OCR Further Maths Unit Test Mock Exam Solutions | OCR 进阶数学单元测试模拟卷解析

📚 OCR Further Maths Unit Test Mock Exam Solutions | OCR 进阶数学单元测试模拟卷解析

This article provides detailed, step-by-step solutions to a mock unit test designed for Year 11 OCR Further Maths. The selected problems cover core topics tested at this level, including quadratic discriminants, the factor theorem, binomial expansion, function transformations, differentiation and tangents, integration and area, exponential equations, and vectors with matrices. Each question is presented with a full explanation in English, immediately followed by its Chinese counterpart, helping students consolidate their understanding and exam technique.

本文为 Year 11 OCR 进阶数学单元模拟卷提供详细的逐题解析。所选题目涵盖了该阶段常考的核心知识点，如二次判别式、因式定理、二项式展开、函数变换、微分与切线、积分与面积、指数方程以及向量与矩阵。每道题目先给出英文解答，紧接着提供中文翻译，帮助你巩固理解并提升应试技巧。

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1. Quadratic Discriminant | 二次判别式

Question: has two distinct real roots. State the value of k for which the equation has a repeated root.

题目：已知二次函数 f(x) = 2x² − 5x + k，求使得方程 f(x) = 0 有两个不等实根的 k 的取值范围，并写出方程有重根时 k 的值。

ax² + bx + c = 0, the nature of the roots is determined by the discriminant Δ = b² − 4ac.

对于二次方程 ax² + bx + c = 0，根的性质由判别式 Δ = b² − 4ac 决定。

Here a = 2, b = −5, c = k. So Δ = (−5)² − 4(2)(k) = 25 − 8k.

此处 a = 2，b = −5，c = k。因此 Δ = (−5)² − 4×2×k = 25 − 8k。

For two distinct real roots we require Δ > 0 → 25 − 8k > 0 → k < 25/8.

有两个不等实根需要 Δ > 0 → 25 − 8k > 0 → k < 25/8。

A repeated root occurs when Δ = 0 → 25 − 8k = 0 → k = 25/8.

重根出现在 Δ = 0 时 → 25 − 8k = 0 → k = 25/8。

Thus the set of values is k < 25/8, and the repeated root happens at k = 25/8.

因此取值范围为 k < 25/8，重根对应的 k = 25/8。

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2. Polynomials and the Factor Theorem | 多项式与因式定理

Question: ives a remainder of 3 when divided by (x − 2), and a remainder of −9 when divided by (x + 1). Determine the values of a and b.

题目：多项式 P(x) = 2x³ − 5x² + ax + b 除以 (x − 2) 时余数为 3，除以 (x + 1) 时余数为 −9。求 a 和 b 的值。

By the Remainder Theorem, P(2) = 3 and P(−1) = −9.

根据余数定理，P(2) = 3，P(−1) = −9。

+ 2a + b = 2a + b − 4. Setting equal to 3 gives: 2a + b − 4 = 3 → 2a + b = 7 (Equation 1).

首先，P(2) = 2×2³ − 5×2² + a×2 + b = 16 − 20 + 2a + b = 2a + b − 4。令其等于 3 得：2a + b − 4 = 3 → 2a + b = 7（方程 1）。

− 5 − a + b = −a + b − 7. Setting equal to −9: −a + b − 7 = −9 → −a + b = −2 (Equation 2).

接着，P(−1) = 2×(−1)³ − 5×(−1)² + a×(−1) + b = −2 − 5 − a + b = −a + b − 7。令其等于 −9：−a + b − 7 = −9 → −a + b = −2（方程 2）。

−a + b) = 7 − (−2) → 3a = 9 → a = 3. Substitute back into Equation 2: −3 + b = −2 → b = 1.

联立求解：方程 1 减方程 2 得 (2a + b) − (−a + b) = 7 − (−2) → 3a = 9 → a = 3。代回方程 2：−3 + b = −2 → b = 1。

Hence a = 3, b = 1.

因此 a = 3，b = 1。

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3. Binomial Expansion | 二项式展开

Question: Expand (2 − 3x)⁴ in ascending powers of x up to the term in x³, and hence find the coefficient of x² in the full expansion.

题目：将 (2 − 3x)⁴ 按 x 的升幂展开至 x³ 项，并写出完全展开式中 x² 的系数。

Using the binomial theorem: (a + b)ⁿ = Σ [from r=0 to n] ⁿCᵣ aⁿ⁻ʳ bʳ. Here a = 2, b = −3x, n = 4.

使用二项式定理：(a + b)ⁿ = Σ [r=0 到 n] ⁿCᵣ aⁿ⁻ʳ bʳ。此处 a = 2，b = −3x，n = 4。

We compute the terms: r = 0: ⁴C₀·2⁴·(−3x)⁰ = 1·16·1 = 16.

我们计算各项：r = 0：⁴C₀·2⁴·(−3x)⁰ = 1·16·1 = 16。

r = 1: ⁴C₁·2³·(−3x)¹ = 4·8·(−3x) = −96x.

r = 1：⁴C₁·2³·(−3x)¹ = 4·8·(−3x) = −96x。

r = 2: ⁴C₂·2²·(−3x)² = 6·4·9x² = 216x².

r = 2：⁴C₂·2²·(−3x)² = 6·4·9x² = 216x²。

r = 3: ⁴C₃·2¹·(−3x)³ = 4·2·(−27x³) = −216x³.

r = 3：⁴C₃·2¹·(−3x)³ = 4·2·(−27x³) = −216x³。

r = 4: ⁴C₄·2⁰·(−3x)⁴ = 1·1·81x⁴ = 81x⁴.

r = 4：⁴C₄·2⁰·(−3x)⁴ = 1·1·81x⁴ = 81x⁴。

The expansion is 16 − 96x + 216x² − 216x³ + 81x⁴. The coefficient of x² is 216.

展开式为 16 − 96x + 216x² − 216x³ + 81x⁴。x² 的系数为 216。

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4. Function Transformations | 函数变换

Question: tate the equations of any asymptotes and the domain and range of the transformed function.

题目：描述将曲线 y = 1/x 变换为 y = 1/(x − 2) + 3 的变换顺序。写出渐近线的方程以及变换后的函数的定义域和值域。

replacing x with (x − 2) gives y = 1/(x − 2). This is a translation 2 units to the right.

从 y = 1/x 出发，将 x 用 (x − 2) 代替得到 y = 1/(x − 2)。这是向右平移 2 个单位。

Then adding 3 to the function gives y = 1/(x − 2) + 3, which is a translation 3 units up.

再对函数加上 3 得到 y = 1/(x − 2) + 3，这是向上平移 3 个单位。

asymptote shifts from x = 0 to x = 2. The horizontal asymptote shifts from y = 0 to y = 3.

垂直渐近线从 x = 0 移动到 x = 2。水平渐近线从 y = 0 移动到 y = 3。

Domain: x ∈ ℝ, x ≠ 2. Range: y ∈ ℝ, y ≠ 3.

定义域：x ∈ ℝ，x ≠ 2。值域：y ∈ ℝ，y ≠ 3。

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5. Differentiation and Tangents | 微分与切线方程

Question: Find the equation of the tangent to the curve y = x³ − 3x² + 2 at the point where x = 2.

题目：求曲线 y = x³ − 3x² + 2 在 x = 2 处的切线方程。

First, find the derivative: dy/dx = 3x² − 6x.

首先，求导数：dy/dx = 3x² − 6x。

At x = 2, the gradient m = 3(2²) − 6(2) = 12 − 12 = 0.

在 x = 2 处，斜率 m = 3×2² − 6×2 = 12 − 12 = 0。

The y-coordinate when x = 2: y = (2)³ − 3(2)² + 2 = 8 − 12 + 2 = −2.

当 x = 2 时的 y 坐标：y = 2³ − 3×2² + 2 = 8 − 12 + 2 = −2。

So the point of tangency is (2, −2). Since the gradient is zero, the tangent is horizontal.

因此切点为 (2, −2)。由于斜率为零，切线是水平的。

Equation of tangent: y = −2.

切线方程为 y = −2。

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6. Integration and Area | 积分与面积

Question: e the definite integral ∫₀² (4x³ − 6x + 1) dx, and interpret its meaning in terms of area.

题目：求不定积分 ∫ (4x³ − 6x + 1) dx。然后计算定积分 ∫₀² (4x³ − 6x + 1) dx，并解释其在面积上的意义。

Integrate term by term: ∫ 4x³ dx = x⁴, ∫ −6x dx = −3x², ∫ 1 dx = x. Don't forget the constant of integration C.

逐项积分：∫ 4x³ dx = x⁴，∫ −6x dx = −3x²，∫ 1 dx = x。不要忘记积分常数 C。

Thus ∫ (4x³ − 6x + 1) dx = x⁴ − 3x² + x + C.

因此 ∫ (4x³ − 6x + 1) dx = x⁴ − 3x² + x + C。

For the definite integral from 0 to 2: [x⁴ − 3x² + x]₀² = (16 − 12 + 2) − (0) = 6.

对于从 0 到 2 的定积分：[x⁴ − 3x² + x]₀² = (16 − 12 + 2) − (0) = 6。

y cross the axis, it accounts for regions above and below, but here the overall net area is 6 square units.

该值表示曲线与 x 轴之间从 x = 0 到 x = 2 的净符号面积。尽管函数可能穿过 x 轴，但总体净面积为 6 平方单位。

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7. Exponential Equations | 指数方程

Question: Solve the equation 2^x+1 = 8^x−2.

题目：解方程 2^x+1 = 8^x−2。

Express both sides with the same base. Since 8 = 2³, the right-hand side becomes (2³)^x−2 = 2^3x−6.

将两边化为同底。因为 8 = 2³，右边化为 (2³)^x−2 = 2^3x−6。

Now the equation is 2^x+1 = 2^3x−6.

现在方程为 2^x+1 = 2^3x−6。

Since the bases are equal, the exponents must be equal: x + 1 = 3x − 6.

由于底数相同，指数必须相等：x + 1 = 3x − 6。

Solve: 1 + 6 = 3x − x → 7 = 2x → x = 7/2.

求解：1 + 6 = 3x − x → 7 = 2x → x = 7/2。

Check: LHS 2^7/2+1 = 2^9/2, RHS 8^7/2−2=8^3/2=(2³)^3/2=2^9/2. Valid solution.

检验：左边 2^7/2+1 = 2^9/2，右边 8^7/2−2 = 8^3/2 = (2³)^3/2 = 2^9/2。解成立。

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8. Vectors and Matrices | 向量与矩阵

Question: Given vectors a = 3i − 2j and b = −i + 4j, find (i) 2a − 3b, (ii) |a + b|. Let matrix M = [[2, 3], [1, 4]]. Calculate M² and the inverse M⁻¹, if it exists.

题目：已知向量 a = 3i − 2j，b = −i + 4j，求 (i) 2a − 3b，(ii) |a + b|。设矩阵 M = [[2, 3], [1, 4]]，计算 M² 以及逆矩阵 M⁻¹（若存在）。

Vectors: 2a = 6i − 4j, 3b = −3i + 12j. So 2a − 3b = (6i − 4j) − (−3i + 12j) = 9i − 16j.

向量：2a = 6i − 4j，3b = −3i + 12j。因此 2a − 3b = (6i − 4j) − (−3i + 12j) = 9i − 16j。

a + b = (3 − 1)i + (−2 + 4)j = 2i + 2j. Its magnitude |a+b| = √2² + 2² = √8 = 2√2.

a + b = (3 − 1)i + (−2 + 4)j = 2i + 2j。其模长 |a+b| = √2² + 2² = √8 = 2√2。

Matrices: 3],[1,4]] × [[2,3],[1,4]] = [[2·2+3·1, 2·3+3·4], [1·2+4·1, 1·3+4·4]] = [[7, 18], [6, 19]].

矩阵： ,3],[1,4]] × [[2,3],[1,4]] = [[2·2+3·1, 2·3+3·4], [1·2+4·1, 1·3+4·4]] = [[7, 18], [6, 19]]。

Determinant of M: det(M) = 2·4 − 3·1 = 8 − 3 = 5 ≠ 0, so inverse exists.

M 的行列式：det(M) = 2·4 − 3·1 = 8 − 3 = 5 ≠ 0，故逆矩阵存在。

M⁻¹ = 1/det(M) × [[4, −3], [−1, 2]] = 1/5 [[4, −3], [−1, 2]] = [[4/5, −3/5], [−1/5, 2/5]].

M⁻¹ = 1/det(M) × [[4, −3], [−1, 2]] = 1/5 [[4, −3], [−1, 2]] = [[4/5, −3/5], [−1/5, 2/5]]。

You can verify M·M⁻¹ = I.

你可验证 M·M⁻¹ = I。

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