📚 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。
1. Discriminant and Nature of Roots | 判别式与根的性质
Question: alues 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。
, 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)。
oots, 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}。
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。
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。
te 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。
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。
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。
ythagorean 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。
ing 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。
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。
. 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。
on: 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)。
). 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。
5. Definite Integration | 定积分计算
Question: Evaluate ∫₀² (6x² - 2x + 3) dx.
题目：计算 ∫₀² (6x² - 2x + 3) dx。
-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。
= 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。
8. 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。
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 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。
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。
7. Factor Theorem and Polynomial Factorisation | 因式定理与多项式因式分解
Question: 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。
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。
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)。
8. Intersection of a Line and a Circle | 直线与圆的交点
Question: he : 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。
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。
- 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。
-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)。
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°。
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。
o 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。
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°。
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。
. 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。
: 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)。
, -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。
11. Coordinate Geometry: Perpendicular Lines | 坐标几何：垂直直线
Question: t (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。
ndicular 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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