不能相邻，一共有多少种排列方式？

Total arrangements without any restriction: 5! = 120.

无任何限制的总排列数为 5! = 120。

re are now 4 items to arrange: the block and the other 3 books. These can be arranged in 4! ways. Within the block, the two books can be ordered in 2! = 2 ways. So number with the two together = 4! × 2! = 24 × 2 = 48.

计算两本书相邻的排列数：将它们视为一个整体块。现在共有 4 个物体（块 + 其他 3 本书），排列方式有 4! 种。在块内，两本书可互换顺序，有 2! = 2 种方式。因此相邻的排列数为 4! × 2! = 24 × 2 = 48。

nce, arrangements where the two books are not together = total - together = 120 - 48 = 72.

因此，两本书不相邻的排列数 = 总排列数 - 相邻排列数 = 120 - 48 = 72。

This subtraction method is often simpler than directly positioning the separated books. Always check for double counting in these types of combinatorial problems.

这种相减的方法通常比直接放置不相邻的书更简单。在这类组合问题中，务必检查是否有重复计算。

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