📚 Exercise I.1: Foundations of Functions and Equations | 练习I.1：函数与方程基础
This exercise set consolidates the essential building blocks of IB Mathematics, focusing on the algebra and function skills that underpin both Analysis & Approaches and Applications & Interpretation. By working through domain and range, composite and inverse functions, polynomial algebra, graph transformations, quadratics, simultaneous equations, exponentials, logarithms, and basic trigonometry, you will strengthen your readiness for more advanced topics in the course.
本练习系列旨在巩固IB数学的基本构建模块，重点关注代数和函数技能，这些技能是分析与方法和应用与解释两门课程的基础。通过处理定义域与值域、复合函数与反函数、多项式代数、图象变换、二次函数、联立方程、指数与对数以及基础三角学，你将增强学习更高级主题的准备程度。
1. Understanding Domain and Range | 理解定义域与值域
Begin with f(x) = √3x + 6. The radicand must be non‑negative: 3x + 6 ≥ 0 ⇒ x ≥ −2. Thus the domain is {x ∈ ℝ | x ≥ −2}. Since a square root yields only non‑negative values, the range is {y ∈ ℝ | y ≥ 0}.
从 f(x) = √3x + 6 开始。被开方式必须非负：3x + 6 ≥ 0 ⇒ x ≥ −2。因此定义域为 {x ∈ ℝ | x ≥ −2}。由于平方根只输出非负值，值域为 {y ∈ ℝ | y ≥ 0}。
take g(x) = 2/(x² − 9). The denominator cannot be zero: x² − 9 ≠ 0, so x ≠ ±3. Domain: {x ∈ ℝ | x ≠ −3}. For the range, observe that g(x) can take any real value except 0, because a non‑zero constant divided by a varying quantity never equals 0. Hence the range is {y ∈ ℝ | y ≠ 0}.
现在考虑 g(x) = 2/(x² − 9)。分母不能为零：x² − 9 ≠ 0，即 x ≠ ±3。定义域：{x ∈ ℝ | x ≠ 3, x ≠ −3}。对于值域，注意到 g(x) 可以取零以外的任何实数值，因为非零常数除以变化的量永远不会等于0。因此值域为 {y ∈ ℝ | y ≠ 0}。
1. Domain: (1, ∞). The range of the natural logarithm is all real numbers, so the range is ℝ.
对于涉及对数的函数 h(x) = ln(x − 1)，自变量必须严格为正：x − 1 > 0，因此 x > 1。定义域：(1, ∞)。自然对数的值域是全体实数，因此值域为 ℝ。
2. Working with Composite Functions | 处理复合函数
Suppose f(x) = 2x + 5 and g(x) = x². To find (f ∘ main of the composite function is all real numbers, since both f and g are defined for all ℝ.
设 f(x) = 2x + 5，g(x) = x²。求 (f ∘ g)(x) = f(g(x))，将 f 中的每个 x 替换为 g(x)：f(g(x)) = 2(x²) + 5 = 2x² + 5。因为 f 和 g 在全体实数上有定义，复合函数的定义域也是全体实数。
Now compute (g ∘ f)(x) = g(f(x)): g(2x + 5) = (2x + 5)² = 4x² + 20x + 25. Again the domain is ℝ Note that f(g(x)) and g(f(x)) are not equal, showing that composition is not commutative.
现在计算 (g ∘ f)(x) = g(f(x))：g(2x + 5) = (2x + 5)² = 4x² + 20x + 25。定义域同样为 ℝ。注意 f(g(x)) 和 g(f(x)) 不相等，这表明复合运算是不可交换的。
If k(x) = √x and m(x) = x − 4, then (k ∘ m)(x) = √x − 4 − 4 ≥ 0, so the composite domain is x ≥ 4. The order of composition affects the domain; (m ∘ k)(x) = √x − 4 has domain x ≥ 0.
如果 k(x) = √x，m(x) = x − 4，那么 (k ∘ m)(x) = √x − 4。这里内层函数 m(x) 可以取任何实数，但平方根要求 x − 4 ≥ 0，因此复合定义域为 x ≥ 4。复合顺序会影响定义域；(m ∘ k)(x) = √x − 4 的定义域为 x ≥ 0。
3. Finding Inverse Functions | 求反函数
2)/3. Verify by checking f(f⁻¹(x)) = 3((x+2)/3) − 2 = x, and f⁻¹(f(x)) = ((3x−2)+2)/3 = x.
为了求 f(x) = 3x − 2 的反函数，写出 y = 3x − 2。交换 x 和 y：x = 3y − 2。解出 y：y = (x + 2)/3。因此 f⁻¹(x) = (x + 2)/3。通过验证 f(f⁻¹(x)) = 3((x+2)/3) − 2 = x 和 f⁻¹(f(x)) = ((3x−2)+2)/3 = x 来检查。
quadratic function such as p(x) = x² + 4x + 1, we must restrict the domain to make it one‑to‑ ose the domain x ≥ −2, the function is strictly increasing. Then inverse: x = (y + 2)² − 3 ⇒ (y + 2)² = x + 3 ⇒ y + 2 = √x + 3 (taking the positive root) ⇒ y = √x + 3 − 2. So p⁻¹(x) = √x + 3 − 2, with domain x ≥ −3.
对于二次函数如 p(x) = x² + 4x + 1，我们必须限制定义域使它成为单射。配方：p(x) = (x + 2)² − 3。若选取定义域 x ≥ −2，函数严格递增。然后求反函数：x = (y + 2)² − 3 ⇒ (y + 2)² = x + 3 ⇒ y + 2 = √x + 3（取正根） ⇒ y = √x + 3 − 2。所以 p⁻¹(x) = √x + 3 − 2，定义域为 x ≥ −3。
ometric property provides a quick way to check sketches: the point (a, b) on f corresponds to (b, a) on f⁻¹.
f 和 f⁻¹ 的图象关于直线 y = x 互相对称。这一几何性质为检查草图提供了快速方法：f 上的点 (a, b) 对应于 f⁻¹ 上的点 (b, a)。
4. Polynomial Division and the Factor Theorem | 多项式除法与因式定理
x + 6) + 7. The Factor Theorem tells us that if the remainder is zero, then the divisor is a factor.
用长除法将 2x³ − 3x² + 4x − 5 除以 x − 2。商为 2x² + x + 6，余数为 7。这可以表示为 2x³ − 3x² + 4x − 5 = (x − 2)(2x² + x + 6) + 7。因式定理告诉我们，若余数为零，则除式为因式。
tor Theorem, (x + 1) is a factor. Dividing yields x³ − x² − 4x − 2 = (x + 1)(x² − 2x − 2).
检查 x + 1 是否为 P(x) = x³ − x² − 4x − 2 的因式。计算 P(−1)：(−1)³ − (−1)² − 4(−1) − 2 = −1 − 1 + 4 − 2 = 0。由于 P(−1) = 0，根据因式定理，(x + 1) 是一个因式。除法得到 x³ − x² − 4x − 2 = (x + 1)(x² − 2x − 2)。
Synthetic division is a shortcut for linear divisors. For the same polynomial divided by x + 1, the coefficients 1, −1, −4, −2 are brought down, multiplied by −1, and added. The result confirms the quotient 1, −2, −2 and remainder 0, giving x² − 2x − 2.
综合除法是线性除式的快捷方法。对于同一个多项式除以 x + 1，将系数 1, −1, −4, −2 写出，依次乘以 −1 并相加。结果确认商为 1, −2, −2，余数为 0，得到 x² − 2x − 2。
5. Sketching Rational Functions | 绘制有理函数图象
Consider R(x) = (2x + 1)/(x − 3). Vertical asymptote: set denominator x − 3 = 0 ⇒ Horizontal asymptote: as x → ±∞, the ratio of leading coefficients is 2/1, so y = 2. The x‑intercept occurs when numerator = 0: 2x + 1 = 0 ⇒ x = −1/2. The y‑intercept is R(0) = 1/(−3) = −1/3.
考虑 R(x) = (2x + 1)/(x − 3)。垂直渐近线：分母 x − 3 = 0 ⇒ x = 3。水平渐近线：当 x → ±∞，首项系数之比为 2/1，故 y = 2。x 截距出现在分子为 0 时：2x + 1 = 0 ⇒ x = −1/2。y 截距为 R(0) = 1/(−3) = −1/3。
ositive → R(x) is positive large. Sketch the curve approaching the asymptotes accordingly.
画出这些关键特征，并在垂直渐近线附近测试区间，以观察函数是趋向 +∞ 还是 −∞。当 x 略小于 3，如 x = 2.9，分子为正，分母为小的负数 → R(x) 为负的大值。当 x 略大于 3，分母为小的正数 → R(x) 为正的大值。据此绘制曲线趋近渐近线。
S(x) = x + 2 + 5/(x − 2). As x → ±∞, 5/(x − 2) → 0, so the oblique asymptote is y = x + 2.
对于具有斜渐近线的函数，如 S(x) = (x² + 1)/(x − 2)，进行除法：x² + 1 除以 x − 2 得到商 x + 2 和余数 5。因此 S(x) = x + 2 + 5/(x − 2)。当 x → ±∞，5/(x − 2) → 0，故斜渐近线为 y = x + 2。
6. Graph Transformations in Action | 图象变换实践
Starting with the parent function y = f(x), common transformations are:
- y = f(x) + d: vertical translation by d units (up if d > 0).
- y = f(x − c): horizontal translation by c units to the right.
- y = a f(x): vertical stretch by factor |a|; if a < 0, reflection in the x‑axis.
- y = f(bx): horizontal stretch by factor 1/|b|; if b < 0, reflection in the y‑axis.
从父函数 y = f(x) 出发，常见的变换有：
- y = f(x) + d：纵向平移 d 个单位（d > 0 时向上）。
- y = f(x − c)：横向平移 c 个单位（向右）。
- y = a f(x)：纵向伸缩，因子为 |a|；若 a < 0，则关于 x 轴反射。
- y = f(bx)：横向伸缩，因子为 1/|b|；若 b < 0，则关于 y 轴反射。
tify: horizontal shift 3 units left (x + 3), vertical stretch by factor 2, reflection in x‑axis, then vertical shift up by 4. The vertex moves from (0,0) to (−3,4).
将这些应用于 f(x) = x²。要绘制 g(x) = −2(x + 3)² + 4 的图象，识别变换：向左平移 3 个单位 (x + 3)，纵向拉伸因子 2，关于 x 轴反射，然后向上平移 4 个单位。顶点从 (0,0) 移至 (−3,4)。
n before the translation when the function is written as f(b(x − c)). For instance, h(x) = √2x − 6 as √(2(x − 3)). The sequence: compress horizontally by factor 1/2, then shift right by 3.
在水平方向组合变换时，若函数写为 f(b(x − c)) 的形式，应先进行伸缩/反射，再进行平移。例如，h(x) = √2x − 6 应改写为 √(2(x − 3))。顺序：横向压缩因子 1/2，然后向右平移 3。
7. Solving Quadratic Equations and the Discriminant | 解二次方程与判别式
The quadratic formula x = [−b ± √b² − 4ac al roots; Δ = 0 gives one repeated real root; Δ < 0 gives a pair of complex conjugates.
二次公式 x = [−b ± √b² − 4ac]/(2a) 用于解 ax² + bx + c = 0。判别式 Δ = b² − 4ac 决定根的性质：Δ > 0 时有两个不同的实根；Δ = 0 时有一个重实根；Δ < 0 时有一对共轭复根。
= 49. Since Δ > 0, there are two real roots: x = (3 ± 7)/4, giving x = 5/2 and x = −1.
解 2x² − 3x − 5 = 0。此处 a = 2, b = −3, c = −5。Δ = (−3)² − 4(2)(−5) = 9 + 40 = 49。由于 Δ > 0，有两个实根：x = (3 ± 7)/4，得到 x = 5/2 和 x = −1。
+ 9 = 0, Δ = 0, so the root is x = −3 (repeated). The vertex of the parabola touches the x‑axis at x = −3, illustrating a perfect square trinomial.
若方程不是标准形式，需先整理。对于 x² + 6x + 9 = 0，Δ = 0，因此根为 x = −3（重根）。抛物线的顶点在 x = −3 处与 x 轴相切，体现完全平方三项式。
8. Simultaneous Equations: Linear and Quadratic | 联立方程组：线性与二次
system y = 2x + 1 and y = x² + x − 3, set the expressions for y equal: x² + x − 3 = 2x + 1 ⇒ x² − x − 4 = 0. Solve using the quadratic formula: x = [1 ± √1 + 16]/2 = [1 ± √17]/2. Then substitute back to find y.
解方程组 y = 2x + 1 和 y = x² + x − 3，令两个 y 表达式相等：x² + x − 3 = 2x + 1 ⇒ x² − x − 4 = 0。使用二次公式求解：x = [1 ± √1 + 16]/2 = [1 ± √17]/2。然后代回求出 y。
y² = 25 and y = x + 1, substitute the linear equation into the circle: x² + (x + 1)² = 25 ⇒ x² + x² + 2x + 1 = 25 ⇒ 2x² + 2x − 24 = 0 ⇒ x² + x − 12 = 0. Factor: (x + 4)(x − 3) = 0, so x = −4 or x = 3. The corresponding y‑values are −3 and 4. Thus the intersection points are (−4, −3) and (3, 4).
对于直线与圆的方程组，如 x² + y² = 25 和 y = x + 1，将线性方程代入圆的方程：x² + (x + 1)² = 25 ⇒ x² + x² + 2x + 1 = 25 ⇒ 2x² + 2x − 24 = 0 ⇒ x² + x − 12 = 0。因式分解：(x + 4)(x − 3) = 0，故 x = −4 或 x = 3。相应的 y 值为 −3 和 4。因此交点坐标为 (−4, −3) 和 (3, 4)。
Always check solutions in both original equations to avoid extraneous results, although with polynomial systems this is less common than with radical equations.
务必在两个原方程中检验解，以避免增根，尽管在多项式方程组中这不如在根式方程中常见。
9. Introduction to Exponentials and Logarithms | 指数与对数入门
Solve 2ˣ = 32. Since 32 = 2⁵, we have 2ˣ = 2⁵ ⇒ x = 5. For a more complex equation, 3²ˣ⁻¹ = 81, note that 81 = 3⁴, so 2x − 1 = 4 ⇒ x = 5/2.
解 2ˣ = 32。由于 32 = 2⁵，有 2ˣ = 2⁵ ⇒ x = 5。对于更复杂的方程 3²ˣ⁻¹ = 81，注意到 81 = 3⁴，因此 2x − 1 = 4 ⇒ x = 5/2。
Logarithms help when bases cannot be easily matched. For 5ˣ = 20, take log base 10: log(5ˣ) = log 20 ⇒ x log 5 = log 20 ⇒ x = log 20 / log 5 ≈ 1.861. Alternatively, use natural logs: x = ln 20 / ln 5.
当底数难以匹配时，对数可提供帮助。对于 5ˣ = 20，取以 10 为底的对数：log(5ˣ) = log 20 ⇒ x log 5 = log 20 ⇒ x = log 20 / log 5 ≈ 1.861。或者使用自然对数：x = ln 20 / ln 5。
logarithmic equation log₂(x + 1) = 3 can be rewritten in exponential form: x + 1 = 2³ = 8 ⇒ x = 7. Always check the domain: x + 1 > 0, which is satisfied.
对数方程 log₂(x + 1) = 3 可改写为指数形式：x + 1 = 2³ = 8 ⇒ x = 7。务必检查定义域：x + 1 > 0，成立。
e essential for simplifying expressions and solving equations where logs appear on both sides.
性质如 logₐ(MN) = logₐM + logₐN 和 logₐ(M/N) = logₐM − logₐN 对于化简表达式和求解两边出现对数的方程至关重要。
10. Trigonometric Equations in a Given Interval | 给定区间上的三角方程
e sine is positive in the second quadrant as well, another solution is x = π − π/6 = 5π/6.
解 sin x = 1/2，其中 0 ≤ x ≤ 2π。主解为 x = π/6。由于正弦在第二象限也为正，另一个解为 x = π − π/6 = 5π/6。
For 2cos²x − cos x − 1 = 0 on [0, 2π], treat as a quadratic in cos x: let
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