Pure Mathematics 4
Pearson Edexcel · International A-Level · Mathematics Topic 4 4:11 English narration · English + 中文 subtitles burned in
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Paper WMA14: 1 hour 30 minutes, 75 marks, with P1 to P3 assumed.
WMA14 试卷：1 小时 30 分钟，75 分，默认已学 P1 到 P3。
Memorise the binomial series for any rational n, valid for x between minus 1 and 1; the volume of revolution; and the vector magnitude and position-vector rules.
记住任意有理数 n 的二项式级数（−1 < x < 1 时成立）、旋转体体积，以及向量的模和位置向量法则。
Suppose root 2 is p over q in lowest terms.
假设 √2 = p/q 且为最简分数。
Squaring gives p squared equals 2q squared, so p is even.
平方得 p² = 2q²，所以 p 是偶数。
Writing p as 2r shows q is even too, which contradicts lowest terms.
令 p = 2r，可知 q 也是偶数，这与最简分数矛盾。
The same pattern proves there are infinitely many primes.
同样的模式可以证明质数有无穷多个。
Suppose some x has the modulus at least 1.
假设某个 x 使绝对值至少为 1。
Squaring gives 1 plus sine 2x at least 1, so sine 2x is at least 0.
平方得 1 + sin 2x ≥ 1，所以 sin 2x ≥ 0。
But 2x lies between 180 and 360 degrees, where sine is negative.
但 2x 在 180° 到 360° 之间，正弦为负。
The graph agrees: it stays below 1, with zero at 135.
图像也印证：它始终低于 1，在 135° 处为零。
Split 3x squared plus 7x plus 6 over x times x plus 1 squared, giving the repeated factor both powers.
把 (3x² + 7x + 6)/ 拆开，重因式要写出两个幂次。
Substituting x equals 0 gives A, x equals minus 1 gives C, and comparing x squared gives B.
代入 x = 0 得 A，x = −1 得 C，比较 x² 系数得 B。
At x equals 1 both sides are 4.
x = 1 时两边都是 4。
As t runs from minus 3 to 3, the point traces a parabola; equal steps in t give equal steps in x here.
t 从 −3 变到 3 时，点描出一条抛物线；这里 t 的等步长对应 x 的等步长。
From x equals 2t, t is x over 2, and substituting gives y equals x squared over 4, minus 1.
由 x = 2t 得 t = x/2，代入得 y = x²/4 − 1。
The tangent gradient at t equals 2 is minus four fifths, so the normal's is five quarters.
t = 2 处切线斜率为 −4/5，所以法线斜率为 5/4。
Through P, minus 7, minus 3, that gives 5x minus 4y plus 23 equals 0.
过 P(−7, −3)，得 5x − 4y + 23 = 0。
Factor out 4 so the bracket starts with 1, then use n equals a half.
提出 4，使括号以 1 开头，再用 n = ½。
Two terms give a straight line; three terms bend it to hug the true curve near 0.
两项给出一条直线；三项使它弯曲，在 0 附近紧贴真实曲线。
The series only works for x between minus 4 and 4.
这个级数只在 −4 < x < 4 时成立。
Differentiate every term with respect to x: y squared gives 4y dy by dx, and minus 6xy needs the product rule.
对每一项关于 x 求导：y² 项得 4y dy/dx，−6xy 要用乘积法则。
Collect the dy by dx terms: at the point the gradient is 38 over 13, giving the tangent.
整理 dy/dx 项：在该点斜率为 38/13，由此得切线。
The volume changes at a constant rate k.
体积以恒定速率 k 变化。
The chain rule links it to the radius through dV by dr, 4 pi r squared.
链式法则通过 dV/dr = 4πr² 把它和半径联系起来。
So dr by dt is k over 4 pi, times 1 over r squared: small spheres shrink fastest.
所以 dr/dt = (k/4π)·(1/r²)：小球缩得最快。
Put the y terms on one side and the x terms on the other, then integrate both: y squared is 3 sec 2x plus a constant.
把 y 的项放一边、x 的项放另一边，两边积分：y² = 3 sec 2x + 常数。
The condition y equals 4 at pi over 6 fixes the constant at 10.
条件 x = π/6 时 y = 4 确定常数为 10。
Integrate by parts with u equal to e to the x.
令 u = eˣ 分部积分。
The new integral needs parts again, and then the original integral reappears.
新的积分还要再分部一次，然后原积分重新出现。
Call it I and collect: 2I is e to the x times sine plus cosine.
记为 I 并移项：2I = eˣ(sin x + cos x)。
The region under y equals 3 cos x spins about the x-axis.
y = 3cos x 下方的区域绕 x 轴旋转。
Every slice is a circle of radius y, area pi y squared, so the volume is pi times the integral of y squared.
每个截面都是半径为 y 的圆，面积 πy²，所以体积是 π 乘以 y² 的积分。
The double angle identity finishes it.
用二倍角恒等式完成计算。
Find A B and A C as differences of position vectors.
用位置向量之差求出 AB 和 AC。
Their scalar product is 40.
它们的数量积是 40。
Setting the cosine of the angle to 1 over root 2 makes the length of A C equal 10, so alpha minus 2 squared is 32.
令夹角余弦等于 1/√2，得 AC 的长度为 10，所以 (α − 2)² = 32。
Check yourself.
自我检测。
A contradiction starts by assuming the opposite.
反证法先假设结论的反面。
A repeated factor needs both powers.
重因式需要两个幂次。
The series for root 4 plus x is valid for x between minus 4 and 4.
√(4 + x) 的级数在 −4 < x < 4 时成立。
The derivative of xy uses the product rule.
xy 的导数用乘积法则。
And the volume is pi times the integral of y squared.
体积是 π 乘以 y² 的积分。