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بيرون إيديكلر · International A-Level · Pure Mathematics الموضوع 1 5:48 سرد باللغة الإنجليزية · ترجمة مدمجة بالإنجليزية + الصينية
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Paper WMA11: 1 hour 30 minutes, 75 marks.
WMA11 试卷：1 小时 30 分钟，75 分。
These formulae are not in the booklet, so memorise them: the quadratic formula, the sine and cosine rules, the triangle area, arc and sector in radians, and the power rules.
这些公式不在公式册里，必须记住：求根公式、正弦和余弦定理、三角形面积、弧度制下的弧长与扇形面积，以及幂法则。
When a question says show all stages, a calculator answer alone scores nothing.
题目要求写出所有步骤时，只给计算器答案不得分。
A power of a half is a square root: x squared over 4.
½ 次幂就是平方根：x²/4。
Multiply before inverting: p q is 5 x cubed over 6, so its reciprocal is six fifths x to the minus 3, and its square is 25 x to the 6 over 36.
先相乘再取倒数：pq = 5x³/6，所以它的倒数是 (6/5)x⁻³，平方是 25x⁶/36。
1 plus root 2 times root 2 minus 1 is exactly 1, so multiply top and bottom by root 2 minus 1.
(1 + √2)(√2 − 1) 恰好等于 1，所以分子分母同乘 √2 − 1。
The denominator disappears, leaving 12 root 2 minus 12, and both forms give 4.9706.
分母消失，得到 12√2 − 12，两种形式都等于 4.9706。
Three parabolas, one picture.
三条抛物线，一幅图。
A positive discriminant: the curve cuts the axis twice.
判别式为正：曲线与 x 轴相交两次。
Zero: it touches it at one repeated root.
为零：在一个重根处相切。
Negative: it never meets it.
为负：从不与 x 轴相交。
With k equal to 4.5, doubling gives 7x to the 6 plus 8x cubed plus 1.
k = 4.5 时，乘以 2 得 7x⁶ + 8x³ + 1 = 0。
Since x to the 6 is x cubed squared, put z equal to x cubed: the quadratic factorises, z is minus a seventh or minus 1, and cube roots give x.
由于 x⁶ = (x³)²，令 z = x³：二次式可因式分解，z = −1/7 或 −1，开立方得 x。
Treat the equation as a quadratic in x cubed.
把方程看作关于 x³ 的二次方程。
Its discriminant simplifies to 4k times 5 minus k, a parabola in k opening downward.
它的判别式化简为 4k(5 − k)，是关于 k 的开口向下的抛物线。
It is negative outside its roots: k below 0 or above 5.
它在两根之外为负：k < 0 或 k > 5。
C1 and C2 meet where their heights are equal.
C₁ 与 C₂ 在高度相等处相交。
Expanding and collecting gives x cubed plus 2x squared minus 48x equals 0, which factorises: x is 0, minus 8 or 6.
展开并合并得 x³ + 2x² − 48x = 0，因式分解：x = 0、−8 或 6。
Three roots, three points on the picture.
三个根，图上三个交点。
Where is y equals x squared below the line y equals 3x plus 10?
y = x² 在直线 y = 3x + 10 下方的范围是什么？
They cross at minus 2 and 5, and the curve is below the line between the crossings.
它们在 −2 和 5 处相交，曲线在两交点之间位于直线下方。
The inequality is strict, so the boundary is a dotted line.
不等式是严格的，所以边界画虚线。
From y equals root x to y equals root 2x: the x-values halve, so the point 4, 2 moves to 2, 2.
从 y = √x 到 y = √(2x)：x 值减半，所以点 (4, 2) 移到 (2, 2)。
That is a stretch parallel to the x-axis, scale factor a half.
这是平行于 x 轴、比例因子 ½ 的伸缩。
And 12 minus root x is minus root x moved up by 12.
而 12 − √x 是 −√x 向上平移 12。
l1 has gradient a half, so the perpendicular l2 through the origin has gradient minus 2.
l₁ 的斜率是 ½，所以过原点的垂线 l₂ 斜率为 −2。
They meet at P.
它们交于 P。
The shortest distance from the origin to l1 runs along the perpendicular: O P.
原点到 l₁ 的最短距离沿着垂线：OP。
Rearrange l1 to read its gradient, a half; the perpendicular gradient is minus 2.
把 l₁ 整理后读出斜率 ½；垂线斜率为 −2。
Substitute y equals minus 2x into l1: x is minus 5, y is 10.
把 y = −2x 代入 l₁：x = −5，y = 10。
Then O P is root 125, which simplifies to 5 root 5.
于是 OP = √125，化简为 5√5。
The pad is a major sector of radius 15 metres joined to triangle A O C, with BC is 2 and angle C is 0.6.
停机坪由半径 15 米的优扇形和三角形 AOC 组成，BC = 2，角 C = 0.6。
The sine rule gives angle C O A as 1.847, so the major sector takes 2 pi minus that.
正弦定理给出角 COA = 1.847，所以优扇形的角是 2π 减去它。
The sector is a half r squared theta, 499.1; the triangle a half a b sine C, 122.7: 622 square metres.
扇形面积 ½r²θ = 499.1；三角形 ½ab sin C = 122.7：共 622 平方米。
The perimeter is the major arc, 66.5, plus AC from the cosine rule, 25.6, plus BC: 94.1 metres.
周长是优弧 66.5，加上由余弦定理得到的 AC = 25.6，再加 BC：94.1 米。
Solve 5 plus 4 cos x equals 1.
解 5 + 4cos x = 1。
The graph of 5 plus 4 cos x dips to 1 once per period, and the line y equals 1 meets those dips: four touches, at minus 540, minus 180, 180 and 540 degrees.
5 + 4cos x 的图像每个周期下降到 1 一次，直线 y = 1 与这些低点相接：四个切点，在 −540°、−180°、180° 和 540°。
The derivative is the gradient of the tangent.
导数是切线的斜率。
The curve passes through P 4, 12 with tangent y equals 10x minus 28, so f dash of 4 is 10, which fixes k as a quarter.
曲线过 P(4, 12)，切线为 y = 10x − 28，所以 f′(4) = 10，由此 k = ¼。
Differentiating again, f double dash of 4 is eleven quarters.
再求导一次，f″(4) = 11/4。
Differentiate term by term: 5x to the three halves minus 4.
逐项求导：5x^{3/2} − 4。
Setting it to 16 gives 2 to the k, to the three halves, equals 4, which is 2 squared.
令其等于 16，得 (2ᵏ)^{3/2} = 4 = 2²。
Compare powers: three k over 2 is 2, so k is four thirds.
比较指数：3k/2 = 2，所以 k = 4/3。
Expand x plus 3 squared and divide each term by 3 root x.
展开 (x + 3)²，每一项除以 3√x。
Then raise each power by one and divide by the new power.
然后每个指数加 1，再除以新指数。
Differentiating the answer returns the integrand exactly.
对答案求导恰好回到被积函数。
Integrating gives a whole family of curves, every vertical shift with the same gradient.
积分得到一整族曲线，所有竖直平移都有相同的斜率。
Substituting P picks the one through P: 16 plus 16 thirds plus d equals 12, so d is minus 28 thirds.
代入 P 选出过 P 的那一条：16 + 16/3 + d = 12，所以 d = −28/3。
Check yourself.
自我检测。
a to the minus a half is one over root a.
a^{−1/2} = 1/√a。
A curve that touches the axis has discriminant zero.
与 x 轴相切的曲线判别式为零。
f of x plus a moves left by a.
f(x + a) 向左平移 a。
Perpendicular gradients multiply to minus 1.
互相垂直的斜率之积为 −1。
And integration raises the power, plus c.
积分把指数加一，再加 c。