AQA MA04 January 2021 Mark Scheme: Question Type Breakdown | AQA MA04 2021年1月评分标准解析：核心题型分析

📚 AQA MA04 January 2021 Mark Scheme: Question Type Breakdown | AQA MA04 2021年1月评分标准解析：核心题型分析

The January 2021 AQA MA04 mark scheme offers invaluable insight into how examiners allocate marks across pure and applied question types. By studying the typical mark allocations and common pitfalls, students can sharpen their exam technique and maximise scores on these high-value tasks.

2021年1月的AQA MA04评分方案为了解考官在纯数学与应用题上的给分方式提供了宝贵视角。通过学习常见的分数分配与易错点，学生可以打磨应试技巧，在这些高分题型中力争满分。

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1. Implicit Differentiation | 隐函数微分

or instance, a Jan 2021 question gave x² + 2xy + 3y² = 4 and asked for the gradient at (1, –1).

一个典型的MA04题目要求通过隐函数微分求梯度或切线方程。例如，2021年1月的一道题给出 x² + 2xy + 3y² = 4，要求求出在 (1, –1) 处的梯度。

Differentiating term‑by‑term: 2x + 2y + 2x(dy/dx) + 6y(dy/dx) = 0. Rearranging gives dy/dx = –(x + y) / (x + 3y). Substituting the point yields 0, so the tangent is horizontal (y = –1).

逐项微分得：2x + 2y + 2x(dy/dx) + 6y(dy/dx) = 0。整理得 dy/dx = –(x + y) / (x + 3y)。代入该点得0，因此切线为水平线 y = –1。

The mark scheme awards M1 for correctly applying the product rule to 2xy, A1 for collecting all dy/dx terms accurately, and B1 for substitution. A common slip is forgetting the dy/dx factor on the y‑terms, costing the A1.

评分方案中，对 2xy 正确使用乘积法则可得 M1，准确合并所有 dy/dx 项得 A1，代入坐标值得 B1。常见失误是遗漏了 y 项上的 dy/dx 因子，从而丢失 A1。

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2. Parametric Equations & Tangents | 参数方程与切线

Questions often give parametric curves such as x = t² + 1, y = t³ – 21 pack, a similar item assessed the chain rule and the ability to form the line equation.

题目常给出参数曲线，如 x = t² + 1, y = t³ – 3t，要求求某特定 t 处的切线。在2021年1月的试卷中，类似题目考查了链式法则和建立直线方程的能力。

First, compute dx/dt = 2t and dy/dt = 3t² – 3, so dy/dx = (3t² – 3) / (2t). At t = 2, the gradient is (12 – 3)/4 = 9/4. The point is (5, 2), leading to the tangent equation y – 2 = (9/4)(x – 5).

首先计算 dx/dt = 2t 和 dy/dt = 3t² – 3，则 dy/dx = (3t² – 3) / (2t)。当 t = 2 时，梯度为 (12 – 3)/4 = 9/4。对应点为 (5, 2)，切线方程为 y – 2 = (9/4)(x – 5)。

Marks: M1 for dy/dt ÷ dx/dt, A1 for simplified gradient, M1 for using the point to form the line. Many candidates lost an A mark by misreading the sign when simplifying the numerator.

分值分配：求出 dy/dt ÷ dx/dt 得 M1，化简梯度得 A1，用点求直线方程得 M1。许多考生在化简分子时因正负号出错而失去准确度分（A 分）。

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3. Integration by Substitution | 换元积分法

Substitution integrals, such as ∫ x √2x – 1 dx with u = 2x – nuary 2021 markscheme highlighted the need to convert both the integrand and dx correctly.

换元积分如 ∫ x √2x – 1 dx 并令 u = 2x – 1 是常考题型。2021年1月的评分标准强调必须正确转换被积函数和 dx。

With u = 2x – 1, we obtain x = (u + 1)/2 and du = 2 dx → dx = du/2. The integral becomes ∫ (u+1)/2 · √u · (du/2) = ¼ ∫ (u^3/2 + u^1/2) du = ¼ [ (2/5)u^5/2 + (2/3)u^3/2 ] + c.

令 u = 2x – 1，得到 x = (u + 1)/2 且 du = 2 dx → dx = du/2。积分变为 ∫ (u+1)/2 · √u · (du/2) = ¼ ∫ (u^3/2 + u^1/2) du = ¼ [ (2/5)u^5/2 + (2/3)u^3/2 ] + c。

The mark scheme awards M1 for choosing u and finding du/dx, A1 for complete substitution, and A1 for the final integrated form. The most frequent error was neglecting to express dx in terms of du, resulting in an extra factor of 2.

评分方案：选择 u 并求出 du/dx 得 M1，完全代换得 A1，最终积分式准确再得一个 A1。最频繁的错误是忘记将 dx 用 du 表出，导致多出一个因子 2。

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4. Integration by Parts | 分部积分法

Items such as ∫ x e^–x dx demand a clear choice of u and dv. The Jan 2021 script exemplified how a single sign error can cascade through the solution.

如 ∫ x e^–x dx 这类题目要求明确选择 u 与 dv。2021年1月的答卷展示了单个正负号错误如何引发连锁反应。

Set u = x → du = dx, and dv = e^–x dx → v = –e^–x. Then ∫ x e^–x dx = –x e^–x – ∫ –e^–x dx = –x e^–x – e^–x + c.

设 u = x → du = dx ，dv = e^–x dx → v = –e^–x。于是 ∫ x e^–x dx = –x e^–x – ∫ –e^–x dx = –x e^–x – e^–x + c。

nd a subsequent A1 for the final simplified expression. A typical mistake is writing v = e^–x instead of –e^–x, which flips the sign of the second term.

考官给分：正确选定 u 和 dv 得 M1，应用分部积分公式得 A1，最终化简表达式再得一个 A1。典型错误是将 v 误写为 e^–x 而非 –e^–x，导致第二项符号反转。

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5. Vectors: Angles Between Lines | 向量：直线间的夹角

acute angle between two given lines. A Jan 2021 item provided L₁: r = (i + 2j + 3k) + λ(i – j + 2k) and L₂: r = (2i – j + 4k) + μ(2i + j – k).

MA04 的向量题常要求两条给定直线的锐角。2021年1月的一道题给出 L₁: r = (i + 2j + 3k) + λ(i – j + 2k) 和 L₂: r = (2i – j + 4k) + μ(2i + j – k)。

Take direction vectors d₁ = (1, –1, 2) and d₂ = (2, 1, –1). Their dot product is 1·2 + (–1)·1 + 2·(–1) = –1. Magnitudes are |d₁| = √1+1+4 = √6, |d₂| = √4+1+1 = √6. Hence cos θ = |–1| / (√6·√6) = 1/6, giving θ ≈ 80.4°.

取方向向量 d₁ = (1, –1, 2) 与 d₂ = (2, 1, –1)。点积为 1·2 + (–1)·1 + 2·(–1) = –1。模为 |d₁| = √6，|d₂| = √6。因此 cos θ = |–1| / (√6·√6) = 1/6，得 θ ≈ 80.4°。

The mark scheme: M1 for a correct dot product, M1 for both magnitudes, A1 for the cosine value, and a final A1 for the acute angle. Candidates who omitted the absolute value in the numerator often obtained an obtuse angle, losing the final accuracy mark.

评分方案：正确点积得 M1，求得两模长得 M1，余弦值对得 A1，最终锐角准确得 A1。遗漏分子绝对值的考生常得到钝角，丢失最后的准确度分。

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6. Solving Differential Equations | 解微分方程

dx = (2y + 1)/x with the condition y = 2 when x = 1, and required the particular solution.

可分离的微分方程定期出现。2021年1月的一道题给出 dy/dx = (2y + 1)/x 且满足 x = 1 时 y = 2，要求求出特解。

Separate: ∫ 1/(2y + 1) dy = ∫ 1/x dx. Integrating yields ½ ln|2y + 1| = ln|x| + c. Using the initial condition gives c = ½ ln 5, so ln|2y + 1| = 2 ln|x| + ln 5 = ln(5x²). Hence 2y + 1 = 5x², giving y = (5x² – 1)/2.

分离变量：∫ 1/(2y + 1) dy = ∫ 1/x dx。积分得 ½ ln|2y + 1| = ln|x| + c。利用初始条件得 c = ½ ln 5，于是 ln|2y + 1| = 2 ln|x| + ln 5 = ln(5x²)，因此 2y + 1 = 5x²，得 y = (5x² – 1)/2。

Marks: M1 for separation, A1 for correct integration of both sides, B1 for finding c, and A1 for the final expression. A very common slip was forgetting the constant of integration entirely, which prevented the recovery of the particular solution.

分值：分离变量得 M1，两边积分正确各得 A1，求出 c 得 B1，最终表达式准确得 A1。十分常见的疏漏是完全忘记积分常数，导致无法求得特解。

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7. Binomial Distribution | 二项分布

robability from discrete distributions is assessed regularly. A typical MA04 item reads: X ~ nd P(X = 3). The Jan 2021 scheme required careful use of the binomial probability formula.

离散分布的概率是常考内容。典型的 MA04 题目如：X ~ B(20, 0.15)，求 P(X = 3)。2021年1月的评分方案要求谨慎使用二项概率公式。

. Candidates could use the formula directly or refer to statistical tables. The mark scheme awards M1 for the binomial coefficient, M1 for correct powers, and A1 for the final value (approx. 0.242).

P(X = 3) = C(20, 3) × (0.15)³ × (0.85)¹⁷。考生可直接用公式或查统计表。评分方案中，二项系数正确得 M1，幂次正确得 M1，最终数值（约 0.242）准确得 A1。

Examiners noted that errors often arose from mixing up p and q, or miscomputing the combination. Writing out the coefficient as 1140 and the powers step‑by‑step can help secure both method marks.

考官指出，错误常源于混淆 p 和 q，或组合数计算失误。将系数写为 1140 并逐步写出幂次，有助于确保两个方法分（M 分）。

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8. Normal Approximation to Binomial | 正态近似

When n is large, the normal approximation is tested. A Jan 2021 question might ask: X ~ B(100, 0.2), approximate P(15 < X < 30). Continuity correction is essential for method marks.

当 n 较大时，会检验正态近似。2021年1月可能这样提问：X ~ B(100, 0.2)，近似求 P(15 < X < 30)。连续性校正是获得方法分的关键。

Mean μ = 100 × 0.2 = 20

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