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TMUA (Test of Mathematics for University Admission)

TMUA · Admissions TestingMathematics30 篇笔记 个文件夹,164 KB

cookie 制作的 Admissions Testing TMUA Mathematics 复习笔记,共 30 篇,分为 2 个章节,按照课程主题编排。

Notes for the Test of Mathematics for University Admission (TMUA), in two folders in the content specification's order: the mathematical knowledge (AS-level pure mathematics, then the Higher GCSE content) and the logic and proof used to construct and analyse arguments. Each note has definitions, methods and worked examples. Follows the content specification of April 2025.

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内容

  • Mathematical knowledge
    • Pure mathematics
      • Indices and surds4 KB
      • Quadratics, inequalities and simultaneous equations4 KB
      • Polynomials and functions5 KB
      • Sequences and series4 KB
      • Binomial expansion and counting5 KB
      • Straight lines and circles5 KB
      • Circle theorems5 KB
      • Trigonometry in triangles and radians6 KB
      • Trigonometric functions and equations6 KB
      • Exponentials and logarithms4 KB
      • Differentiation5 KB
      • Integration6 KB
      • Graphs and transformations6 KB
    • Number, geometry, statistics and probability
      • Units and number7 KB
      • Rounding, bounds and estimation5 KB
      • Ratio, proportion and percentages7 KB
      • Algebraic manipulation and rearranging6 KB
      • Graphs in context6 KB
      • Angles, polygons, congruence and similarity6 KB
      • Transformations and vectors6 KB
      • Perimeter, area, volume, arcs and sectors5 KB
      • Pythagoras, right-angled trigonometry, bearings and plans5 KB
      • Statistics7 KB
      • Probability7 KB
  • Mathematical reasoning
    • Statements, converses and contrapositives6 KB
    • Necessary and sufficient conditions, quantifiers and negation6 KB
    • Direct proof and proof by cases6 KB
    • Proof by contradiction and counterexample5 KB
    • Conjecture, implication and ordering a proof6 KB
    • Errors in proofs6 KB

第一篇笔记

Mathematical knowledge / Pure mathematics / Indices and surds

Laws of indices

An index (or exponent) says how many copies of a base are multiplied. For a positive base aa and any rational exponents mm and nn the laws are

am×an=am+naman=am−n(am)n=amn a^m \times a^n = a^{m+n} \qquad \frac{a^m}{a^n} = a^{m-n} \qquad (a^m)^n = a^{mn}

and two further results extend the idea of a power beyond positive whole numbers:

a0=1a−n=1an a^0 = 1 \qquad a^{-n} = \frac{1}{a^n}

The zero power is 11 because an÷an=an−n=a0a^n \div a^n = a^{n-n} = a^0 and anything divided by itself is 11. A negative power is a reciprocal, so 2−3=182^{-3} = \tfrac{1}{8} and (23)−2=94\left(\tfrac{2}{3}\right)^{-2} = \tfrac{9}{4}: the fraction is turned upside down and the exponent made positive.

Fractional exponents

A power of 1n\tfrac{1}{n} is an nnth root, and a general fractional power combines a root with a power:

a1/n=anam/n=(an)m=amn a^{1/n} = \sqrt[n]{a} \qquad a^{m/n} = \left(\sqrt[n]{a}\right)^m = \sqrt[n]{a^m}

Take the root first when the numbers are whole, because it keeps them small.

272/3=(273)2=32=916−3/4=1(164)3=123=18 \begin{aligned} 27^{2/3} &= \left(\sqrt[3]{27}\right)^2 = 3^2 = 9 \\ 16^{-3/4} &= \frac{1}{\left(\sqrt[4]{16}\right)^3} = \frac{1}{2^3} = \frac{1}{8} \end{aligned}
Worked example

Write 82/3×2−141/2\dfrac{8^{2/3} \times 2^{-1}}{4^{1/2}} as a single power of 22.

Express every base as a power of 22: 8=238 = 2^3 and 4=224 = 2^2. Then

(23)2/3×2−1(22)1/2=22×2−121=2121=20=1 \begin{aligned} \frac{(2^3)^{2/3} \times 2^{-1}}{(2^2)^{1/2}} &= \frac{2^{2} \times 2^{-1}}{2^{1}} \\ &= \frac{2^{1}}{2^{1}} \\ &= 2^0 = 1 \end{aligned}

Converting to a common base turns products and quotients into sums and differences of exponents, which is the quickest way to handle mixed powers. The same idea solves equations such as 4x=84^x = 8: writing 22x=232^{2x} = 2^3 gives x=32x = \tfrac{3}{2}.

Conditions

The laws need a positive base. A negative base with a fractional exponent can be undefined in the real numbers: (−8)1/3=−2(-8)^{1/3} = -2 is a real number, but (−4)1/2(-4)^{1/2} is not, and ((−1)2)1/2=1\left((-1)^2\right)^{1/2} = 1 is not the same as (−1)2×1/2=−1(-1)^{2 \times 1/2} = -1. So (am)n=amn(a^m)^n = a^{mn} is used freely only when a>0a > 0.

Surds

A surd is an irrational root, such as 2\sqrt{2} or 5\sqrt{5}. Its decimal never ends and never repeats, so an exact answer keeps the root symbol. The two rules for square roots of positive numbers are

a b=abab=ab \sqrt{a}\,\sqrt{b} = \sqrt{ab} \qquad \frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}

Note that a+b\sqrt{a+b} is not a+b\sqrt{a}+\sqrt{b}: for instance 9+16=5\sqrt{9+16} = 5 but 9+16=7\sqrt{9}+\sqrt{16} = 7.

Simplifying

Pull out the largest square factor. Then 75=25×3=53\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3} and 450=225×2=152\sqrt{450} = \sqrt{225 \times 2} = 15\sqrt{2}. Surds can be added or subtracted only when they are the same root after simplifying:

12+27−3=23+33−3=43 \sqrt{12} + \sqrt{27} - \sqrt{3} = 2\sqrt{3} + 3\sqrt{3} - \sqrt{3} = 4\sqrt{3}
Expanding brackets

Expand as normal and use a×a=a\sqrt{a}\times\sqrt{a} = a.

(3+2)2=9+62+2=11+62(4+5)(4−5)=16−5=11 \begin{aligned} (3 + \sqrt{2})^2 &= 9 + 6\sqrt{2} + 2 = 11 + 6\sqrt{2} \\ (4 + \sqrt{5})(4 - \sqrt{5}) &= 16 - 5 = 11 \end{aligned}

The second is a difference of two squares, and the answer is rational because the cross terms cancel.

Rationalising the denominator

A fraction with a surd in the denominator is rewritten so that the denominator is a whole number.

For a single surd, multiply top and bottom by that surd:

63=6333=633=23 \frac{6}{\sqrt{3}} = \frac{6\sqrt{3}}{\sqrt{3}\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}

For a denominator of the form a+bca + b\sqrt{c}, multiply top and bottom by its conjugate a−bca - b\sqrt{c}, so that the denominator becomes the difference of two squares a2−b2ca^2 - b^2c.

53+25=5 (3−25)(3+25)(3−25)=15−1059−20=105−1511 \begin{aligned} \frac{5}{3+2\sqrt{5}} &= \frac{5\,(3-2\sqrt{5})}{(3+2\sqrt{5})(3-2\sqrt{5})} \\ &= \frac{15 - 10\sqrt{5}}{9 - 20} \\ &= \frac{10\sqrt{5} - 15}{11} \end{aligned}

A second example, where both parts of the denominator contain surds:

37−23=3 (7+23)49−12=21+6337 \begin{aligned} \frac{3}{7-2\sqrt{3}} &= \frac{3\,(7+2\sqrt{3})}{49 - 12} \\ &= \frac{21 + 6\sqrt{3}}{37} \end{aligned}
Surds inside larger problems

Rationalising is also the route to simplifying a sum of fractions. For example

12+3+12−3=(2−3)+(2+3)4−3=4 \frac{1}{2+\sqrt{3}} + \frac{1}{2-\sqrt{3}} = \frac{(2-\sqrt{3}) + (2+\sqrt{3})}{4-3} = 4

and to showing that an expression is rational: (1+2)3+(1−2)3(1+\sqrt{2})^3 + (1-\sqrt{2})^3 is unchanged by swapping the sign of 2\sqrt{2}, so every odd power of 2\sqrt{2} cancels and only a rational number remains. Expanding gives 2(1+3⋅2)=142(1 + 3\cdot 2) = 14.

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