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

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

This mock paper analysis breaks down a typical unit test for GCSE OCR Further Mathematics, covering core topics such as algebra, functions, matrices, calculus, vectors, and probability. Each section walks through a model answer and highlights common pitfalls to reinforce key exam skills.

本模拟卷解析针对 GCSE OCR 进阶数学的典型单元测试，覆盖代数、函数、矩阵、微积分、向量和概率等核心主题。每个部分都逐步展示标准答案，并指出常见易错点，以强化关键应试技能。

1. Exam Structure and Tips | 试卷结构与应试策略

The unit test usually lasts 60 minutes and includes short-answer and structured questions worth 50–60 marks. A non-calculator section tests fluency in algebra and number, while a calculator section allows complex computations and graphing.

单元测试通常持续 60 分钟，包含简答题与结构化题目，总分 50–60 分。非计算器部分考查代数与数的熟练度，计算器部分则允许进行复杂计算和作图。

Read each question carefully, show all working, and allocate time based on marks. For proof-style questions, logical steps with notation are essential. Always check answers where possible, especially for matrix multiplication or differentiation.

仔细阅读每一道题，展示全部解题步骤，并根据分值分配时间。对于证明类问题，带有符号的逻辑步骤至关重要。在可能的情况下务必检查答案，尤其是矩阵乘法或求导运算。

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2. Solving a Quadratic Inequality | 解二次不等式

Question: Find the set of values of x for which x² − 4x − 5 ≤ 0.

题目：求满足 x² − 4x − 5 ≤ 0 的 x 的取值范围。

Factorise the quadratic to obtain (x − 5)(x + 1) ≤ 0.

将二次式因式分解为 (x − 5)(x + 1) ≤ 0。

The critical values are x = 5 and x = −1. Sketch the graph of y = (x − 5)(x + 1): a positive parabola crossing the x-axis at −1 and 5.

临界值为 x = 5 和 x = −1。画出 y = (x − 5)(x + 1) 的图像：一条开口向上的抛物线，与 x 轴交于 −1 和 5。

The inequality is ≤ 0, so we take the interval where the curve is on or below the x-axis. This occurs between the roots.

不等式要求 ≤ 0，因此我们取曲线位于 x 轴上或下方的区间，即两个根之间。

Hence the solution set is −1 ≤ x ≤ 5.

因此解集为 −1 ≤ x ≤ 5。

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3. Differentiation by First Principles | 从第一原理求导

Question: Using first principles, find the derivative of f(x) = x² + 3x.

题目：使用第一原理求 f(x) = x² + 3x 的导数。

First principles formula: f'(x) = lim_h→0 [f(x+h) − f(x)] / h.

第一原理公式：f'(x) = lim_h→0 [f(x+h) − f(x)] / h。

Calculate f(x+h) = (x+h)² + 3(x+h) = x² + 2xh + h² + 3x + 3h.

计算 f(x+h) = (x+h)² + 3(x+h) = x² + 2xh + h² + 3x + 3h。

Subtract f(x) = x² + 3x to get 2xh + h² + 3h.

减去 f(x) = x² + 3x 得到 2xh + h² + 3h。

Divide by h: (2xh + h² + 3h) / h = 2x + h + 3.

除以 h：(2xh + h² + 3h) / h = 2x + h + 3。

Take the limit as h → 0, the term h vanishes, leaving f'(x) = 2x + 3.

令 h → 0 取极限，含 h 的项消失，得到 f'(x) = 2x + 3。

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4. Finding the Inverse of a 2×2 Matrix | 求 2×2 逆矩阵

Question: Given matrix A =

2	3
1	4

, find A⁻¹ and use it to solve 2x + 3y = 5, x + 4y = 6.

题目：已知矩阵 A =

2	3
1	4

，求 A⁻¹ 并用它解方程组 2x + 3y = 5, x + 4y = 6。

Determinant det(A) = (2)(4) − (3)(1) = 8 − 3 = 5.

行列式 det(A) = (2)(4) − (3)(1) = 8 − 3 = 5。

For a matrix

a	b
c	d

, the inverse is (1/det)

d	−b
−c	a

. Therefore A⁻¹ = 1/5

4	−3
−1	2

=

4/5	−3/5
−1/5	2/5

.

对矩阵

a	b
c	d

，其逆矩阵为 (1/det)

d	−b
−c	a

。因此 A⁻¹ = 1/5

4	−3
−1	2

=

4/5	−3/5
−1/5	2/5

。

Write the system as A

x

y

=

5

6

. Multiply both sides by A⁻¹:

x

y

= A⁻¹

5

6

.

将方程组写为 A

x

y

=

5

6

。两边左乘 A⁻¹ 得：

x

y

= A⁻¹

5

6

。

Compute: x = (4/5)×5 + (−3/5)×6 = 4 − 18/5 = 2/5; y = (−1/5)×5 + (2/5)×6 = −1 + 12/5 = 7/5.

计算得：x = (4/5)×5 + (−3/5)×6 = 4 − 18/5 = 2/5；y = (−1/5)×5 + (2/5)×6 = −1 + 12/5 = 7/5。

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5. Trigonometric Equation in a Given Interval | 给定区间内的三角方程

Question: Solve sin 2θ = 0.5 for 0° ≤ θ ≤ 360°.

题目：在 0° ≤ θ ≤ 360° 范围内解方程 sin 2θ = 0.5。

Let 2θ = α, then sin α = 0.5. The principal solutions for α are 30° and 180°−30° = 150°.

令 2θ = α，则 sin α = 0.5。α 的基本解为 30° 和 180°−30° = 150°。

Because θ is up to 360°, α ranges up to 720°. Add multiples of 360°: α = 30°, 150°, 390°, 510°.

因为 θ 最大 360°，α 可到 720°。加 360° 的整数倍得 α = 30°, 150°, 390°, 510°。

Divide by 2 to find θ: 15°, 75°, 195°, 255°.

除以 2 得到 θ：15°, 75°, 195°, 255°。

Always verify each angle lies within the required interval – all four are valid here.

务必验证每个角是否在指定区间内——此处四个角均有效。

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6. Binomial Expansion of (1+ax)ⁿ | 二项式展开 (1+ax)ⁿ

Question: Expand (1 + 2x)⁵ in ascending powers of x up to the term in x³.

题目：将 (1 + 2x)⁵ 按 x 的升幂展开，写到 x³ 项。

Use the binomial expansion: (1 + u)ⁿ = 1 + n u + [n(n−1)/2!] u² + [n(n−1)(n−2)/3!] u³ + …, with u = 2x.

使用二项式展开公式：(1 + u)ⁿ = 1 + n u + [n(n−1)/2!] u² + [n(n−1)(n−2)/3!] u³ + …，其中 u = 2x。

Here n = 5. Compute term by term: T₁ = 1, T₂ = 5×(2x) = 10x.

这里 n = 5。逐项计算：T₁ = 1，T₂ = 5×(2x) = 10x。

T₃ = [5×4/2] × (2x)² = 10 × 4x² = 40x².

T₃ = [5×4/2] × (2x)² = 10 × 4x² = 40x²。

T₄ = [5×4×3/6] × (2x)³ = 10 × 8x³ = 80x³.

T₄ = [5×4×3/6] × (2x)³ = 10 × 8x³ = 80x³。

Therefore the expansion is 1 + 10x + 40x² + 80x³.

因此展开式为 1 + 10x + 40x² + 80x³。

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7. Vector Proof of Collinearity | 共线向量证明

Question: Points A(1,2), B(4,8), C(7,14) are given. Prove that A, B and C are collinear.

题目：给定点 A(1,2), B(4,8), C(7,14)。证明 A, B, C 三点共线。

Find vectors AB and AC: AB = (4−1, 8−2) = (3, 6); AC = (7−1, 14−2) = (6, 12).

求向量 AB 和 AC：AB = (4−1, 8−2) = (3, 6)；AC = (7−1, 14−2) = (6, 12)。

Observe that AC = 2 × AB, meaning they are parallel and share point A.

观察到 AC = 2 × AB，这意味着两向量平行且拥有公共点 A。

Since two parallel vectors emanating from the same point lie on the same straight line, A, B, C are collinear.

由于从同一点出发的两平行向量位于同一直线上，因此 A, B, C 三点共线。

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8. Composite and Inverse Functions | 复合函数与反函数

Question: Given f(x) = 2x + 1 and g(x) = x², find (a) fg(x) and (b) f⁻¹(x).

题目：已知 f(x) = 2x + 1，g(x) = x²，求 (a) fg(x) 和 (b) f⁻¹(x)。

For fg(x), substitute g(x) into f: fg(x) = f(x²) = 2(x²) + 1 = 2x² + 1.

对于 fg(x)，将 g(x) 代入 f：fg(x) = f(x²) = 2(x²) + 1 = 2x² + 1。

To find the inverse f⁻¹(x), let y = 2x + 1. Swap x and y: x = 2y + 1, then solve for y.

为求反函数 f⁻¹(x)，令 y = 2x + 1。交换 x 与 y 得 x = 2y + 1，然后解出 y。

y = (x − 1)/2, thus f⁻¹(x) = (x − 1)/2.

y = (x − 1)/2，因此 f⁻¹(x) = (x − 1)/2。

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9. Applying Circle Theorem to Find Angles | 应用圆定理求角度

Question: O is the centre of a circle. Points A, B and C lie on the circumference. ∠AOB = 110°. Find ∠ACB.

题目：O 是圆心，A、B、C 均在圆周上。∠AOB = 110°，求 ∠ACB。

The angle at the centre is twice any angle subtended at the circumference by the same arc. Thus ∠AOB = 2 × ∠ACB.

圆心角等于相同圆弧所对圆周角的两倍。因此 ∠AOB = 2 × ∠ACB。

So ∠ACB = 110° / 2 = 55°.

所以 ∠ACB = 110° / 2 = 55°。

Always state the theorem clearly: ‘The angle at the centre is twice the angle at the circumference when both stand on the same arc.’

务必清晰表述定理：“当圆心角与圆周角对同一条弧时，圆心角是圆周角的两倍。”

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10. Conditional Probability with Tree Diagrams | 条件概率与树状图

Question: A bag contains 3 red and 5 blue balls. Two balls are drawn without replacement. Find the probability that the second ball is blue given that the first ball was red.

题目：袋中有 3 个红球和 5 个蓝球，每次取一球不放回。已知第一次取到红球，求第二次取到蓝球的概率。

After drawing a red ball first, the bag has 2 red and 5 blue left, total 7 balls.

第一次取出红球后，袋中剩余 2 红 5 蓝，共 7 个球。

The desired probability is simply P(second blue | first red) = number of blue left / total left = 5/7.

所求概率即为 P(第二次蓝 | 第一次红) = 剩余蓝球数 / 剩余总数 = 5/7。

A tree diagram would show the first branch P(Red)=3/8, then from Red node a branch to Blue with probability 5/7, confirming the calculation.

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