Year 12 OCR Maths Unit Test Mock Paper Walkthrough | Year 12 OCR 数学单元测试模拟卷解析

📚 Year 12 OCR Maths Unit Test Mock Paper Walkthrough | Year 12 OCR 数学单元测试模拟卷解析

This walkthrough article takes you through a complete mock unit test designed for Year 12 students following the OCR A-Level Mathematics specification. The paper covers key Pure Mathematics topics typically assessed in the first-year unit tests, including quadratics, coordinate geometry, trigonometry, differentiation, integration, exponentials, polynomials, and applications of calculus. Each question is presented with a full, step-by-step solution that mirrors the reasoning expected in an actual OCR examination. Use this resource to identify gaps in your understanding, sharpen your problem-solving technique, and build confidence ahead of your real assessment.

本文为遵循 OCR A-Level 数学大纲的 Year 12 学生提供一份完整的单元测试模拟卷解析。试卷涵盖了通常在一年级单元测试中考查的核心纯数学主题，包括二次方程、坐标几何、三角学、微分、积分、指数函数、多项式以及微积分应用。每道题目都配有完整的分步解答，呈现了真实 OCR 考试中所期望的推理过程。请利用这份资源来发现理解上的薄弱点、打磨解题技巧，并在真正考试前建立信心。

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1. Discriminant and Nature of Roots | 判别式与根的性质

Question: Find the range of values of k for which the equation x² + 2kx + (k² + k – 2) = 0 has two distinct real roots.

题目：求 k 的取值范围，使得方程 x² + 2kx + (k² + k – 2) = 0 有两个不相等的实数根。

For a quadratic equation ax² + bx + c = 0 to have two distinct real roots, the discriminant Δ = b² – 4ac must be strictly greater than zero. Here, a = 1, b = 2k, and c = k² + k – 2. Substituting these into the discriminant formula gives Δ = (2k)² – 4(1)(k² + k – 2).

对于二次方程 ax² + bx + c = 0，要有两个不相等的实数根，判别式 Δ = b² – 4ac 必须严格大于零。此处 a = 1，b = 2k，c = k² + k – 2。代入判别式公式得 Δ = (2k)² – 4(1)(k² + k – 2)。

Simplify the expression: Δ = 4k² – 4k² – 4k + 8 = -4k + 8. For two distinct real roots, we require Δ > 0, so -4k + 8 > 0. Solving this inequality gives -4k > -8, and dividing both sides by -4 (remembering to flip the inequality sign) yields k < 2.

化简表达式：Δ = 4k² – 4k² – 4k + 8 = -4k + 8。要得到两个不相等的实数根，需要 Δ > 0，即 -4k + 8 > 0。解此不等式得 -4k > -8，两边同时除以 -4（注意不等号方向改变）得 k < 2。

Therefore, the equation has two distinct real roots for all real values of k strictly less than 2. In set notation, this is {k ∈ ℝ : k < 2}.

因此，当 k 严格小于 2 时，方程有两个不相等的实数根。用集合符号表示为 {k ∈ ℝ : k < 2}。

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2. Circle Geometry: Centre and Radius | 圆的几何：圆心与半径

Question: Find the centre and radius of the circle with equation x² + y² – 6x + 4y – 12 = 0.

题目：求圆 x² + y² – 6x + 4y – 12 = 0 的圆心和半径。

The general form of a circle equation is x² + y² + 2gx + 2fy + c = 0, where the centre is (-g, -f) and the radius r satisfies r² = g² + f² – c. Comparing coefficients, we have 2g = -6, so g = -3, and 2f = 4, so f = 2. The constant term c = -12.

圆的一般方程为 x² + y² + 2gx + 2fy + c = 0，其中圆心为 (-g, -f)，半径 r 满足 r² = g² + f² – c。比较系数得 2g = -6，即 g = -3，2f = 4，即 f = 2。常数项 c = -12。

The centre coordinates are therefore (-g, -f) = (3, -2). To find the radius, compute g² + f² – c = (-3)² + 2² – (-12) = 9 + 4 + 12 = 25. Since r² = 25, the radius is r = 5.

因此圆心坐标为 (-g, -f) = (3, -2)。求半径时，计算 g² + f² – c = (-3)² + 2² – (-12) = 9 + 4 + 12 = 25。由 r² = 25 得半径 r = 5。

As an alternative method, complete the square: rewrite the equation as (x – 3)² – 9 + (y + 2)² – 4 – 12 = 0, which simplifies to (x – 3)² + (y + 2)² = 25, confirming the centre (3, -2) and radius 5.

另一种方法是配方：将方程改写为 (x – 3)² – 9 + (y + 2)² – 4 – 12 = 0，化简得 (x – 3)² + (y + 2)² = 25，确认圆心为 (3, -2)，半径为 5。

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3. Trigonometric Identity Proof | 三角恒等式证明

Question: Prove that (sin θ + cos θ)² = 1 + sin 2θ.

题目：证明 (sin θ + cos θ)² = 1 + sin 2θ。

Begin by expanding the left-hand side. The expression (sin θ + cos θ)² expands to sin²θ + 2 sin θ cos θ + cos²θ. By the fundamental Pythagorean identity, sin²θ + cos²θ = 1, so the expression simplifies to 1 + 2 sin θ cos θ.

首先展开左边。(sin θ + cos θ)² 展开得 sin²θ + 2 sin θ cos θ + cos²θ。由基本勾股恒等式 sin²θ + cos²θ = 1，表达式化简为 1 + 2 sin θ cos θ。

Recall the double-angle identity for sine: sin 2θ = 2 sin θ cos θ. Substituting this into our expression yields 1 + sin 2θ, which is exactly the right-hand side. The proof is complete, and the identity holds for all real values of θ.

回顾正弦的二倍角公式：sin 2θ = 2 sin θ cos θ。将其代入我们的表达式得到 1 + sin 2θ，这正是右边。证明完成，该恒等式对所有实数 θ 均成立。

This type of proof frequently appears on OCR papers, often as a precursor to solving a trigonometric equation. Students should be comfortable manipulating identities in both directions.

这类证明题经常出现在 OCR 试卷中，通常作为解三角方程的前置步骤。学生应熟练掌握双向运用恒等式的技巧。

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4. Differentiation and Tangent Lines | 微分与切线方程

Question: Find the equation of the tangent to the curve y = 3x² – 4x + 1 at the point where x = 2.

题目：求曲线 y = 3x² – 4x + 1 在 x = 2 处的切线方程。

First, differentiate the function to find the gradient function. Given y = 3x² – 4x + 1, the derivative is dy/dx = 6x – 4. Evaluate this at x = 2 to obtain the gradient of the tangent: m = 6(2) – 4 = 12 – 4 = 8.

首先对函数求导以获得梯度函数。已知 y = 3x² – 4x + 1，导数为 dy/dx = 6x – 4。在 x = 2 处求值得到切线的梯度：m = 6(2) – 4 = 12 – 4 = 8。

Next, find the y-coordinate of the point on the curve where x = 2. Substitute into the original equation: y = 3(2)² – 4(2) + 1 = 3(4) – 8 + 1 = 12 – 8 + 1 = 5. The point of tangency is (2, 5).

接下来求曲线上 x = 2 处的 y 坐标。代入原方程：y = 3(2)² – 4(2) + 1 = 3(4) – 8 + 1 = 12 – 8 + 1 = 5。切点坐标为 (2, 5)。

Using the point-slope form y – y₁ = m(x – x₁) with m = 8 and (x₁, y₁) = (2, 5), we obtain y – 5 = 8(x – 2). Expanding gives y – 5 = 8x – 16, and rearranging yields the final equation y = 8x – 11.

使用点斜式 y – y₁ = m(x – x₁)，代入 m = 8 和 (x₁, y₁) = (2, 5)，得 y – 5 = 8(x – 2)。展开得 y – 5 = 8x – 16，整理得最终方程 y = 8x – 11。

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5. Definite Integration | 定积分计算

Question: Evaluate ∫₀² (6x² – 2x + 3) dx.

题目：计算 ∫₀² (6x² – 2x + 3) dx。

Integrate term by term using the power rule: the integral of xⁿ is xⁿ⁺¹/(n+1) for n ≠ -1. For 6x², integrate to get 6x³/3 = 2x³. For -2x, integrate to get -2x²/2 = -x². For the constant 3, integrate to get 3x. The indefinite integral is therefore 2x³ – x² + 3x + C.

使用幂法则逐项积分：xⁿ 的积分为 xⁿ⁺¹/(n+1)（n ≠ -1）。对 6x² 积分得 6x³/3 = 2x³。对 -2x 积分得 -2x²/2 = -x²。对常数 3 积分得 3x。因此不定积分为 2x³ – x² + 3x + C。

Apply the limits of integration from 0 to 2. First, evaluate the antiderivative at the upper limit x = 2: F(2) = 2(2)³ – (2)² + 3(2) = 2(8) – 4 + 6 = 16 – 4 + 6 = 18. Then evaluate at the lower limit x = 0: F(0) = 2(0)³ – (0)² + 3(0) = 0.

代入积分上下限 0 到 2。首先计算原函数在上限 x = 2 处的值：F(2) = 2(2)³ – (2)² + 3(2) = 2(8) – 4 + 6 = 16 – 4 + 6 = 18。然后计算下限 x = 0 处的值：F(0) = 2(0)³ – (0)² + 3(0) = 0。

The value of the definite integral is F(2) – F(0) = 18 – 0 = 18. Always remember to subtract the lower limit evaluation from the upper limit evaluation.

定积分的值为 F(2) – F(0) = 18 – 0 = 18。务必记住用上限处的值减去下限处的值。

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6. Exponential Equations and Hidden Quadratics | 指数方程与隐藏二次型

Question: Solve 2e²ˣ – 5eˣ – 3 = 0, giving your answer in exact form.

题目：解方程 2e²ˣ – 5eˣ – 3 = 0，答案以精确形式给出。

Recognise that this equation is a ‘hidden quadratic’ in eˣ. Let u = eˣ, which means e²ˣ = (eˣ)² = u². Substituting transforms the equation into 2u² – 5u – 3 = 0, a standard quadratic in u.

识别出该方程是关于 eˣ 的”隐藏二次方程”。令 u = eˣ，则 e²ˣ = (eˣ)² = u²。代入后方程变为 2u² – 5u – 3 = 0，一个关于 u 的标准二次方程。

Factorise the quadratic: 2u² – 5u – 3 = (2u + 1)(u – 3) = 0. Setting each factor to zero gives two possible solutions: 2u + 1 = 0 yields u = -1/2, and u – 3 = 0 yields u = 3.

对二次式进行因式分解：2u² – 5u – 3 = (2u + 1)(u – 3) = 0。令每个因式为零得到两个可能的解：2u + 1 = 0 得 u = -1/2，u – 3 = 0 得 u = 3。

Recall that u = eˣ, and the exponential function eˣ is strictly positive for all real x. Therefore, u = -1/2 is invalid because eˣ can never be negative. The only valid solution is u = 3, which gives eˣ = 3. Taking the natural logarithm of both sides yields x = ln 3 as the exact solution.

回顾 u = eˣ，指数函数 eˣ 对所有实数 x 均为严格正值。因此 u = -1/2 无效，因为 eˣ 不可能为负。唯一有效解为 u = 3，即 eˣ = 3。两边取自然对数得精确解 x = ln 3。

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7. Factor Theorem and Polynomial Factorisation | 因式定理与多项式因式分解

Question: Given f(x) = 2x³ – 3x² – 8x + 12, show that (x – 2) is a factor and hence fully factorise f(x).

题目：已知 f(x) = 2x³ – 3x² – 8x + 12，证明 (x – 2) 是一个因式，并由此对 f(x) 进行完全因式分解。

By the Factor Theorem, (x – a) is a factor of f(x) if and only if f(a) = 0. Evaluate f(2): f(2) = 2(2)³ – 3(2)² – 8(2) + 12 = 2(8) – 3(4) – 16 + 12 = 16 – 12 – 16 + 12 = 0. Since f(2) = 0, (x – 2) is indeed a factor.

根据因式定理，(x – a) 是 f(x) 的因式当且仅当 f(a) = 0。计算 f(2)：f(2) = 2(2)³ – 3(2)² – 8(2) + 12 = 2(8) – 3(4) – 16 + 12 = 16 – 12 – 16 + 12 = 0。由于 f(2) = 0，(x – 2) 确实是一个因式。

Now perform polynomial division. Dividing 2x³ – 3x² – 8x + 12 by (x – 2) yields a quadratic quotient. Using synthetic division or long division: the first term is 2x², then multiply (x – 2) by 2x² to get 2x³ – 4x², subtract to get x², bring down -8x, next term is x, multiply to get x² – 2x, subtract to get -6x, bring down 12, final term is -6, multiply to get -6x + 12, remainder 0. The quotient is 2x² + x – 6.

现在进行多项式除法。用 2x³ – 3x² – 8x + 12 除以 (x – 2) 得到一个二次商式。使用综合除法或长除法：首项为 2x²，将 (x – 2) 乘以 2x² 得 2x³ – 4x²，相减得 x²，带下 -8x，下一项为 x，相乘得 x² – 2x，相减得 -6x，带下 12，最后一项为 -6，相乘得 -6x + 12，余数为 0。商式为 2x² + x – 6。

Now factorise the quadratic: 2x² + x – 6 = (2x – 3)(x + 2). Therefore, the complete factorisation of f(x) is f(x) = (x – 2)(2x – 3)(x + 2).

现在对二次式进行因式分解：2x² + x – 6 = (2x – 3)(x + 2)。因此 f(x) 的完全因式分解为 f(x) = (x – 2)(2x – 3)(x + 2)。

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8. Intersection of a Line and a Circle | 直线与圆的交点

Question: Find the coordinates of the points where the line y = 2x – 1 intersects the circle x² + y² = 10.

题目：求直线 y = 2x – 1 与圆 x² + y² = 10 的交点坐标。

Substitute the expression for y from the line equation into the circle equation. Replace y with (2x – 1) in x² + y² = 10 to obtain x² + (2x – 1)² = 10. Expand the squared term: (2x – 1)² = 4x² – 4x + 1, so the equation becomes x² + 4x² – 4x + 1 = 10, which simplifies to 5x² – 4x + 1 = 10.

将直线方程中 y 的表达式代入圆的方程。在 x² + y² = 10 中用 (2x – 1) 替换 y，得到 x² + (2x – 1)² = 10。展开平方项：(2x – 1)² = 4x² – 4x + 1，因此方程变为 x² + 4x² – 4x + 1 = 10，化简为 5x² – 4x + 1 = 10。

Subtract 10 from both sides to set the equation to zero: 5x² – 4x – 9 = 0. Factorise this quadratic: (5x – 9)(x + 1) = 0. This gives x = 9/5 or x = -1.

两边减去 10 使方程等于零：5x² – 4x – 9 = 0。对该二次式进行因式分解：(5x – 9)(x + 1) = 0。解得 x = 9/5 或 x = -1。

Find the corresponding y-coordinates using y = 2x – 1. For x = 9/5, y = 2(9/5) – 1 = 18/5 – 5/5 = 13/5. For x = -1, y = 2(-1) – 1 = -2 – 1 = -3. The two intersection points are (9/5, 13/5) and (-1, -3).

使用 y = 2x – 1 求对应的 y 坐标。当 x = 9/5 时，y = 2(9/5) – 1 = 18/5 – 5/5 = 13/5。当 x = -1 时，y = 2(-1) – 1 = -2 – 1 = -3。两个交点坐标为 (9/5, 13/5) 和 (-1, -3)。

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9. Solving Trigonometric Equations | 解三角方程

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

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

Use the identity sin² θ = 1 – cos² θ to express everything in terms of cos θ. Substitute into the equation: 2(1 – cos² θ) – cos θ – 1 = 0. Expand to get 2 – 2 cos² θ – cos θ – 1 = 0, which simplifies to -2 cos² θ – cos θ + 1 = 0. Multiply through by -1 to obtain a more familiar form: 2 cos² θ + cos θ – 1 = 0.

使用恒等式 sin² θ = 1 – cos² θ 将所有项用 cos θ 表示。代入方程：2(1 – cos² θ) – cos θ – 1 = 0。展开得 2 – 2 cos² θ – cos θ – 1 = 0，化简为 -2 cos² θ – cos θ + 1 = 0。两边乘以 -1 得到更熟悉的形式：2 cos² θ + cos θ – 1 = 0。

Factorise the quadratic in cos θ: 2 cos² θ + cos θ – 1 = (2 cos θ – 1)(cos θ + 1) = 0. Setting each factor to zero gives two cases: 2 cos θ – 1 = 0, so cos θ = 1/2, and cos θ + 1 = 0, so cos θ = -1.

对关于 cos θ 的二次式进行因式分解：2 cos² θ + cos θ – 1 = (2 cos θ – 1)(cos θ + 1) = 0。令每个因式为零得到两种情况：2 cos θ – 1 = 0，即 cos θ = 1/2；以及 cos θ + 1 = 0，即 cos θ = -1。

For cos θ = 1/2 in the interval 0° ≤ θ ≤ 360°, the solutions are θ = 60° and θ = 300° (cosine is positive in the first and fourth quadrants). For cos θ = -1, the only solution in the given interval is θ = 180°. Thus, the complete solution set is θ = 60°, 180°, 300°.

对于 cos θ = 1/2 在区间 0° ≤ θ ≤ 360° 内，解为 θ = 60° 和 θ = 300°（余弦在第一和第四象限为正）。对于 cos θ = -1，给定区间内的唯一解为 θ = 180°。因此完整解集为 θ = 60°、180°、300°。

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10. Stationary Points and Their Nature | 驻点及其性质判断

Question: Find the stationary points of the curve y = x³ – 3x² – 9x + 2 and determine their nature.

题目：求曲线 y = x³ – 3x² – 9x + 2 的驻点并判断其性质。

Stationary points occur where the first derivative equals zero. Differentiate the function: dy/dx = 3x² – 6x – 9. Set this equal to zero: 3x² – 6x – 9 = 0. Divide through by 3 to simplify: x² – 2x – 3 = 0. Factorise to get (x – 3)(x + 1) = 0, giving x = 3 and x = -1 as the x-coordinates of the stationary points.

驻点出现在一阶导数为零处。对函数求导：dy/dx = 3x² – 6x – 9。令其为零：3x² – 6x – 9 = 0。除以 3 化简：x² – 2x – 3 = 0。因式分解得 (x – 3)(x + 1) = 0，得到驻点的 x 坐标为 x = 3 和 x = -1。

Find the corresponding y-coordinates by substituting back into the original equation. For x = 3: y = (3)³ – 3(3)² – 9(3) + 2 = 27 – 27 – 27 + 2 = -25. For x = -1: y = (-1)³ – 3(-1)² – 9(-1) + 2 = -1 – 3 + 9 + 2 = 7. The stationary points are (3, -25) and (-1, 7).

代回原方程求对应的 y 坐标。当 x = 3 时：y = (3)³ – 3(3)² – 9(3) + 2 = 27 – 27 – 27 + 2 = -25。当 x = -1 时：y = (-1)³ – 3(-1)² – 9(-1) + 2 = -1 – 3 + 9 + 2 = 7。驻点为 (3, -25) 和 (-1, 7)。

To determine the nature, find the second derivative: d²y/dx² = 6x – 6. Evaluate at x = 3: d²y/dx² = 6(3) – 6 = 18 – 6 = 12 > 0, indicating a local minimum at (3, -25). Evaluate at x = -1: d²y/dx² = 6(-1) – 6 = -6 – 6 = -12 < 0, indicating a local maximum at (-1, 7).

为判断性质，求二阶导数：d²y/dx² = 6x – 6。在 x = 3 处求值：d²y/dx² = 6(3) – 6 = 18 – 6 = 12 > 0，表明 (3, -25) 处为局部极小值。在 x = -1 处求值：d²y/dx² = 6(-1) – 6 = -6 – 6 = -12 < 0，表明 (-1, 7) 处为局部极大值。

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11. Coordinate Geometry: Perpendicular Lines | 坐标几何：垂直直线

Question: The line L₁ has equation 3x + 4y – 7 = 0. A second line L₂ passes through the point (2, -1) and is perpendicular to L₁. Find the equation of L₂ in the form ax + by + c = 0.

题目：直线 L₁ 的方程为 3x + 4y – 7 = 0。第二条直线 L₂ 经过点 (2, -1) 且与 L₁ 垂直。求 L₂ 的方程，结果以 ax + by + c = 0 的形式表示。

First, find the gradient of L₁. Rearrange 3x + 4y – 7 = 0 into slope-intercept form: 4y = -3x + 7, so y = -3x/4 + 7/4. The gradient of L₁ is m₁ = -3/4. For perpendicular lines, the product of their gradients is -1, so m₁ × m₂ = -1. Therefore, m₂ = -1 / m₁ = -1 / (-3/4) = 4/3.

首先求 L₁ 的梯度。将 3x + 4y – 7 = 0 转化为斜截式：4y = -3x + 7，即 y = -3x/4 + 7/4。L

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