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NSW HSC Mathematics Advanced

HSC · NESAMathematics Advanced53 notas em 10 pastas, 276 KB

53 notas de revisão de NESA HSC Mathematics Advanced, criadas por cookie, em 7 seções que seguem os tópicos do curso.

Notes for NSW HSC Mathematics Advanced, in folders for the syllabus's content: functions, trigonometric functions, exponential and logarithmic functions, sequences and series, calculus, statistical analysis and financial mathematics. Each note has definitions, methods, proofs the syllabus names and worked examples. They follow the Mathematics Advanced 11-12 Syllabus (2024), which is first examined in the 2027 HSC.

Ao adicioná-las, você coloca uma cópia nas suas anotações, em uma pasta própria com as pastas abaixo. Você pode alterá-las e transformá-las em cartões de estudo ou em um conjunto de questões. Ao baixar, você recebe um arquivo zip com arquivos Markdown, que podem ser abertos em qualquer aplicativo de anotações.

O que contém

  • Functions
    • Algebraic techniques6 KB
    • Quadratic equations and the discriminant5 KB
    • Functions and relations5 KB
    • Even, odd and composite functions5 KB
    • Straight lines and linear inequalities5 KB
    • Modelling with linear functions and simultaneous equations5 KB
    • Parabolas and quadratic inequalities6 KB
    • Cubic and reciprocal functions4 KB
    • Circles and semicircles4 KB
    • Direct and inverse variation4 KB
    • Piecewise functions and continuity4 KB
    • Absolute value4 KB
    • Graph transformations6 KB
  • Trigonometric functions
    • Right-angled triangles and angles of any magnitude7 KB
    • Sine rule, cosine rule and area of a triangle5 KB
    • Radians, arc length and sectors5 KB
    • Reciprocal ratios and trigonometric identities7 KB
    • Solving trigonometric equations6 KB
    • Graphs of trigonometric functions and their transformations6 KB
    • Modelling periodic phenomena5 KB
  • Exponential and logarithmic functions
    • Exponential functions and the number e5 KB
    • Logarithms and their laws5 KB
    • Exponential and logarithmic equations, change of base and graphs5 KB
    • Logarithmic scales5 KB
  • Sequences and series
    • Sequences, series and sigma notation5 KB
    • Arithmetic sequences and series5 KB
    • Geometric sequences and series5 KB
  • Calculus
    • Differentiation
      • Average and instantaneous rates of change5 KB
      • Introduction to differentiation, first principles and the power rule5 KB
      • Product, quotient and chain rules5 KB
      • Tangents, normals and the angle of inclination4 KB
      • Derivatives of exponential, logarithmic and trigonometric functions5 KB
    • Applications of differentiation
      • Increasing, decreasing and stationary points6 KB
      • Concavity, inflections, differentiability and curve sketching6 KB
      • Optimisation and global maxima and minima5 KB
      • Motion in a straight line5 KB
      • Rates of change, exponential growth and decay5 KB
    • Integration
      • Primitive functions and indefinite integrals5 KB
      • Integrating exponential, logarithmic and trigonometric functions5 KB
      • The definite integral and the Fundamental Theorem of Calculus6 KB
      • Areas and the definite integral6 KB
      • Approximating integrals with rectangles and trapezia5 KB
  • Statistical analysis
    • Sets and Venn diagrams5 KB
    • Probability and multistage events6 KB
    • Conditional probability and independence5 KB
    • Data and random variables6 KB
    • Discrete random variables5 KB
    • Probability density and cumulative distribution functions5 KB
    • Mean, variance and median of a continuous random variable5 KB
    • The normal distribution5 KB
    • z-scores and normal probabilities5 KB
  • Financial mathematics
    • Reducing balance loans6 KB
    • Annuities6 KB

A primeira anotação

Functions / Algebraic techniques

Index laws

For a base a≠0a \neq 0 and whole-number or fractional indices, the laws follow from what an index counts: ama^m is mm copies of aa multiplied together, so combining them just counts copies.

am×an=am+n,aman=am−n,(am)n=amn,(ab)n=anbn,(ab)n=anbn a^m \times a^n = a^{m+n}, \qquad \frac{a^m}{a^n} = a^{m-n}, \qquad (a^m)^n = a^{mn}, \qquad (ab)^n = a^n b^n, \qquad \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}

Dividing ana^n by itself gives 11 by the second law and an−n=a0a^{n-n} = a^0, so a0=1a^0 = 1 for a≠0a \neq 0. Dividing a0a^0 by ana^n in the same way gives a negative index:

a−n=1an,(ab)−n=(ba)n a^{-n} = \frac{1}{a^n}, \qquad \left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^{n}

A fractional index is a root. Since (a1/n)n=a\left(a^{1/n}\right)^n = a, the number a1/na^{1/n} must be the nnth root of aa, and combining this with the power law gives

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

Taking the root first keeps the numbers small, so 82/3=(83)2=22=48^{2/3} = \left(\sqrt[3]{8}\right)^2 = 2^2 = 4, and 16−3/4=1(164)3=1816^{-3/4} = \dfrac{1}{\left(\sqrt[4]{16}\right)^3} = \dfrac{1}{8}.

Worked example: simplify (2x3)2×x−5x\dfrac{(2x^3)^2 \times x^{-5}}{\sqrt{x}}.

(2x3)2×x−5x1/2=4x6×x−5x1/2=4x1x1/2=4x1/2 \begin{aligned} \frac{(2x^3)^2 \times x^{-5}}{x^{1/2}} &= \frac{4x^6 \times x^{-5}}{x^{1/2}} \\ &= \frac{4x^{1}}{x^{1/2}} \\ &= 4x^{1/2} \end{aligned}

An equation with the unknown in the index is solved by writing both sides with the same base. To solve 4x=324^{x} = 32, write 4=224 = 2^2 and 32=2532 = 2^5, so 22x=252^{2x} = 2^5, which gives 2x=52x = 5 and x=52x = \tfrac{5}{2}.

Expanding, factorising and simplifying

Expanding multiplies every term in one bracket by every term in the other. The special products come up so often that they are worth recognising on sight:

(a+b)2=a2+2ab+b2,(a−b)2=a2−2ab+b2,(a+b)(a−b)=a2−b2 (a+b)^2 = a^2 + 2ab + b^2, \qquad (a-b)^2 = a^2 - 2ab + b^2, \qquad (a+b)(a-b) = a^2 - b^2

Factorising reverses expanding, and there is a fixed order to try. Take out any common factor first; then look for a difference of two squares; then factorise a quadratic trinomial; and for four terms try grouping in pairs.

For a trinomial ax2+bx+cax^2 + bx + c with a≠1a \ne 1, find two numbers whose product is acac and whose sum is bb, split the middle term with them, and group. For 6x2+7x−36x^2 + 7x - 3 the product acac is −18-18 and the pair that sums to 77 is 99 and −2-2, so

6x2+7x−3=6x2+9x−2x−3=3x(2x+3)−(2x+3)=(2x+3)(3x−1) \begin{aligned} 6x^2 + 7x - 3 &= 6x^2 + 9x - 2x - 3 \\ &= 3x(2x + 3) - (2x + 3) \\ &= (2x+3)(3x-1) \end{aligned}

Sums and differences of cubes also factorise, and are worth knowing for simplifying fractions:

a3+b3=(a+b)(a2−ab+b2),a3−b3=(a−b)(a2+ab+b2) a^3 + b^3 = (a+b)(a^2 - ab + b^2), \qquad a^3 - b^3 = (a-b)(a^2 + ab + b^2)

Algebraic fractions

An algebraic fraction is handled exactly like a numerical one, with factorising doing the work that finding common factors does for numbers. Two points about validity matter. A denominator can never be zero, so any value of the variable that makes it zero is excluded, and cancelling a common factor is legitimate only when that factor is non-zero.

Simplifying, multiplying and dividing

Factorise every numerator and denominator, then cancel common factors. To divide, multiply by the reciprocal of the second fraction.

x2−9x+1÷x+3x2−1=(x−3)(x+3)x+1×(x−1)(x+1)x+3=(x−3)(x−1) \begin{aligned} \frac{x^2-9}{x+1} \div \frac{x+3}{x^2-1} &= \frac{(x-3)(x+3)}{x+1} \times \frac{(x-1)(x+1)}{x+3} \\ &= (x-3)(x-1) \end{aligned}

Here x≠−1, −3, 1x \neq -1, \, -3, \, 1, because each of these makes a denominator, or the divisor, zero at some stage.

Adding and subtracting

Factorise the denominators, build the lowest common denominator from every distinct factor, rewrite each fraction over it, and combine the numerators.

x2x2+3x+2−xx+1=x2(x+1)(x+2)−x(x+2)(x+1)(x+2)=x2−x2−2x(x+1)(x+2)=−2x(x+1)(x+2) \begin{aligned} \frac{x^2}{x^2+3x+2} - \frac{x}{x+1} &= \frac{x^2}{(x+1)(x+2)} - \frac{x(x+2)}{(x+1)(x+2)} \\ &= \frac{x^2 - x^2 - 2x}{(x+1)(x+2)} \\ &= \frac{-2x}{(x+1)(x+2)} \end{aligned}

Expressions such as a3b−ab3a2+2ab+b2\dfrac{a^3b - ab^3}{a^2 + 2ab + b^2} simplify by taking out abab in the numerator and recognising a square in the denominator:

ab(a2−b2)(a+b)2=ab(a−b)(a+b)(a+b)2=ab(a−b)a+b \frac{ab(a^2 - b^2)}{(a+b)^2} = \frac{ab(a-b)(a+b)}{(a+b)^2} = \frac{ab(a-b)}{a+b}

Surds

A surd is an irrational root that cannot be written without a root sign, such as 2\sqrt{2} or 53\sqrt[3]{5}; 9\sqrt{9} is not a surd because it equals 33. The rules for square roots of non-negative numbers are

a b=ab,ab=ab,(a)2=a \sqrt{a}\,\sqrt{b} = \sqrt{ab}, \qquad \frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}, \qquad (\sqrt{a})^2 = a

and there is no rule for a+b\sqrt{a+b}: it is not a+b\sqrt{a}+\sqrt{b}. To simplify a surd, pull out the largest square factor, so 72=36×2=62\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}. Like surds add and subtract as like terms do: 32+52=823\sqrt{2} + 5\sqrt{2} = 8\sqrt{2}.

Brackets containing surds expand as usual, and a surd multiplied by itself becomes rational:

(3+5)(2−5)=6−35+25−5=1−5 (3 + \sqrt{5})(2 - \sqrt{5}) = 6 - 3\sqrt{5} + 2\sqrt{5} - 5 = 1 - \sqrt{5}
Conjugates and rationalising the denominator

The conjugate of a+b\sqrt{a} + \sqrt{b} is a−b\sqrt{a} - \sqrt{b}, and the other way round. Their product is a difference of two squares, so the roots disappear:

(a+b)(a−b)=a−b (\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b

To rationalise a denominator, multiply the numerator and denominator by the conjugate of the denominator. When the denominator is a single surd, multiply by that surd instead.

67−2=6(7+2)(7−2)(7+2)=6(7+2)5 \frac{6}{\sqrt{7} - \sqrt{2}} = \frac{6(\sqrt{7} + \sqrt{2})}{(\sqrt{7} - \sqrt{2})(\sqrt{7} + \sqrt{2})} = \frac{6(\sqrt{7} + \sqrt{2})}{5}

When the numerator also contains surds, the same multiplication applies and the numerator is expanded:

3+13−1=(3+1)(3+1)(3−1)(3+1)=3+23+13−1=4+232=2+3 \begin{aligned} \frac{\sqrt{3} + 1}{\sqrt{3} - 1} &= \frac{(\sqrt{3}+1)(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} \\ &= \frac{3 + 2\sqrt{3} + 1}{3 - 1} \\ &= \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3} \end{aligned}

The same method works when a fraction has two terms in its denominator and the numbers are not whole: the conjugate is formed from the same two terms with the sign between them changed.

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