📚 Further Maths: ENGAA 2021 S1 Answer Key | 进阶数学：ENGAA 2021 Section 1 答案解析
The Engineering Admissions Assessment (ENGAA) Section 1 challenges candidates with fast-paced multiple-choice questions that blend core mathematics and physics. This article provides a full answer key with detailed, bilingual explanations for the 20 mathematics questions from the 2021 paper, helping you master the advanced problem-solving techniques required for success.
工程入学评估（ENGAA）第一部分通过快节奏的多选题考查核心数学与物理的综合能力。本文针对2021年试卷中的20道数学题，提供完整的答案与中英双语详细解析，帮助你掌握拿下高分所需的高级解题技巧。
1. Exponential Equations and Logarithms | 指数方程与对数
This section tests the laws of indices and the ability to rewrite exponential expressions with a common base. Candidates must also be comfortable converting between exponential and logarithmic forms.
本部分考查指数运算法则以及将指数式化为同底的能力。考生还需熟练在指数形式与对数形式之间进行转换。
Q1: Solve 2x+1 = 82x-3. Express 8 as 23, giving 2x+1 = (23)2x-3 = 26x-9. Equating exponents yields x + 1 = 6x – 9, so 5x = 10, x = 2. The correct option is D.
Q1：解方程 2x+1 = 82x-3。 将 8 写成 23，得 2x+1 = (23)2x-3 = 26x-9。令指数相等得 x + 1 = 6x – 9，故 5x = 10，x = 2。正确选项为 D。
Q2: Simplify log28 + log327 – log51. Evaluate each term: log28 = 3, log327 = 3, log51 = 0. The sum is 3 + 3 – 0 = 6. The correct option is A.
Q2：化简 log28 + log327 – log51。 分别计算：log28 = 3，log327 = 3，log51 = 0。求和得 3 + 3 – 0 = 6。正确选项为 A。
2. Quadratic and Polynomial Functions | 二次与多项式函数
ENGAA frequently includes questions on completing the square, discriminant analysis, and polynomial identities. A solid grasp of factor theorems and remainder theorems can save valuable time.
ENGAA 经常出现配方法、判别式分析以及多项式恒等式。扎实掌握因式定理与余数定理能显著节省时间。
Q3: Find the range of values of k for which x2 + kx + 9 = 0 has no real roots. The discriminant Δ = k2 – 36. For no real roots, Δ < 0, so k2 < 36, giving -6 < k < 6. The correct option is B.
Q3：求使 x2 + kx + 9 = 0 无实数根的 k 的取值范围。 判别式 Δ = k2 – 36。无实根要求 Δ < 0，即 k2 < 36，得 -6 < k < 6。正确选项为 B。
Q4: When the polynomial P(x) = 2x3 – x2 + ax + b is divided by (x – 1), the remainder is 5. When divided by (x + 2), the remainder is -10. Find a + b. By the remainder theorem: P(1) = 2 – 1 + a + b = a + b + 1 = 5, so a + b = 4. Also P(-2) = 2(-8) – 4 – 2a + b = -20 – 2a + b = -10 ⇒ b – 2a = 10. Solving, from a + b = 4, b = 4 – a, substitute: 4 – a – 2a = 10 ⇒ -3a = 6, a = -2, b = 6. Hence a + b = 4. There is no need for further solving; the first condition already gives a + b = 4. The correct option is C.
Q4：多项式 P(x) = 2x3 – x2 + ax + b 除以 (x – 1) 余数为 5，除以 (x + 2) 余数为 -10。求 a + b。 由余数定理：P(1) = 2 – 1 + a + b = a + b + 1 = 5，得 a + b = 4。第二个条件 P(-2) = -20 – 2a + b = -10 进一步验证，无需重新计算 a+b。正确选项为 C。
3. Trigonometric Equations and Identities | 三角方程与恒等式
This part demands fluency in basic identities and the ability to solve trigonometric equations within a given interval. Recognizing symmetries on the unit circle is often the key to finding all solutions quickly.
本部分要求熟练掌握基本恒等式，并能在指定区间内解三角方程。识别单位圆上的对称性是快速找到所有解的关键。
Q5: Solve 2 sin2θ – cos θ = 1 for 0 ≤ θ ≤ 360°. Replace sin2θ with 1 – cos2θ: 2(1 – cos2θ) – cos θ = 1 ⇒ 2 – 2 cos2θ – cos θ = 1 ⇒ 2 cos2θ + cos θ – 1 = 0. Factorise: (2 cos θ – 1)(cos θ + 1) = 0. Thus cos θ = ½ or cos θ = -1. Solutions: cos θ = ½ ⇒ θ = 60°, 300°; cos θ = -1 ⇒ θ = 180°. The three solutions match option E.
Q5：解 2 sin2θ – cos θ = 1，其中 0 ≤ θ ≤ 360°。 将 sin2θ 代换为 1 – cos2θ：2(1 – cos2θ) – cos θ = 1 ⇒ 2 – 2 cos2θ – cos θ = 1 ⇒ 2 cos2θ + cos θ – 1 = 0。因式分解得 (2 cos θ – 1)(cos θ + 1) = 0。因此 cos θ = ½ 或 cos θ = -1。解为：cos θ = ½ ⇒ θ = 60°, 300°；cos θ = -1 ⇒ θ = 180°。三个解对应选项 E。
Q6: Simplify (sin x + cos x)2 + (sin x – cos x)2. Expand: sin2x + 2 sin x cos x + cos2x + sin2x – 2 sin x cos x + cos2x = 2(sin2x + cos2x) = 2(1) = 2. The expression is independent of x. Correct option is B.
Q6：化简 (sin x + cos x)2 + (sin x – cos x)2。 展开：sin2x + 2 sin x cos x + cos2x + sin2x – 2 sin x cos x + cos2x = 2(sin2x + cos2x) = 2。该表达式与 x 无关。正确选项为 B。
4. Coordinate Geometry and Graphs | 坐标几何与图像
Questions here involve straight-line properties, conic sections, and graph transformations. Careful reading of intercepts, gradients, and asymptotes is vital.
这部分题目涉及直线性质、圆锥曲线以及图像变换。准确解读截距、斜率和渐近线至关重要。
Q7: A line passes through (2,5) and has gradient -½. Find its y-intercept. Equation: y – 5 = -½(x – 2) ⇒ y = -½x + 1 + 5 = -½x + 6. The y-intercept is 6. Correct option is D.
Q7：一直线过点 (2,5) 且斜率为 -½，求其 y 轴截距。 方程为 y – 5 = -½(x – 2) ⇒ y = -½x + 1 + 5 = -½x + 6。y 轴截距为 6。正确选项为 D。
Q8: The graph of y = 1/(x – 3)2 + 2 is obtained from y = 1/x2 by which transformations? The expression replaces x with (x – 3), shifting the graph 3 units to the right, and then adds 2, shifting it 2 units up. The asymptotes move to x = 3 and y = 2. The correct option is A.
Q8：y = 1/(x – 3)2 + 2 的图像可由 y = 1/x2 经过哪些变换得到？ 解析式将 x 替换为 (x – 3)，图像向右平移 3 个单位，再加上 2，向上平移 2 个单位。渐近线移至 x = 3 和 y = 2。正确选项为 A。
5. Differentiation Techniques and Applications | 微分技巧及其应用
Calculus questions in ENGAA test not only standard derivatives but also the ability to find stationary points, gradients of tangents, and rates of change from first principles.
ENGAA 的微积分题目不仅考标准导数，还考查求驻点、切线斜率以及用第一原理处理变化率的能力。
Q9: Differentiate y = e2x ln(3x). Use the product rule: dy/dx = 2e2x ln(3x) + e2x · (1/x) = e2x (2 ln(3x) + 1/x). The simplest form matches option C.
Q9：求导 y = e2x ln(3x)。 使用乘法法则：dy/dx = 2e2x ln(3x) + e2x · (1/x) = e2x (2 ln(3x) + 1/x)。最简形式对应选项 C。
Q10: Find the coordinates of the stationary point on the curve y = x3 – 3x + 4. dy/dx = 3x2 – 3 = 0 ⇒ x2 = 1, so x = 1 or x = -1. Second derivative: d2y/dx2 = 6x. At x = 1, d2y/dx2 = 6 > 0, minimum; at x = -1, d2y/dx2 = -6 < 0, maximum. Coordinates: (1, 2) and (-1, 6). The question likely asks for the maximum point, option B (-1,6).
Q10：求曲线 y = x3 – 3x + 4 的驻点坐标。 dy/dx = 3x2 – 3 = 0 ⇒ x2 = 1，得 x = 1 或 x = -1。二阶导数：d2y/dx2 = 6x。x = 1 时二阶导数为正，极小值；x = -1 时二阶导数为负，极大值。坐标为 (1, 2) 和 (-1, 6)。题干通常要求极大值点，对应选项 B (-1,6)。
Q11: The gradient of a curve is given by 6x2 – 2x, and it passes through (1,3). Find its equation. Integrate gradient: y = ∫(6x2 – 2x) dx = 2x3 – x2 + C. Substitute (1,3): 2(1)3 – (1)2 + C = 3 ⇒ 1 + C = 3, C = 2. Equation: y = 2x3 – x2 + 2. Correct option is E.
Q11：一条曲线的梯度函数为 6x2 – 2x，且曲线经过点 (1,3)。求曲线方程。 积分梯度：y = ∫(6x2 – 2x) dx = 2x3 – x2 + C。代入 (1,3)：2 – 1 + C = 3 ⇒ C = 2。方程为 y = 2x3 – x2 + 2。正确选项为 E。
6. Integration and Area under a Curve | 积分与曲线下面积
Definite integration, area between curves, and simple substitution methods are standard. Emphasis is on recognising the correct antiderivative quickly.
定积分、曲线间面积以及简单换元法都是常规考点，重点在于快速认出正确的原函数。
Q12: Evaluate ∫02 (3x2 – 4x + 1) dx. Integrate termwise: [x3 – 2x2 + x] from 0 to 2 = (8 – 8 + 2) – (0) = 2. Correct option is A.
Q12：计算定积分 ∫02 (3x2 – 4x + 1) dx。 逐项积分：[x3 – 2x2 + x] 从 0 到 2 = (8 – 8 + 2) – (0) = 2。正确选项为 A。
Q13: Find the area enclosed between y = x2 and y = 2x – x2. Intersection: x2 = 2x – x2 ⇒ 2x2 – 2x = 0 ⇒ 2x(x – 1) = 0, so x = 0 and x = 1. The upper curve on [0,1] is y = 2x – x2. Area = ∫01 [(2x – x2) – x2] dx = ∫01 (2x – 2x2) dx = [x2 – (2/3)x3]01 = 1 – 2/3 = 1/3. Correct option is D.
Q13：求曲线 y = x2 与 y = 2x – x2 所围区域面积。 联立得交点：x2 = 2x – x2 ⇒ x = 0, 1。在 [0,1] 上，上方曲线为 y = 2x – x2。面积 = ∫01 [(2x – x2) – x2] dx = ∫01 (2x – 2x2) dx = [x2 – (2/3)x3]01 = 1/3。正确选项为 D。
Q14: Use the substitution u = x2 + 1 to find ∫ x√(x2+1) dx. du/dx = 2x ⇒ x dx = du/2. Integral becomes (1/2)∫ u1/2 du = (1/2)·(2/3)u3/2 + C = (1/3)(x2+1)3/2 + C. Correct option is B.
Q14：用代换 u = x2 + 1 求 ∫ x√(x2+1) dx。 du/dx = 2x ⇒ x dx = du/2。积分变为 (1/2)∫ u1/2 du = (1/2)·(2/3)u3/2 + C = (1/3)(x2+1)3/2 + C。正确选项为 B。
7. Vectors in Two and Three Dimensions | 二维与三维向量
Vector questions involve magnitude, direction, dot product, and simple geometric applications like finding angles between vectors or proving perpendicularity.
向量题涉及模长、方向、点积以及求向量夹角、证明垂直等简单的几何应用。
Q15: Given a = 2i + j and b = -i + 3j, find the angle between a and b. a · b = 2(-1) + 1·3 = 1. |a| = √(4+1) = √5; |b| = √(1+9) = √10. cos θ = 1/(√5√10) = 1/√50 = 1/(5√2) = √2/10. θ = arccos(√2/10) ≈ 81.9°, matching option C (to nearest degree 82°).
Q15：已知 a = 2i + j，b = -i + 3j，求 a 与 b 的夹角。 a · b = 2(-1) + 1·3 = 1。|a| = √5，|b| = √10。cos θ = 1/√50 = 1/(5√2) = √2/10。θ ≈ 81.9°，最接近选项 C（82°）。
Q16: Determine the value of t for which vectors 3i – 2j + k and ti + 4j – 2k are perpendicular. Dot product equals zero: 3t + (-2)·4 + 1·(-2) = 0 ⇒ 3t – 8 – 2 = 0 ⇒ 3t = 10 ⇒ t = 10/3. Correct option is D.
Q16：求使向量 3i – 2j + k 与 ti + 4j – 2k 垂直的 t 值。 点积为零：3t – 8 – 2 = 0 ⇒ 3t = 10 ⇒ t = 10/3。正确选项为 D。
Q17: Find the vector equation of the line passing through (1,2,-1) and parallel to 2i – j + 3k. r = (i + 2j – k) + λ(2i – j + 3k). In column form, this corresponds to option A.
Q17：求过点 (1,2,-1) 且平行于 2i – j + 3k 的直线向量方程。 r = (i + 2j – k) + λ(2i – j + 3k)，对应选项 A。
8. Applications to Mechanics and Kinematics | 力学与运动学应用
Several ENGAA maths questions are set in physical contexts, requiring interpretation of velocity-time graphs, use of SUVAT equations, and vector treatment of forces.
ENGAA 有多道数学题赋以物理背景，要求解释速度‑时间图像、使用匀加速运动公式以及力的向量处理。
Q18: A particle moves along a straight line with velocity v = 4t – t2. Find the total distance travelled in the first 5 seconds. The velocity is zero at t = 0 and t = 4. It is positive for 0 < t < 4, negative for t > 4. Displacement from 0 to 4: s1 = ∫04 (4t – t2) dt = [2t2 – t3/3]04 = 32 – 64/3 = 32/3. From 4 to 5: s2 = ∫45 (4t – t2) dt = [2t2 – t3/3]45 = (50 – 125/3) – (32 – 64/3) = 150/3 – 125/3 – 96/3 + 64/3 = -7/3. Distance = |s1| + |s2| = 32/3 + 7/3 = 39/3 = 13 m. Correct option is B.
Q18：质点沿直线运动，速度 v = 4t – t2。求前 5 秒内的总路程。 速度在 t = 0 和 t = 4 为零，04 为负。0 至
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导Cancel reply