进阶纯数学2
Pearson Edexcel · International A-Level · 进阶数学 知识点 2 7:13 英语旁白 · 英中内嵌字幕
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Paper WFM02: 1 hour 30 minutes, 75 marks, building on P1 to P4 and FP1.
WFM02 试卷：1 小时 30 分钟，75 分，建立在 P1 至 P4 和 FP1 之上。
Its tools: modulus inequalities, the method of differences, Euler and de Moivre, loci and transformations, first and second order differential equations, Maclaurin and Taylor series, and polar coordinates.
工具包括：含绝对值的不等式、差分法、欧拉公式与棣莫弗定理、轨迹与变换、一阶和二阶微分方程、麦克劳林和泰勒级数，以及极坐标。
Two safe moves: split into cases from the definition of the modulus, or square both sides when both are moduli.
两种安全做法：按绝对值的定义分情况讨论，或在两边都是绝对值时两边平方。
One banned move: multiplying by an expression whose sign you have not fixed.
一种禁止做法：乘以一个符号未确定的表达式。
Split on that sign instead.
应当按它的符号分情况。
Where x squared is at least 1: x squared minus 2x minus 3 is positive, so x is greater than 3 or less than minus 1.
当 x² ≥ 1 时：x² − 2x − 3 > 0，所以 x > 3 或 x < −1。
Where x squared is below 1: the inequality becomes x plus 1 squared less than 0, impossible.
当 x² < 1 时：不等式变为 (x + 1)² < 0，不可能。
The branch dies, and the graph shows why: between minus 1 and 3 the curve stays below the line.
这一支无解，图像说明了原因：在 −1 和 3 之间，曲线一直在直线下方。
Solve x minus 2 greater than 15x over modulus x plus 4.
解 x − 2 > 15x/(|x| + 4)。
The denominator is always positive, so multiply through.
分母总是正的，所以直接相乘。
For x non-negative, x squared minus 13x minus 8 is positive.
x ≥ 0 时，x² − 13x − 8 > 0。
For x negative, x plus 1 times x plus 8 is negative, so x lies between minus 8 and minus 1.
x < 0 时，(x + 1)(x + 8) < 0，所以 x 在 −8 和 −1 之间。
Split the term by partial fractions into 1 over r plus 1, minus 1 over r plus 2.
用部分分式把通项拆成 1/(r + 1) − 1/(r + 2)。
Writing out the terms, every fraction from a third to 1 over n plus 1 cancels.
写出各项，从 1/3 到 1/(n + 1) 的每个分数都相互抵消。
Only a half and minus 1 over n plus 2 survive: n over 2 times n plus 2.
只剩下 1/2 和 −1/(n + 2)：等于 n/ 。
Euler's relation, e to the i theta equals cos theta plus i sine theta, turns the polar form into an exponential, so powers become arithmetic: raise the modulus to the n, multiply the argument by n.
欧拉公式 e^{iθ} = cos θ + i sin θ 把极坐标形式变成指数形式，于是幂运算变成算术：模取 n 次方，辐角乘以 n。
That is de Moivre's theorem.
这就是棣莫弗定理。
Expand cos theta plus i sine theta to the fifth by the binomial theorem and keep the imaginary part.
用二项式定理展开 (cos θ + i sin θ)⁵，保留虚部。
De Moivre says it equals sine 5 theta.
棣莫弗定理说它等于 sin 5θ。
Replacing cos squared by 1 minus sine squared gives 16 s to the 5 minus 20 s cubed plus 5s.
把 cos² 换成 1 − sin²，得 16s⁵ − 20s³ + 5s。
Setting 5 theta equal to pi gives sine of pi over 5 as the smaller root.
令 5θ = π，sin(π/5) 取较小的根。
Solve z to the fifth equals minus 2 root 2 minus 2 root 6 i.
解 z⁵ = −2√2 − 2√6 i。
The modulus is 4 root 2, whose fifth root is root 2; the argument is minus 2 pi over 3, divided by 5.
模是 4√2，五次方根是 √2；辐角是 −2π/3，除以 5。
Starting from k equals 0, the five roots sit on the circle of radius root 2, spaced 2 pi over 5 apart.
从 k = 0 开始，五个根落在半径为 √2 的圆上，彼此相隔 2π/5。
Modulus z minus a equals b is a circle.
|z − a| = b 是圆。
Equal distances from a and b give the perpendicular bisector.
到 a 和 b 距离相等，是垂直平分线。
A ratio k other than 1 gives a circle: square both sides and collect terms.
比值 k 不等于 1 时是圆：两边平方再整理。
An argument fixes a half-line from a.
辐角固定时是从 a 出发的半直线。
For w equals 9i minus i z over 1 plus z, with modulus z equals 1: make z the subject, z equals 9i minus w over w plus i.
对于 w = (9i − iz)/(1 + z)，|z| = 1：先把 z 表示出来，z = (9i − w)/(w + i)。
Modulus z equals 1 forces w to be equidistant from 9i and minus i: the line v equals 4.
|z| = 1 迫使 w 到 9i 和 −i 的距离相等：即直线 v = 4。
Check: z equals i maps to 5 plus 4i, and minus i maps to minus 5 plus 4i.
检验：z = i 映到 5 + 4i，−i 映到 −5 + 4i。
Put the equation in the form dy by dx plus P y equals Q, and multiply by e to the integral of P.
把方程化成 dy/dx + Py = Q 的形式，再乘以 e^{∫P dx}。
Here that is e to the 2x, which turns the left side into the derivative of y e to the 2x.
这里是 e^{2x}，它把左边变成 ye^{2x} 的导数。
Integrate once: y equals e to the minus x plus C e to the minus 2x.
积分一次：y = e^{−x} + Ce^{−2x}。
With z equals 1 over y squared, the chain rule turns the equation into dz by dx minus z cot x equals tan x sec x.
令 z = 1/y²，由链式法则方程变为 dz/dx − z cot x = tan x sec x。
The integrating factor is cosec x, and the right side becomes sec squared x, so z cosec x equals tan x plus C.
积分因子是 csc x，右边变成 sec² x，所以 z csc x = tan x + C。
The condition fixes C; at pi over 3, y squared is 4 over 7, so y is plus or minus 2 over root 7.
条件确定 C；在 π/3 处 y² = 4/7，所以 y = ±2/√7。
The complementary function solves the equation with zero on the right, through the auxiliary equation.
余函数通过辅助方程求解右边为零的方程。
Distinct roots give two exponentials; a repeated root gives A plus B x times one exponential; complex roots p plus or minus q i give e to the p x times cos and sine of q x.
不同实根给出两个指数项；重根给出 (A + Bx) 乘一个指数；复根 p ± qi 给出 e^{px} 乘 qx 的余弦和正弦。
The particular integral copies the shape of the right side.
特积分模仿右边的形式。
If the exponent p is already a root of the auxiliary equation, the plain trial gives zero, so multiply by x.
若指数 p 已是辅助方程的根，普通试解会得到零，所以要乘以 x。
Constants come last, from the full general solution.
常数最后用完整的通解来定。
Substituting lambda x e to the 3x, the left side collapses to 13 lambda e to the 3x, so lambda is 6.
代入 λxe^{3x}，左边化为 13λe^{3x}，所以 λ = 6。
The auxiliary equation factors as 4m plus 1 times m minus 3, giving the complementary function.
辅助方程分解为 (4m + 1)(m − 3)，得到余函数。
Then y is 9 over 2 and y dash is 0 at x equals 0: A is 6 and B is minus 3 over 2.
再由 x = 0 时 y = 9/2、y′ = 0，得 A = 6，B = −3/2。
A Maclaurin series matches a function's value and derivatives at 0.
麦克劳林级数在 0 处匹配函数的值和各阶导数。
Near 0, e to the x and its first three terms, 1 plus x plus x squared over 2, are almost the same curve; further away they part.
在 0 附近，eˣ 和它的前三项 1 + x + x²/2 几乎是同一条曲线；离得远了就分开。
A Taylor series does the same about any point a.
泰勒级数在任意点 a 处做同样的事。
The equation gives y double dash at 3 from y and y dash: 6 plus e to the two thirds.
由方程和 y、y′ 得到 y″(3) = 6 + e^{2/3}。
Differentiate the equation itself to get y triple dash: 19 plus 38 over 9 e to the two thirds.
对方程本身求导得 y‴(3) = 19 + (38/9)e^{2/3}。
Divide by 2 factorial and 3 factorial to build the series in powers of x minus 3.
分别除以 2! 和 3!，构成关于 (x − 3) 的幂级数。
In polar coordinates a point is r, theta from the pole.
在极坐标中，点用到极点的 (r, θ) 表示。
The cardioid r equals 2 times 1 plus cos theta touches the pole and reaches 4 on the initial line.
心形线 r = 2(1 + cos θ) 经过极点，在初始线上达到 4。
Its area is a half the integral of r squared: 2 times 2 pi plus 0 plus pi, which is 6 pi.
它的面积是 r² 积分的一半：2 × (2π + 0 + π) = 6π。
The lemniscate r squared equals 8 cos 2 theta exists only where cos 2 theta is non-negative.
双纽线 r² = 8cos 2θ 只在 cos 2θ ≥ 0 处存在。
Its right loop runs from minus pi over 4 to pi over 4: a half times the integral of 8 cos 2 theta is 4.
右边一瓣从 −π/4 到 π/4：8cos 2θ 积分的一半等于 4。
Two loops: total area 8.
两瓣：总面积 8。
Check yourself.
自我检测。
Never multiply by an expression of unknown sign.
不要乘以符号未知的表达式。
Equal distances give the perpendicular bisector.
距离相等给出垂直平分线。
When p is a root, the trial is lambda x e to the p x.
p 是根时，试解为 λxe^{px}。
Polar area is a half the integral of r squared.
极坐标面积是 r² 积分的一半。
And fifth roots are spaced 2 pi over 5 apart.
五次方根彼此相隔 2π/5。