Дополнительная чистая математика 1
Pearson Edexcel · International A-Level · Pure Mathematics Тема 5 8:25 Английское озвучивание · Английский + китайские субтитры (встроенные)
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Paper WFM01: 1 hour 30 minutes, 75 marks, with P1 and P2 assumed.
WFM01 试卷：1 小时 30 分钟，75 分，默认已学 P1 和 P2。
Five families of formulae are not in the booklet: the root sum and product, the sum of r, the modulus and conjugate, the two conics, and the two-by-two determinant and inverse.
有五类公式不在公式册中：根的和与积、r 的求和公式、模与共轭、两种圆锥曲线，以及二阶行列式和逆矩阵。
On an Argand diagram, z one is 3 plus 2i, plotted at 3, 2, and z two is minus 1 plus 2i.
在阿甘特图上，z₁ = 3 + 2i 画在 (3, 2)，z₂ = −1 + 2i。
Adding them is vector addition: complete the parallelogram, and the diagonal is the sum, 2 plus 4i.
相加就是向量相加：补全平行四边形，对角线就是和 2 + 4i。
For z equals 3 plus 4i, the modulus is the distance from the origin: root of 9 plus 16, which is 5.
z = 3 + 4i 的模是它到原点的距离：√(9 + 16) = 5。
The conjugate, 3 minus 4i, reflects it in the real axis.
共轭 3 − 4i 是它关于实轴的对称点。
Since z times its conjugate is the modulus squared, multiplying by the conjugate makes any denominator real.
因为 z 乘它的共轭等于模的平方，所以乘以共轭能把任何分母化为实数。
z one has argument 7 pi over 6 and z two has modulus 5, with the product's modulus 15.
z₁ 的辐角是 7π/6，z₂ 的模是 5，乘积的模是 15。
Moduli multiply, so 5r is 15 and r is 3.
模相乘，所以 5r = 15，r = 3。
If z one plus z two is real, the imaginary parts cancel, so z two's imaginary part is 3 over 2.
若 z₁ + z₂ 是实数，虚部抵消，所以 z₂ 的虚部是 3/2。
Then a squared plus 9 over 4 is 25, giving a equals plus or minus root 91 over 2.
于是 a² + 9/4 = 25，a = ±√91/2。
x squared minus 6x plus 13 completes to x minus 3 squared plus 4, so the roots are 3 plus 2i and 3 minus 2i: mirror images in the real axis.
x² − 6x + 13 配方得 (x − 3)² + 4，所以根是 3 + 2i 和 3 − 2i：关于实轴对称。
With real coefficients, every complex root brings its conjugate, and the pair multiplies to a real quadratic: sum 6, product 13.
系数为实数时，每个复根都带着它的共轭，这一对相乘得到实二次式：和为 6，积为 13。
f of z is P z to the 4 minus 36 z cubed plus Q z squared plus 192 z plus 68, with root 3 plus 5i.
f(z) = Pz⁴ − 36z³ + Qz² + 192z + 68，有根 3 + 5i。
Its conjugate is a root too, so z squared minus 6z plus 34 is a factor.
它的共轭也是根，所以 z² − 6z + 34 是因式。
Comparing coefficients, the constant gives c equals 2, then b equals 6, P equals 7 and Q equals 204.
比较系数：常数项得 c = 2，然后 b = 6，P = 7，Q = 204。
The remaining quadratic gives minus 3 plus or minus i root 5, over 7.
剩下的二次式给出 (−3 ± i√5)/7。
For roots alpha and beta, the sum is minus b over a and the product is c over a.
对于根 α、β，和为 −b/a，积为 c/a。
Every symmetric expression rewrites in those two, so you never solve for the roots.
每个对称式都能用这两个量改写，所以永远不必解出根。
A new quadratic is x squared minus the new sum times x, plus the new product.
新的二次方程是 x² 减去新的和乘 x，再加上新的积。
For new roots alpha squared plus beta and alpha plus beta squared: the new sum is the sum of squares plus the old sum, 640 over 3.
新根是 α² + β 和 α + β²：新的和是平方和加上原来的和，等于 640/3。
The new product expands to the sum of cubes plus alpha beta plus alpha beta squared, minus 31040 over 9.
新的积展开为立方和加 αβ 再加 (αβ)²，等于 −31040/9。
Multiply by 9 to clear fractions.
乘以 9 消去分母。
Three locators, all needing f continuous.
三种求根方法，都要求 f 连续。
Bisection halves an interval with a sign change.
二分法把有变号的区间对半分。
Linear interpolation takes the zero of the chord.
线性插值取弦与 x 轴的交点。
Newton Raphson follows the tangent.
牛顿-拉夫逊法沿切线前进。
Show every evaluation: calculator-only answers score nothing.
每次计算都要写出来：只给计算器答案不得分。
For x squared plus 5 minus 8 to the 5x on 0 to 1: f of 0 is 4, positive; f of a half is negative, so halve again; f of a quarter is minus 8.392, still negative.
对于区间 上的 x² + 5 − 8⁵ˣ：f(0) = 4 为正；f(0.5) 为负，再对半；f(0.25) = −8.392 仍为负。
The root is in 0 to a quarter.
根在 内。
Then interpolating g between 4 and 5 gives 4.827.
然后在 4 与 5 之间对 g 做线性插值，得 4.827。
Solve x cubed minus 2x minus 5 from x nought equals 2.
从 x₀ = 2 开始解 x³ − 2x − 5 = 0。
f of 2 is minus 1 and the gradient is 10, so the tangent meets the axis at x one is 2.1.
f(2) = −1，斜率为 10，所以切线与 x 轴交于 x₁ = 2.1。
One more step gives 2.0946, the root to 4 decimal places.
再走一步得到 2.0946，即精确到四位小数的根。
The parabola y squared equals 4x has focus at 1, 0 and directrix x equals minus 1.
抛物线 y² = 4x 的焦点在 (1, 0)，准线是 x = −1。
Take P at 4, 4: PF is 5 and PM is 5.
取 P(4, 4)：PF = 5，PM = 5。
Equal distances define the curve, and they turn geometry questions into arithmetic.
距离相等定义了这条曲线，也把几何题变成了算术。
P is 5 from the focus, so it is 5 from the directrix too.
P 到焦点的距离是 5，所以到准线的距离也是 5。
Each right triangle PQM has hypotenuse 10 and leg 5, so QM is root 75, 5 root 3.
每个直角三角形 PQM 的斜边是 10，直角边是 5，所以 QM = √75 = 5√3。
M lies between Q and R, so QR is 10 root 3.
M 在 Q 和 R 之间，所以 QR = 10√3。
For xy equals 81 at P, 9t, 9 over t: the tangent gradient is minus 1 over t squared, so the normal's is t squared.
对于 xy = 81 上的点 P(9t, 9/t)：切线斜率是 −1/t²，所以法线斜率是 t²。
Its equation is t y equals t cubed x plus 9 times 1 minus t to the 4.
法线方程为 ty = t³x + 9(1 − t⁴)。
With t a third, P is 3, 27 and A is 0, 80 over 3, so triangle O P A has area 40.
取 t = 1/3，P = (3, 27)，A = (0, 80/3)，三角形 OPA 的面积是 40。
Swap the leading diagonal, negate the other one, and divide by the determinant, ad minus bc.
交换主对角线元素，副对角线取相反数，再除以行列式 ad − bc。
If the determinant is zero the matrix is singular: no inverse.
行列式为零时矩阵是奇异的：没有逆矩阵。
And the inverse of A B is B inverse A inverse: undo the last transformation first.
AB 的逆是 B⁻¹A⁻¹：先撤销最后一个变换。
The determinant of M is minus 5 minus 3a.
M 的行列式是 −5 − 3a。
Compare the inverse with 2M plus 8I entry by entry: the bottom-left gives 3 over 5 plus 3a equals 6, so 5 plus 3a is a half and a is minus 3 over 2.
把逆矩阵与 2M + 8I 逐项比较：左下角给出 3/(5 + 3a) = 6，所以 5 + 3a = 1/2，a = −3/2。
Check the other entries agree.
再检查其他元素是否一致。
The first column, 2, 0, is where i goes; the second column, 1, 1, is where j goes.
第一列 (2, 0) 是 i 的像；第二列 (1, 1) 是 j 的像。
So the unit square becomes a parallelogram of area 2: the determinant.
所以单位正方形变成面积为 2 的平行四边形：面积等于行列式。
Each standard matrix follows from where i and j go.
每个标准矩阵都由 i 和 j 的像得出。
Reflection in the x-axis keeps i and flips j.
关于 x 轴的反射保持 i 不变，翻转 j。
Reflection in y equals x swaps them.
关于 y = x 的反射交换二者。
Rotation sends i to cos theta, sin theta.
旋转把 i 送到 (cos θ, sin θ)。
Enlargement scales both by k.
放大把二者都乘以 k。
P sends i to 0, minus 1 and j to 1, 0: a quarter-turn clockwise.
P 把 i 送到 (0, −1)，把 j 送到 (1, 0)：顺时针旋转四分之一圈。
Q is an enlargement by 5.
Q 是放大 5 倍。
A is 4, 3 reflected in y equals minus x: minus 3, minus 4.
A 是 (4, 3) 关于 y = −x 的对称点：(−3, −4)。
To find B, undo in reverse order: Q inverse first, then P inverse, giving 0.8, minus 0.6.
求 B 要按相反顺序撤销：先 Q⁻¹，再 P⁻¹，得到 (0.8, −0.6)。
To sum 2 r squared minus 1 up to 2n, split it into twice the sum of r squared minus 2n, and substitute 2n into the r squared formula.
求 2r² − 1 从 1 到 2n 的和，拆成 2 倍的 r² 之和减 2n，并把 2n 代入 r² 求和公式。
Simplifying gives 4n over 3 times 4 n squared plus 3n minus 1, which factors as n plus 1 times 4n minus 1.
化简得 (4n/3)(4n² + 3n − 1)，因式分解为 (n + 1)(4n − 1)。
Check n equals 1: 1 plus 7 is 8.
检验 n = 1：1 + 7 = 8。
Induction runs in three moves: verify n equals 1; assume true for n equals k; deduce it for k plus 1.
归纳法分三步：验证 n = 1；假设 n = k 时成立；推出 k + 1 时成立。
Then close: true for 1, and k implies k plus 1, so true for all positive integers.
然后总结：n = 1 成立，且 k 成立推出 k + 1 成立，所以对所有正整数成立。
The closing sentence earns its mark.
总结句本身就得分。
For n equals 1, n cubed plus 2n is 3.
n = 1 时，n³ + 2n = 3。
Assume k cubed plus 2k is 3m.
假设 k³ + 2k = 3m。
Expanding k plus 1 cubed plus 2 times k plus 1 gives the old expression plus 3k squared plus 3k plus 3: 3 times m plus k squared plus k plus 1, a multiple of 3.
展开 (k + 1)³ + 2(k + 1) 得到原式加上 3k² + 3k + 3，即 3(m + k² + k + 1)，是 3 的倍数。
So by induction it holds for all positive integers.
所以由归纳法，对所有正整数成立。
Check yourself.
自我检测。
With real coefficients, 3 plus 5i brings 3 minus 5i.
系数为实数时，3 + 5i 带来 3 − 5i。
Alpha squared plus beta squared is the sum squared minus twice the product.
α² + β² 等于和的平方减两倍的积。
On a parabola PM equals PF.
抛物线上 PM 等于 PF。
The area scale factor is the modulus of the determinant.
面积比例因子是行列式的绝对值。
And the inverse of A B is B inverse A inverse.
AB 的逆是 B⁻¹A⁻¹。