Miscellaneous Exercises 2 | 综合练习二

📚 Miscellaneous Exercises 2 | 综合练习二

This set of miscellaneous exercises brings together key skills from algebra, trigonometry, differentiation, and coordinate geometry. Each question is designed to test not only your ability to recall formulas, but also your fluency in applying them under exam-style conditions.

本组综合练习整合了代数、三角函数、微分以及坐标几何中的核心技能。每道题不仅考查你对公式的记忆能力，更检验你在模拟考试环境下灵活运用这些公式的熟练程度。

1. Simplifying Algebraic Expressions | 化简代数式

Simplify the expression 3x²(2x³ − 5x) + 4x( x⁴ − 2x² ). Start by expanding each term carefully, then collect like terms.

化简表达式 3x²(2x³ − 5x) + 4x( x⁴ − 2x² )。请先仔细展开每一项，再合并同类项。

Expanding gives 6x⁵ − 15x³ + 4x⁵ − 8x³. Combining like terms, the final simplified form is 10x⁵ − 23x³.

展开后得到 6x⁵ − 15x³ + 4x⁵ − 8x³。合并同类项后，最简形式为 10x⁵ − 23x³。

10x⁵ − 23x³

2. Solving Quadratic Equations by Factorisation | 用因式分解法解二次方程

Solve the equation 2x² − 5x − 3 = 0. Look for two numbers whose product is 2 × (−3) = −6 and whose sum is −5.

解方程 2x² − 5x − 3 = 0。寻找两个数，使其乘积为 2 × (−3) = −6，且和为 −5。

These numbers are −6 and 1. Rewrite the middle term: 2x² − 6x + x − 3 = 0. Factor by grouping: 2x(x − 3) + 1(x − 3) = 0, giving (2x + 1)(x − 3) = 0.

这两个数为 −6 和 1。改写中间项：2x² − 6x + x − 3 = 0。分组因式分解：2x(x − 3) + 1(x − 3) = 0，得到 (2x + 1)(x − 3) = 0。

Therefore x = −½ or x = 3.

因此 x = −½ 或 x = 3。

3. Simultaneous Equations with a Quadratic | 含二次方程的联立方程组

Solve the simultaneous equations y = 3x + 2 and y = x² + 2x − 4. Since both expressions equal y, set them equal to each other.

解联立方程组 y = 3x + 2 和 y = x² + 2x − 4。由于两个表达式都等于 y，令它们相等。

This gives 3x + 2 = x² + 2x − 4. Rearranging: x² − x − 6 = 0, which factorises as (x − 3)(x + 2) = 0. Hence x = 3 or x = −2.

由此得 3x + 2 = x² + 2x − 4。整理得 x² − x − 6 = 0，因式分解为 (x − 3)(x + 2) = 0。因此 x = 3 或 x = −2。

Substituting back: when x = 3, y = 11; when x = −2, y = −4. The coordinate pairs are (3, 11) and (−2, −4).

代回原式：当 x = 3 时，y = 11；当 x = −2 时，y = −4。交点坐标为 (3, 11) 和 (−2, −4)。

4. Inequalities and Number Lines | 不等式与数轴

Solve the inequality 2x − 3 ≤ 7 < 3x + 10. Split this compound inequality into two separate parts and solve each one.

解不等式 2x − 3 ≤ 7 < 3x + 10。将复合不等式拆分为两个独立部分并分别求解。

First part: 2x − 3 ≤ 7 gives 2x ≤ 10, so x ≤ 5. Second part: 7 < 3x + 10 gives −3 < 3x, so x > −1.

第一部分：2x − 3 ≤ 7 得 2x ≤ 10，即 x ≤ 5。第二部分：7 < 3x + 10 得 −3 < 3x，即 x > −1。

Combining these results, the solution is −1 < x ≤ 5. On a number line, draw an open circle at −1 and a closed circle at 5, with a solid line joining them.

综合结果，解集为 −1 < x ≤ 5。在数轴上，−1 处画空心圆圈，5 处画实心圆圈，并用实线连接。

−1 < x ≤ 5

5. Coordinate Geometry: Midpoint and Distance | 坐标几何：中点与距离

Given two points A(2, −3) and B(8, 5), find the midpoint of AB and the exact distance AB.

已知两点 A(2, −3) 和 B(8, 5)，求 AB 的中点坐标以及 AB 的精确距离。

The midpoint M has coordinates ((2 + 8)/2, (−3 + 5)/2) = (5, 1).

中点 M 的坐标为 ((2 + 8)/2, (−3 + 5)/2) = (5, 1)。

The distance AB = √[(8 − 2)² + (5 − (−3))²] = √[6² + 8²] = √(36 + 64) = √100 = 10.

距离 AB = √[(8 − 2)² + (5 − (−3))²] = √[6² + 8²] = √(36 + 64) = √100 = 10。

M(5, 1), AB = 10

6. Gradient and Equation of a Straight Line | 斜率与直线方程

Find the equation of the line passing through the point (4, −1) with gradient 2. Write your answer in the form y = mx + c.

求经过点 (4, −1) 且斜率为 2 的直线方程，并将答案写成 y = mx + c 的形式。

Using y − y₁ = m(x − x₁), we have y + 1 = 2(x − 4). Expanding: y + 1 = 2x − 8, so y = 2x − 9.

利用点斜式 y − y₁ = m(x − x₁)，得 y + 1 = 2(x − 4)。展开：y + 1 = 2x − 8，因此 y = 2x − 9。

The gradient is 2 and the y-intercept is −9, confirming the line is steep and crosses the y-axis below the origin.

斜率为 2，y 轴截距为 −9，说明该直线较陡且与 y 轴交于原点下方。

y = 2x − 9

7. Differentiation from First Principles | 用极限定义求导

Use the definition of the derivative as a limit to find dy/dx for y = 4x².

利用导数作为极限的定义，求 y = 4x² 的导数 dy/dx。

The derivative is lim(h→0) [4(x + h)² − 4x²] / h. Expand (x + h)² = x² + 2xh + h², giving lim(h→0) [4x² + 8xh + 4h² − 4x²] / h.

导数为 lim(h→0) [4(x + h)² − 4x²] / h。展开 (x + h)² = x² + 2xh + h²，得 lim(h→0) [4x² + 8xh + 4h² − 4x²] / h。

Simplifying to lim(h→0) [h(8x + 4h)] / h = lim(h→0) (8x + 4h) = 8x.

化简为 lim(h→0) [h(8x + 4h)] / h = lim(h→0) (8x + 4h) = 8x。

dy/dx = 8x

8. Using Differentiation to Find Tangents and Normals | 用微分求切线与法线

For the curve y = x³ − 6x² + 9x + 2, find the equation of the tangent at the point where x = 2.

对于曲线 y = x³ − 6x² + 9x + 2，求在 x = 2 处切线的方程。

First find the derivative: dy/dx = 3x² − 12x + 9. At x = 2, the gradient is 3(4) − 24 + 9 = 12 − 24 + 9 = −3.

首先求导：dy/dx = 3x² − 12x + 9。当 x = 2 时，斜率为 3(4) − 24 + 9 = 12 − 24 + 9 = −3。

The y-coordinate at x = 2 is y = 8 − 24 + 18 + 2 = 4. Using the point (2, 4) and gradient −3: y − 4 = −3(x − 2), so y = −3x + 10.

x = 2 时的 y 坐标为 y = 8 − 24 + 18 + 2 = 4。利用点 (2, 4) 和斜率 −3：y − 4 = −3(x − 2)，因此 y = −3x + 10。

The normal has gradient 1/3, giving y − 4 = (1/3)(x − 2), or y = (1/3)x + 10/3.

法线的斜率为 1/3，因此 y − 4 = (1/3)(x − 2)，即 y = (1/3)x + 10/3。

Tangent: y = −3x + 10, Normal: y = (1/3)x + 10/3

9. Stationary Points and Curve Sketching | 驻点与曲线草图

Find the coordinates and nature of the stationary points of the curve y = 2x³ − 9x² + 12x + 1.

求曲线 y = 2x³ − 9x² + 12x + 1 的驻点坐标及其性质。

Set dy/dx = 0: 6x² − 18x + 12 = 0. Dividing by 6 gives x² − 3x + 2 = 0, so (x − 1)(x − 2) = 0, hence x = 1 or x = 2.

令 dy/dx = 0：6x² − 18x + 12 = 0。两边除以 6 得 x² − 3x + 2 = 0，所以 (x − 1)(x − 2) = 0，因此 x = 1 或 x = 2。

When x = 1, y = 2 − 9 + 12 + 1 = 6. When x = 2, y = 16 − 36 + 24 + 1 = 5.

当 x = 1 时，y = 2 − 9 + 12 + 1 = 6。当 x = 2 时，y = 16 − 36 + 24 + 1 = 5。

Use the second derivative: d²y/dx² = 12x − 18. At x = 1, d²y/dx² = −6 < 0, so (1, 6) is a local maximum. At x = 2, d²y/dx² = 6 > 0, so (2, 5) is a local minimum.

使用二阶导数：d²y/dx² = 12x − 18。当 x = 1 时，d²y/dx² = −6 < 0，因此 (1, 6) 为局部极大值点。当 x = 2 时，d²y/dx² = 6 > 0，因此 (2, 5) 为局部极小值点。

Local maximum (1, 6), Local minimum (2, 5)

10. Trigonometric Equations in a Given Interval | 给定区间内的三角方程

Solve 2 sin θ − 1 = 0 for 0° ≤ θ ≤ 360°. Begin by isolating sin θ.

在 0° ≤ θ ≤ 360° 范围内解方程 2 sin θ − 1 = 0。首先分离 sin θ。

2 sin θ = 1, so sin θ = 1/2. The reference angle is 30°. Since sine is positive in the first and second quadrants, θ = 30° or θ = 180° − 30° = 150°.

2 sin θ = 1，因此 sin θ = 1/2。参考角为 30°。由于正弦在第一和第二象限为正，所以 θ = 30° 或 θ = 180° − 30° = 150°。

For a complete solution in the interval 0° to 360°, the answers are 30° and 150°.

在 0° 到 360° 区间内，完整解为 30° 和 150°。

θ = 30°, 150°

11. Using Trigonometric Identities | 使用三角恒等式

Prove that sin²θ cos²θ + cos⁴θ = cos²θ for all values of θ, and hence solve cos²θ = 1/4 for 0° ≤ θ ≤ 360°.

证明对于所有 θ 值，sin²θ cos²θ + cos⁴θ = cos²θ 成立，并由此在 0° ≤ θ ≤ 360° 范围内解方程 cos²θ = 1/4。

Factor the left-hand side: cos²θ(sin²θ + cos²θ). Using the Pythagorean identity sin²θ + cos²θ = 1, this simplifies to cos²θ.

将左边因式分解：cos²θ(sin²θ + cos²θ)。利用毕达哥拉斯恒等式 sin²θ + cos²θ = 1，该式简化为 cos²θ。

Now solve cos²θ = 1/4. Take square roots: cos θ = ±1/2. For cos θ = 1/2, θ = 60° or 300°. For cos θ = −1/2, θ = 120° or 240°.

现在解 cos²θ = 1/4。两边开平方：cos θ = ±1/2。当 cos θ = 1/2 时，θ = 60° 或 300°。当 cos θ = −1/2 时，θ = 120° 或 240°。

θ = 60°, 120°, 240°, 300°

12. Problem Solving with Area and Perimeter | 面积与周长综合问题

A rectangle has a perimeter of 40 cm and an area of 96 cm². Find the length and width of the rectangle.

一个长方形的周长为 40 cm，面积为 96 cm²。求该长方形的长和宽。

Let length be L and width be W. From the perimeter, 2L + 2W = 40, so L + W = 20. From the area, LW = 96.

设长为 L，宽为 W。由周长得 2L + 2W = 40，即 L + W = 20。由面积得 LW = 96。

Substitute W = 20 − L into LW = 96: L(20 − L) = 96, giving L² − 20L + 96 = 0. Factorise: (L − 12)(L − 8) = 0, so L = 12 or L = 8.

将 W = 20 − L 代入 LW = 96：L(20 − L) = 96，得 L² − 20L + 96 = 0。因式分解：(L − 12)(L − 8) = 0，因此 L = 12 或 L = 8。

Thus the rectangle is 12 cm by 8 cm.

因此该长方形的尺寸为 12 cm × 8 cm。

Length 12 cm, Width 8 cm

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Miscellaneous Exercises 2 | 综合练习二