pure-mathematics-notes

Page |0 Mathematics at your fingertips. Tons of worksheets and videos at sirhunte.teachable.com. Page |1 DIFFERENTIATION DIFFERENTIATION.............................................................. ............................................................................................ ............................................................. .............................................................. ...................................... ....... 4 IMPLICIT DIFFERENTIATION DIFFERENTIATION ............................................................. ............................................................................................ ............................................................. ........................................... ............. 5 PARAMETRIC PARAMETRIC DIFFERENTIATION DIFFERENTIATION ......................................................... ........................................................................................ .............................................................. ...................................... ....... 7 TRIGONOMETRIC TRIGONOMETRIC DIFFERENTIATION DIFFERENTIATION ............................................................ .......................................................................................... .......................................................... ............................ 9 DIFFERENTIATION DIFFERENTIATION OF INVERSE TRIGONOMETRIC TRIGONOMETRIC FUNCTIONS FUNCTIONS ....................................... ................................................................ ......................... 10 DIFFERENTIATION DIFFERENTIATION OF EXPONENTIAL EXPONENTIAL FUNCTIONS FUNCTIONS .......................................................... ......................................................................................... ...............................11 DIFFERENTIATION DIFFERENTIATION OF NATURAL LOGARITHMS LOGARITHMS ........................................................... .......................................................................................... .................................... ..... 13 PARTIAL DERIVATIVES DERIVATIVES .......................................................... ......................................................................................... ............................................................. ........................................................ ..........................15 First Partial Derivative ............................................................ ........................................................................................... ............................................................. ................................................... ..................... 15 Second PARTIAL DERIVATIVE DERIVATIVE ............................................................ ........................................................................................... .............................................................. .................................... .....16 INTEGRATION INTEGRATION RESULTS.................................................................... .................................................................................................. ............................................................. .............................................. ............... 17 Even powers of Odd powers on Even powers of    and ............................................................ ........................................................................................... .............................................................. .................................... ..... 19 and ............................................................ ........................................................................................... .............................................................. .................................... ..... 20 ........................................................... .......................................................................................... ............................................................. ........................................................ ..........................21 INTEGRATION INTEGRATION BY PARTS....................................................................... ..................................................................................................... ............................................................. ......................................... ..........22 Reduction Formulae ............................................................ ........................................................................................... ............................................................. ........................................................ ..........................25 PARTIAL FRACTIONS FRACTIONS.......................................................... ......................................................................................... ............................................................. ............................................................. ...............................28 Denominator with Linear Factors ....................................... ...................................................................... .............................................................. ................................................... .................... 28 Denominator with unfactorizable quadratic factor............................................................. ...................................................................................... ..........................29 Denominator with a repeated factor ........................................................... ......................................................................................... ........................................................ ..........................30 Improper Fractions Fractions (degree of numerator ≥ degree of denominator)................................. denominator)................................................ ............... 31 TRAPEZIUM RULE (NUMERICAL INTEGRATION)................. INTEGRATION)................................................ .............................................................. .............................................. ............... 33 COMPLEX NUMBERS........................................................... .......................................................................................... ............................................................. ............................................................. ...............................34 SQUARE ROOT OF NEGATIVE NUMBERS ................................................................................ .......................................................................................................... ..........................34 Operations on Complex Numbers ...................................................... ..................................................................................... .............................................................. .................................... ..... 34 Adding and Subtracting Complex Numbers.............................. .......................................................................................... 34 Multiplying Complex Numbers ................................................................................................................................................. 34 34 Dividing C omplex Numb ers ....................................................................................................................................................... 34 Square Roots of Complex Numbers ..................................... ................................................................... ............................................................. ................................................... .................... 35 Quadratic Equations ............................................................ ........................................................................................... ............................................................. ........................................................ ..........................35 Equations with Real Coefficients .................................................................................................................................. ...........35 Equations with Complex Coeffic ients .................................................................................................................................... 35 Page |2 Roots of Equations...................................... ...................................................................... .............................................................. ............................................................. ................................................... ....................36 Argand Diagram...................................... Diagram..................................................................... .............................................................. .............................................................. ........................................................ .........................37 – – De Moivre’s Theorem Representing Sums and Differences on Argand Diagrams .......................... ................................................................ 37 Modulus Argument Form .................................................................... .................................................................................................. ............................................................. .................................... ..... 37 The Modulus of a Complex Number N umber ....................................................................................................................................... 37 The Argument Ar gument of Complex Number N umber ........................................................................................................................................ 38 Modulus Argument Form ...................................................................................................................... .................................. 39 .......................................................... ......................................................................................... ............................................................. ........................................................ ..........................41 Multiples of Sine and a nd Cosine ...................................................................................................................................................... 43 The Exponential Exponential Form of a Complex Number ............................................... ............................................................................. ................................................... ..................... 43 Locus on the Argand diagram ................................ .............................................................. ............................................................. ............................................................. .................................... ...... 44 SEQUENCES............................................................. ............................................................................................ .............................................................. ............................................................. ................................................... ..................... 49 SEQUENCES.............................................................. ............................................................................................ ............................................................. ............................................................. .............................................. ................ 50 Types of Sequences ......................................................................... ....................................................................................................... ............................................................. .............................................. ............... 51 Convergent Sequences S equences ...................................................................................................................................................... ...........51 Divergent Sequences ..................................................................................................................................................................... 51 Convergence of a Sequence ....................................................................................... ..................................................................................................................... .............................................. ................ 52 Recurrence Relations .......................................................... ......................................................................................... ............................................................. ........................................................ ..........................52 SERIES ............................................................ ............................................................................................ ............................................................... ............................................................. ............................................................. ...............................54 SERIES ............................................................ ............................................................................................ ............................................................... ............................................................. ........................................................ ..........................55 Using Sigma Notation.......................................................................... ........................................................................................................ ............................................................. ......................................... ..........55 Sum of a Series .................................................................. ................................................................................................. ............................................................. ............................................................. ...............................55 Mathematical Induction .......................................................... ......................................................................................... ............................................................. ................................................... .....................56 Method of Differences .................................................... ................................................................................... ............................................................. ............................................................. ............................... 57 ARITHMETIC ARITHMETIC PROGRESSIONS PROGRESSIONS............................................................. ............................................................................................ .............................................................. .................................... ..... 58 GEOMETRIC PROGRESSIONS PROGRESSIONS.................................................................................... .................................................................................................................. .............................................. ................ 60 MACLAURIN’S SERIES PASCAL’S TRIANGLE ............................................................. ............................................................................................ ............................................................. ........................................................ ..........................63 TAYLOR SERIES ........................................................... ......................................................................................... ............................................................. .............................................................. .................................... ..... 66 BINOMIAL THEOREM.............................................................. ............................................................................................ ............................................................. .............................................................. ...............................68 ........................................................... .......................................................................................... ............................................................. ............................................................. ...............................69 FACTORIALS ............................................................ .......................................................................................... ............................................................. ............................................................. .............................................. ................ 69 THE BINOMIAL THEOREM ......................................................................... ....................................................................................................... ............................................................. .................................... ..... 71 Extension of the Binomial Expansion ................................. ............................................................... ............................................................ ................................................... ..................... 72 ROOTS OF EQUATIONS EQUATIONS ........................................................... .......................................................................................... ............................................................. ............................................................. ...............................76 Page |3 THE INTERMEDIATE INTERMEDIATE VALUE THEOREM ............................................................. ........................................................................................... ................................................... ..................... 77 DETERMINING DETERMINING THE ROOTS OF AN EQUATION EQUATION .................................................................... ................................................................................................... ...............................78 BISECTION METHOD................................................................................ .............................................................................................................. ............................................................. .................................... ..... 78 LINEAR INTERPOLATION INTERPOLATION ........................................................... .......................................................................................... ............................................................. .............................................. ................78 NEWTON RAPHSON................................................................... ................................................................................................. ............................................................ ................................................... ..................... 79 DERIVING AN ITERATIVE ITERATIVE FORMULA ................................................ .............................................................................. ............................................................. .................................... ..... 80 MATRICES........................................................... .......................................................................................... .............................................................. ............................................................. ........................................................ ..........................81 MATRICES........................................................... .......................................................................................... .............................................................. ............................................................. ................................................... ..................... 82 Matrix Multiplication................................................................. ............................................................................................... ............................................................ ................................................... .....................82 THE DETERMINANT OF A × MATRIX ........................................................... ......................................................................................... .............................................. ................ 83 The Transpose of a Matrix........................................................... .......................................................................................... ............................................................. .............................................. ................ 84 Finding the inverse of A Matrix (Cofactor Method) ............................................................................ ...................................................................................... .......... 85 SYSTEMS OF EQUATIONS EQUATIONS............................................................ ........................................................................................... ............................................................. .............................................. ................86 ROW REDUCTION REDUCTION ............................................................ ........................................................................................... ............................................................. ............................................................. ...............................87 Row Reduction and Systems of Equations .......................................................................................................................... 87 DIFFERENTIAL EQUATIONS EQUATIONS........................................................... .......................................................................................... ............................................................. ................................................... ..................... 90 DIFFERENTIAL DIFFERENTIAL EQUATIONS EQUATIONS ........................................................... .......................................................................................... ............................................................. .............................................. ................ 91 Separable Differential Equations.................................... Equations................................................................... .............................................................. ........................................................ .........................91 The Integrating Integrating Factor.................................................................................. ................................................................................................................ ............................................................. ...............................92 Linear Differential Equations with Constant Coefficients........................................................... ........................................................................... ................ 93 – Homogenous Differential Equations ...................................................................................................................................... 93 Non Homogeneous Differential Equations ...................................................................................................................... 94 Differential Equations Requiring a Substitution......................................................... ........................................................................................ .................................... ..... 98 Mathematical Modelling.......................................................... ......................................................................................... ............................................................. ................................................ .................. 101 DIFFERENTIATION Page |4 At the end of this section students should be able to: ln  –     ,,   1. find the derivative of , where is a differentiable function of ; 2. find the derivative of (to include functions of polynomials or trigonometric); trigonometric); 3. apply the chain rule to obtain gradients and equations of tangents and normals to curves given by their parametric equations; 4. use the concept of implicit differentiation, with the assumption that one of the variables is a function of the other; 5. differentiate differentiate any combinations of polynomials, trigonometric, trigonometric, exponential and logarithmic functions; 6. differentiate differentiate inverse trigonometric functions 7. obtain second derivatives, , of the functions in 3, 4, 5 above; . 8. find the first and second partial derivatives of Page |5 IMPLICIT DIFFERENTIATION       0     5    A function which is written in the form is called an explicit function: is stated explicitly in terms of . However, functions such as or are implicit LESSON 4 functions. LESSON 1 Differentiate            2  SOLUTION LESSON 2 determine SOLUTION  SOLUTION Determine   3 3  5 22   3 3  0  333322    SOLUTION   3  3  5   6    6 6 2  3    66 2 3 2  2 62   0 2  2 2  6  2 2 262 26 2 6  2   2 62 6 2 2 663602 2  3 31,13, 3 1 1  0 1   2611262  32         1 LESSON 5 Find the equations of the tangents at the points where Use implicit differentiation to       1 for Determine for with respect to . . SOLUTION .    1         1 2  2   0 2  22    2         5 5  0    5 5  0 4  1110   10  0 1    4   LESSON 3   410  1 1 When for . Gradient at (6, Equation of line: Using (6, ) ) on the curve . Page |6 1  32 6   103    2  10   66232262   32 3   32 6   12  3    2  12 22      6 22     6 2  2   2  2  0 2 1212 2  2   2 1 1   0 2221  0 2 2  0 6    2         22 60 6 6  0 33,33, 2 2 2  0 3,3,3,333,2 2,2, 2   22 2  2221 1122  222 Gradient at (6, 3) Equation of line Using (6, 3) LESSON 6 Find and classify the stationary points on the curve . SOLUTION Stationary points occur when Sub. into and . We have already stated that 2 222220 1 2 3,3,2331 1  2 1 1   2332  1    25 → Maximum  2,2, 2 2   222 1 1   5 → Minimum For For   0 and Page |7           ×      5     3   2    5   12 2   2  3    ×   22 3 31  22 3 3   2   22 3 3  1    44 1 1   √2 7 7   1    4 11  4     122 7 27 2 √   221  7    2 2 7 7−−2  22 7 7  PARAMETRIC DIFFERENTIATION Given that a parameter, then and where is called LESSON 1 Find the gradient of the stated curve at the point defined. ; where SOLUTION when LESSON 2 Find the gradient of the stated curve at the point defined. ; SOLUTION when    ÷   122  272− 2  266  1√2  7   1   26261  112 1 7 7  2  301    ,   2211  8  8−− 24   24    24 1 1      4××   24    6   1   88 18  2   2221211 1  7 when LESSON 3 to the curve Find the equation of the normal at the point where the curve crosses the line SOLUTION when when (1, 7) . Page |8    6 2 At (1, 7): 6 16  3  7  31   10916 16    163   10916     44 1 1     5 5 444 1 1    355      3×  4   32   332 × 14 8      3 1  1     6 6  3 36 1 1     6 6  3 Gradient of normal is LESSON 4 Find and for the parametric equations and SOLUTION LESSON 5 Find and for the parametric equations and . Hence find and classify the stationary point(s). SOLUTION   2  6    ×   23636  11333   3 33    3  33399× 1  1 6 9 6   0  33 3  0  3 3  0   333366131 3 32626 1212 26,26,1212   613  1621 Stationary points occur when Minimum point Page |9 TRIGONOMETRIC DIFFERENTIATION  sec   22  33 sec tan       sec s ec  22  3 tan  se c   ′cos cos      sec s ec   os     2  3 t a n   sec  si n    sseinc     t a n             ≡ 2     1        s e c       csccot   cscsect cotacotn   11 tan    sec     secccsc tann    sec   1 sec   2sec 2secsect sectann      2se 2 sec c    2sec 2 sec  tan  tan     csc   1tan  2se 2 sec c     1t  s e c 1 t a n   cotcosec4t 5   21tan  1tan 1 tan  1tan   3cs+ 3+csc−−12 12     2   tan 1   tan    n      22  tan tan1   tann  22   1   sec4 4sec4t 4sec4 tan4 an4 cot5csc5  5 3cs33csc6612 12 csc12  cot12  1 2 1 2    18 csc12 1 2  cot12 1 2  −− 4  +112csc 2csc4 −2 csc   2csc4 − −         4 4csc 4 csc 4  4 cot 4  4     2csc4 4 4 cot44  4   4csc2csc (e) LESSON 2 SOLUTION LESSON 1 (a) (b) (c) (d) (e) SOLUTION (a) (b) (c) (d) Differentiate the following w.r.t Recall: Given that , show that P a g e | 10 DIFFERENTIATION OF INVERSE TRIGONOMETRIC FUNCTIONS LESSON 1 PROOF Let then Differentiate sin−  . sisinsin−, –  <  <   sisin      1 cos   cos1   cos  sisi  n   1   coscos11si sinn  1     √1 1   sin−   √1 1  −  cos coscoscos,− , 0 ≤  ≤   cos cos    si n 1 1  cossinsisi n  1 ssiinn11cos  cos   1    coscos−    √1 1  −  t a n tantan−,   <  <   tann    sec  1  1  sescec  1  tan  11  tan 1 1   But LESSON 2 PROOF Let then Differentiate But LESSON 3 PROOF Let then But Differentiate .    −   cos− 3  3  cos LESSON 4 Show that if   cos− 3 , then SOLUTION iff cos   33 sincos   3     si3n  sin  111cos    3 3   coscos− 33   1 3    1  3 3    √193      sin−    √1 1  . ′′   cos−     √1 1  . ′′   tan−    1 1 . ′′ In general, given that is a function of , we have P a g e | 11 DIFFERENTIATION OF EXPONENTIAL FUNCTIONS FUNCTION DERIVATIVE     LESSON 1 (a) (b) (c) (d) (e) Differentiate the following   2−   +   11   2−−   223cos3      663cos3  cos3 1 [2] SOLUTION (a) (b) (c) (d)     2 2   12 1 2  +2− 1 1  −1   2−2 121−   −     1  1 2 1  −−2221121    1 13 1131 55   15 151         5     5   1 3 2   15  3 3  2 2   15 33 2 2   15     15153 22     sin  2 2   2  0    sisin    coscos    sisin  c cos    sisin  co coss   coscos  s sin  2  cos (e) LESSON 2 Determine for SOLUTION LESSON 3 that Given that , prove SOLUTION  2 2  2 2  20  coscos  2 2coscos  sisi n   2 sin P a g e | 12 LESSON 4 Determine defined parametrically by  for the equation     1 1   sin− 2     1 1   2    sin− 2   1 2 1  22    ×   √142  × 21 1   √14 and SOLUTION . P a g e | 13 DIFFERENTIATION OF NATURAL LOGARITHMS 2×2 ×+1 1   2l n  1    1   222212122llnn n  2211lnn  1            l n 2  1  1 3 3   2   llnn22 11 11 2l lnn33332222   221 1  2 332 2  222321 1223662222  1 66 24 1 11233 6 2  221911331 2  22  13 2 2 +−   l n  1    ≡ 2  ln11  1ln1     111 1 1   1    1 2 1     1 2 . 1   2 (vii)    LESSON 1 following. (ii) (iii) (iv) (v) (vi) 1  Differentiate each of the   l3lnn377112   llnnsisin4411    ln +   ln22  133  2 ln33 1 1  3  1 3l33ln7772 2   7217 2 2 (vii) (viii) SOLUTION (ii) (iii) (iv) (v) (vi) ln2 1 11     1 ln4co4siscosins444   4co4sicotnt44  22lnlnn   1   22 lnn       2l2ln  1 (viii) LESSON 2 that SOLUTION Given that , show P a g e | 14 LESSON 3 Differentiate SOLUTION   3  l1n  ln 3   ln3   3l nln33 3  w.r.t       ln    −  ln  ln √3 3 l1n  ln1  l1ln2 3 3   1  2  3 3     3 3   3  33 3    3 3    3 3 3 Sub    3 3    3 33 3 3    33 3 In general, if LESSON 4 SOLUTION then If , find P a g e | 15 PARTIAL DERIVATIVES For partial derivatives we differentiate with respect to one variable and treat the other variable(s) as constants. ,,   4ln    ,,   44 ln   2   4ln    4     4ln    8  ,,   442ln      48               LESSON 3 Given that , determine (i) (ii) SOLUTION LESSON 1 Given that ,,    2 2          ,,      2 ,,    2 2  5    3 2 2  52  ,,   22 2   −   4  25 −   4  5   33 2 2      33 22     8843333 2233 2 266 22   33 22     88433 2232222 22 , evaluate (i) (ii) SOLUTION (i) For we differentiate , treating as a constant. (ii) LESSON 2 with respect to (i) (ii) LESSON 4 determine (i) (ii) (iii) Given that , determine    SOLUTION (i) (i) (ii) (ii) SOLUTION (i) (ii) Given that (iii)               2 4 4       3    , P a g e | 16   2      ,,    2,2, 1 ,, 3  2 2 2,2, 1  321  12 LESSON 5 For determine . , SOLUTION LESSON 1 Given that ,,    4 4      ,, 3 4442  2 −−   6  6   ,,     4 42     3  4 2    3 8 8    ,, 8844−       3 8  8  2    , determine (i) (ii) (iii) SOLUTION (i) (ii) (iii) NB: LESSON 2 Given that    l ln s sinn (i)  , determine (ii) (iii) (iv) (v) (vi)      SOLUTION    l ln1 s sin  48       l ln12s sin 2  2223−coscos3−       44  322 cos 23  32   322 sin 23  4   32 cos23  94 sin23    l ln2 s sin  2−   2  2 3cos2 3  2     cos3  3   22   − cos −   2   34 cos23  32  32sin23  22  34 cos23  49 sin 23 (i) (ii) (iii) (iv) (v) (vi) P a g e | 17 INTEGRATION RESULTS SOLUTION    +        ′ √′     FUNCTION LESSON 1 (i) (ii) (v) (vi) ln||   ln||   1  1 ++   1 +      llnn||cos| csec| ossec||  1 lnsecsec     ln|sesectanttan|   lntan2 ln|−sisin||   sin    1 tan−    (vii)  (i) (ii) (iii) or Determine ∫   ∫ −  ∫ −+ ∫∫ tanan+−  ∫ 12∫ 332l2l 5n5  (iii) (iv) INTEGRAL (iv) (v) (vi) ∫  1  22     l2ln n||   ∫−ln|44 1|   ∫ −5  2   2  12   +52 ln|12| 1 2|   ∫ 1+−46  2  2 6 6  9   12 ln|22 6 6  9|   ∫ tansin   cos     sicosn  ∫ 12ln|cos| c33os 5| 5   3335 5   2 ln  ∫ 12ln  33   2 ln  2 ln4     332ln (vii) P a g e | 18 LESSON 2 (i) (ii) (iii) (iv) (v) (vi) Evaluate each of the following. ∫∫ 2− ∫ 2   ∫ ++  ∫∫ 3 √ 6 6  +   ∫∫ csc (v) (vii) (viii) (vi) SOLUTION (i) (ii) (iii) (iv) ∫ 1   2 −   ∫ 2 −  2 2    22 11−−    2   ∫ 222   4  ∫ ++   5−1  −−−   25 2−1 −−−−−    52 1  1      52 1 5−−−    21  −   ∫ √ 6 6      6 6   14  46 6  1 4  636+    2  16  6 6   ∫3  ∫  +          ∫csc    ∫   −   ∫∫ −− +  − ∫   −  1    √2     2 sin− 2   ∫ −− 1 +  3 2   22  1   23 3  222   2 1 1−−   2  3  22  (vii) (viii) LESSON 3 (a) (b) (c) SOLUTION (a) (b) Determine P a g e | 19 (c)  12 1222 1−−  32113 tan−223     4 22  1−−  2 tan−  3    ∫−   Even powers of   and sin   sin   csions211cos2 2 sisin  2 2  sin 1 cos2   2sin22   2 4  cos   cos    cos co s 2  2co 2 cos s  1  1 1 cos2 cos  2  2 cos 1 cos2   2sin22   2 4  LESSON 1 Determine SOLUTION Using the identity LESSON 2 Determine SOLUTION Using the identity P a g e | 20   sin   sinsin  1si ncossisi n   sinsin     sin  1cos 1cos     s sinn    sincn cos    sinn cos  sisin cos    co coss   3   cos   cos     co coss  cos    cos cos  1si 1sin     c cos sin  cos  sin    sin   3   sin 2  sin 2    sin2sisi2 n 2   sisi n 22 1cos 1cos 22   1sisi n 2 cos sin22cos 2cos 2   2 cos2  6   LESSON 1 Determine SOLUTION Since LESSON 2 Determine SOLUTION LESSON 3 SOLUTION Determine cos sin   cos sin     co coss  coscos sin    cos cos  1si 1si n  sin    scoincos s  sisisnin  cos  sin    5  7  LESSON 4 SOLUTION Determine P a g e | 21  tan   tansec  1sec se tcan 1 1   t a n      1si n   cos1 sin    cos  cos   lsen|seseseccttttaann| ln|sec| | s ec   |   |  ln sec sec sec  t tan    tan 2  tan 2   1se secc 2 1   2 tan 2    LESSON 1 Determine SOLUTION Since LESSON 2 SOLUTION Determine P a g e | 22                                      –– –– – INTEGRATION BY PARTS When choosing we use the following acronym. Logarithms Inverses Algebra Trigonometric Ratios Exponentials LESSON 1 SOLUTION Determine         1                LESSON 2 SOLUTION Determine  121 2 3 3  1681 2 3 3    1681 2 3 31414 2 2  3    1681 2 3 31212 3 3    56 2 3 34 1 1    √312 1   √312 1   33  11  22−   3  1 −  3     1  2 12   312 2  √121    3 1 11 2 2   3312 1 2    3 1 11 2 2  1 2 2  11 2 23  1  1  2  12 1 2 2 2   c  cos  cos      1  coscos    sisin  c  cos     sin   sin    s sin  cos cos    LESSON 3 SOLUTION LESSON 4 ∫ 22  3   2 3 3       1 1    22 3 3   12 2 3 3  2 3 3    121 2 3 3   121 22  3  Evaluate SOLUTION Evaluate P a g e | 23 LESSON 5 Determine ln  SOLUTION  lnn   1.ln   ln   1 ln1      lnn 1  1    lnn    LESSON 6 SOLUTION SOLUTION Evaluate       sisin     Determine   lnn    ln   1      3   lnn    3 ln   3   13 ln  19     9 3l3ln  1   LESSON 7   lnn        sin        2     sisin     coscos      sin      coscos    2   cos    22     2  22cosco s cos   sisin   2222   ssisiin— nn — 22sicosconns        sisin   2cos s0sisin   2 coscos   2cos 2cos— — cocoss 022 2co  244 2 2   sisin    sisisisinn        coscos  1 12  1    coscosisisn  2 sisin   2   cos      sisin    12  LESSON 8 SOLUTION Evaluate P a g e | 24  1sin  1 1  1   2 sisin   2 12  coscos  12  sisin 1  5  sisin    2 1 sisin   4 1 coscos   4   sisin   4 sisinsisin 2 2 sisinsisin 1 4 cocoss    5 5 P a g e | 25 LESSON 9 Establish a reduction formula that could be used to find ∫   and use it  4      −             −      −  4   −−    44 44 3 3   44 44 1212  1212 2 2  24   44 44 12 12 12   24        12   24  24          4 4  12  12 24 24 2 24    4 4 12  12 24 2 24      44 4 412 12 12 12 24 24  24      24  24 2 24       ≡ cos ∫     sincos∫ cos−  − −− when . SOLUTION LESSON 10 If show that . Hence, find . SOLUTION STEP 1: Write the integral in the product form   co coss cos−     cos−     1 1 cos−  sisin STEP 2: Integrate by parts or an appropriate method   coscos    sisin    sin  coscos−−    1 1 cos− sin   STEP 3: Simplify   sincos−    1 1 cos−  1 co coss    sin  coscos−−    1 1 cos−  1cos 1cos     sincos−−   1 1 cos−     1 1 cos     sincos   sincos 1−1−−111−1−  1sin  coscos−−  1111−−    sincos    −− STEP 4: Apply the derived formula to the rest of the question   51  4   5 sincos   5    31  2   3 sincos   3  STEP 5: When you have reduced your integral to its lowest form, go back to the original integral &   1   c cos    sisin      15 sincos   45 13 sincos   23 sin   ∫ 1 1  2   1 1  −−    1 1  plug in the final value of . LESSON 11 that SOLUTION If , show P a g e | 26   1 1  1   11−−  1    2   −− 1    2  1 1       11 . 2     12  1 1  2   11−−    12  1 1  2 −− 2   1 1   −−  ∫lnln   ,  ≥ 0.0.  ≥ 1,1,   12  12 −−  ln   22ln   ln −−. 1  ln   22ln   ln −− 22ln    ln   ln −− .  .        2  ln      ln    −−   ln     1   1  ln    2  ln  1  2 −−   112   32 −−   12  32 1 1 3 3  2 1  2 32 1  1  2   4   2    4   2 2   2  LESSON 12 It is given that By considering or otherwise, show that for Hence, find SOLUTION in terms of METHOD 1 Rearrange the equation Take integrals of both sides   14   34  34   12   34  ln    121  34312  3 12  12   83   8  8   8      ln    lln   1ln −−. 1   1   2     2 ln  1  2  ln −−    112 ln 1  2 −−   2   2 −−    4 sisin   1  √16  METHOD 2 is found the same way LESSON 13 By using the substitution , find the exact value of  −16  16√16   1−  √16    Show that  ∫1616 − 11−,− 22≥√23 44cos4cosisin s  2;2;4cos 4cos   6 Deduce, or prove otherwise, that if for SOLUTION , then . Hence find  . P a g e | 27  0;   0  √161    1   16  4 cos    16 4si 4 si n    4cos   √1616sin    4cos   16     16 1 sisi  n  cos   √1sin    cos   cos    1 1  0  6  −16  −. 12 −16 1 6. 22  16 1 1−   2√216     1 116 −− 1616      √16    1 1√16   √16    1 1√161616−−    1 1  √16 1 1   √16 − 1616√16 16 √16 16 1 1  √16    −16  1616√16   1−  √16       √16    −   1616  1  √16     −16  2   1616 1 1−−  −−16 0   1616 1 1−− 2 2 1√12   1616 1 1−− 2 2. 2 . √4.4. √3 2  1616162121√−3− 2 2 √2 2  16 4 4√3    √161    sin− 4 20  6 2 483  4√3   3  2√3 P a g e | 28 PARTIAL FRACTIONS LESSON 2 Denominator with Linear Factors LESSON 1 fractions. Express −−− in partial 2  27   2227 1 1 ≡  2 2   1 1 2  7  1 1 1   2 2   2 3  3 9 3 13 1 2   13 1  2 27    2  272      21 2     13 1   ln|112|2|3l n| 1 1|    ln   2 2    233 1 111333    3 1   3 3 2 11 113 33 33  1  3   3 3  33  1 24   112 2  0 3 13   33  333 2 2 2 1 2 233 1 111333  1  2311 1  213 3 SOLUTION When (a) Express (b) Hence determine SOLUTION (a) Comparing coefficients of in partial fractions ∫ +−−   ++ −− 13 2195195  6    1113132219 3 3 13  2 After factorizing the denominator we get When 13 19       1313 1 19 119 223232 1 331  22 1133  3 121 2 2 453 15   3 202 10   1 616 13 2195195  6    11 1   23 2   32 3 13  2 +−− ∫ ++  ln|11111|3l n|223 22| 2l n|33233|  When When When When When  ++ −− + (b) P a g e | 29 LESSON 3   ++ +++++ + Express ++  + +++ ++ in partial  5   2 2662 22 5   6  5 5  6   2      5   4      5353413 3 333313333  1 5 2626 22 25 2 2225 22 5 2 2   2  2    5  2            3           3  3 3   3   2  5   6  3 3 1   5 254   6  3 321  5 5  2  2 5  8  10 36  4 4   2  5   2  3   3  1   5  5   4 3 2 2 5 6  6   2     3 3  1 3 3  3 3  1  3 3  222 232 4 5        2  2  2  2   5 −+ +   3  1  1  7     l n  2  2       −+   2  2   5 ∫ +    3  1  1 7      l n  2  2             2  2   5  2  2   5 25 5 2  25511 2 211 5     2 27  5    32  22    l n  2  2 25511 2     11  ln 2 2  32 ln| 2 2  5|  7   1 11 4 4  2 5 5 2  11   ln 2 2  32 ln| 2 2  5|  7   1 11 2 2  3 7   −   | |  1 1    l n  2  2  l n  2  2  5| 5  t a n   25 2 2 2    22   0  2 5 5 2    2 51 1−  ln   5t5 tan    LESSON 1 Express fractions. fractions and hence determine in partial SOLUTION SOLUTION LESSON 2 Express and hence determine SOLUTION in partial fractions . P a g e | 30 LESSON 1 Express fractions −−−+ in partial 6363 9 99  3636 999  3 3339 3 3339     3 3   33  3 393  3 3   330  93  3 91  9  11  1 26363 9 99   1   32 3   333 SOLUTION When When Equating coefficients of LESSON 2 Express +++ −− fractions and hence determine in partial  5   2 2662 22 5   9  2  9  24 2  9 9  24  24  2 4 94933218   2 2 3 3 2 2929324323233322333 3   2 2   3   2 1535 502 25  22  2 02 9 9 24  2  3  4 4 3 3 18  2 2  3 3  2   4 499332184  SOLUTION When Equating coefficients of When    22 2  3  3 3 3−−   2ln| 2 2|   33 3   P a g e | 31 ≥      +−− + +  2 3311 4  2 3311 4         1 1 2 3 3    4   1   1 1   1 1                           24  4 4  1111   3    3 23 3 2  3 2 3311 4  22  3  4   12 1  2 3311 4   2 2    3    4−     12 1   22 l ln  4 2l 2ln| 1 1|   +  + +− −+ ∫ ++−−  If for , has degree and has degree , then quotient has degree LESSON 1 Express in partial fractions and hence determine SOLUTION LESSON 2  2 3  2       32 21 122222  12221 1 2222  2 1 1    2    1414  5 11 5    4 2 5  311102 →     14  2 3  2 3 11 2 1 312222  2  101022  1  5 2 2  2 1 1 2 2    32   1120  221 1   145   21 2   32  1120 ln|22  1|  145 ln| 2 2|   SOLUTION Express fractions and hence determine in partial When When Equating coefficients of −−− ++ ++ ++ +      19  19 4 44  21  LESSON 3 Express in partial fractions and hence determine  3 3 2 2 11   33119224441 121       3 3   2 2   1 1    1 19 4 44 21       3 3   2 2   1 1   1193 344242211 1   2 2 1 1  3 33 1 1   3 3222 63  2 11    1 42  2 SOLUTION When When When P a g e | 32   1  11  666 5    3319 19 2244 44 1 1221    5   33 3   21 2   12 1 19 224441 121    3319   5 5  3 3  1  1 22 1  Equating coefficients of Equating coefficients of  3 3  2 2  1 1  2 5 5  3ln| 3 3|  ln| 2 2| 2l n| 1 1|   P a g e | 33 TRAPEZIUM RULE (NUMERICAL INTEGRATION) Introduction    1 1 The area under the curve equal width (as shown above) can be estimated by finding the sum of the areas of trapeziums of −   ℎ2     ℎ2     ⋯  ℎ2 −    ℎℎ2        ⋯−    2   2    ⋯ ⋯ − The area of a trapezium with parallel sides The width of each trapezium is where and and width    ≈ 2     2   ⋯−    1  1  −  0.4 Using 5 trapeziums, estimate   0.00.4 00.141111 1.1616   0.1.01..82  0.1.82 11 11  1.2.64446444   1.12.6 21.16 1115 3.5656    1 1  ≈ 222501 5 5 2 21.1.161.642.443.566 ≈ 4.72 SOLUTION Width of each trapezium is is given by the formula is the number of trapeziums. Thus we have area under curve is LESSON 1 ℎ      P a g e | 34 COMPLEX NUMBERS INTRODUCTION          ̅     If then is its conjugate and vice versa. It is also important to note that the product of a complex number and its conjugate is which is always a Real number. The conjugate is also denoted .         ∗    √1   1 + :       2   → ̅  22  ,,        3 2        26 32    62131  3 5 6 5 5 √16  111616  √1√1√16  4 +      − √18  9211  3√2 37 52  552 2  3151355 7272414152  5 2   11  4512 214     ±      ±±    ±±  1 2929 4129  62i 6 2i  54i 5 4i 116 5 25  2  4 1 18     1  82 8 2  18  12  12  9 22334 368861 6166 6  8 32 3 254 5 4  15151515  81212 12121010 10 810 8   23 23  2 2                Complex numbers are written in the form where and are real numbers and is the imaginary unit such that or Sometimes the letter is used to denote a complex number, . A complex number can also be written as an ordered pair of its real numbers, . is also known as the real part i.e. (i) First we find the conjugate of the denominator Multiply the numerator and the denominator by the conjugate is also known as the imaginary part i.e. SQUARE ROOT OF NEGATIVE NUMBERS Thus complex numbers can be used to find the square roots of negative numbers. Examples (ii) Express With this extension of the number system we can now solve equations which we once unsolvable. For example, (i) (ii) (i) (ii) (iii) (iv) in the form P a g e | 35 SOLUTION  1 11 0   ±√±±√1  3333±33 40 413   3±   2 1 3 ±2√3   3± 4 232±2√311 4 22221 ± 022 4 441   2±2 ± 12 24 √8   2± 48311   2±22 ±4   1 ±8√2√33  4± 4     22 2 2  1515 10 10  0      22 222 2    1510 1 510  0  222 2   1 5  1 0   0 1510   2 2 ± 2221 44111510 2 2 ± 22 2 222 2 2 6 60  40   22 2 2 ± √28 6 60  40   22 2 2 ±2 √6032   22 √606060323232   (a) LESSON 1 Find √1515  88 SOLUTION We assume that the square root of a complex number is a complex number √1515  8      15  8     15  8     2 2     15 2 4 8  4    15 16    15 15  15 16 16  0  16 16 1 1  0   16   1   ±1   ±4 √15  8  4   or  4    ±±4   Invalid since is real Thus we see that a complex number has 2 square roots, which are complex numbers. LESSON 1 (a) (b) (c) Solve the following equations  31 1330  0 4 2 2  1 (b) (c) LESSON 2 SOLUTION Determine such that P a g e | 36 6060  3232 60   2 2   16 →    16 25616    60  6025 60 2566  0 60  64464 4 4  0   ±2   2 82   162  8   2 82  216  8 2 2 ±2 82 82   22 2 2 2 82 82  5   22 2 2 2 82 82  3  2   22         0    ±√± √2  44    4 4,, > 0   −±−±√−    2 ± 2  When When For the equation Letting , we have As a result we can conclude that if a quadratic equation has complex roots they occur in conjugates. 2   2 In general, if a polynomial has complex roots they occur in conjugate pairs. For example, if is the root of a polynomial equation then is also a root of the same equation. Recall:      0           sum of root s  prproduct of roots  0          0                    ,,  andand  543   5   5 5555555525  0  25 25  0   4  3    4 3  3 43 4344 34334 3 25  8  8 8 25 25  0 12        15  0 If , then where and and are the roots of the equation i.e. Also, if then , where and are the roots of the equation. LESSON 1 Given one root find the equation (i) (ii) SOLUTION (i) Let then Equation is (ii) Let then Equation is LESSON 2 equation roots. Given that is a root of the , find the other 2 SOLUTION Since complex roots occur in conjugate pairs and a cubic polynomial has 3 roots one root must be real. 21,   1  2,  ∈ ℝ 12 1 2 112 2  3 1 1 2    1 Let P a g e | 37 INTRODUCTION      ,,  A complex number can be represented on a diagram called an Argand diagram as (i) a point with coordinates (ii) a vector              5 8   2   58        58 58  2    7  9  58  2    3  7 LESSON 1 Find and and and for . Hence, represent on Argand diagrams. SOLUTION (i) (ii)        | |   |  |  √       13 4   511  √3 || 11 1 1  √2 The modulus of a complex number, , is a measure of the magnitude of , and is written as . Thus modulus LESSON 1 (a) (b) (c) (d) SOLUTION (a) . Determine the modulus of P a g e | 38 || |334 344 |  33 4 4  5   11  √3    || 51 1  √3  11   √√3  2 ||  5 |  |   3  4   22    3,3,4 2,2,11  (b) SOLUTION (a) (c) (d) LESSON 2 what is If ? and , (b) SOLUTION We are trying to find the distance between and . In other words, what is the distance between the points and on the Argand Diagram? (c) (d) |  |  |34  2  |  |32 32  41  41|   32  32  41  41  √50   arg t–  <an≤    113  4   511  √3 The angle is called the argument of ( ) where is the angle the vector representing the complex number on the Argand diagram makes with the positive real axis. Thus . To avoid complications we use and this is known as the principal argument of . LESSON 1 (a) (b) (c) (d) Determine the argument of arg 1    tan− 11 4 arg  34  2. 2t1 an− 43 arg 1 1√3  2 tan− √13  3 arg  5  P a g e | 39      5c 5 c os o s   s  s i n n    2 3 2 3  cosco s     sisin   6  5c5 cosos6si n  ∴     cosco s    sisin   || 5c55cos os 6   sisin 6   cos    sinn  ar g    6 1 –   co c o s    s i n   √2cos4  sinn4   c o s    s i n   –   coscoscoscossinsin      113  4   511  √3 coscoscossinncosscions   sinn  sisinncos  si n   coscocscossisincoscos sisisisin ncossisin   √112cos    sin  4 4  coscos     sin     53coscos42.2.211   sin2.2.211     +  +           2cos 11  √3 2   sinn 2   c+ o+         s  s  s i n  c o s     s i n         c o s     s i n n  c o s    s i n n    55cos    s3inn 3 scocoss co  cosssisicoscosnnsisi sisi nnco ncos s  sinsin  cosccoosco   os coss  sisi n  sisinn  sisincos cossin   coscos     sin     c  c os o s    sisi n      cos5cosisisnsi 5cos sinn    ̅       c o s     s i n n           cos c os     s i n   || cosco1s  si  sinn   | | | | |               3 | |       ar g 5sisin55cos arg ||    argarg a argg    arg      arg   arg   5sin 3cos3 If Therefore, is has modulus and and argument in modulus then argument form LESSON 3 and LESSON 1 Write the following in modulus argument form (a) (a) (b) SOLUTION (c) (d) (a) Prove that for (b) SOLUTION (i) (ii) (iii) SOLUTION (b) (iv) LESSON 2 Find the modulus and argument of the following NB: If (i) (ii) SOLUTION (i) Furthermore, we can conclude (a) (b) (ii) (c) (d) then P a g e | 40 LESSON 4   11    1  √3   Given that , determine (i) argument form. and (ii) and in modulus SOLUTION ||  1  √3  2 arg  tan− √13  3 ||  1 1 1 1√2  arg  t tan− 1  4   2√2cos7  si7n     2√ 2 2 c os    s i n  √ 12 12   √ cos  si n     √2cos 12 sin 12  (i) (ii) – P a g e | 41 INTRODUCTION  ccosos   sisin     coscos  sisi n  2 2 sincn coss   coscos 2 s sin22 then This can be extended to give   cos cos sin Use De Moivre’s Theorem to prove the fol owingidentities coscos 4 ≡ 8cos   8 coscos  1 1 LESSON 1 SOLUTION   4 c  cos4 sin4  cos    sinn        cos   4 coscos   sinn 6co  6coss   sin  4c4 coss sin   sin   cos  4 4cosos  sin  6 6 cos sin   4cosos  sisin  sisi n  cos4  cos  6 cossin  sisi n     cos  6 6 cos  1 co coss   1cos 1cos 1  cosos   cos  6 6 cos   6co6 coss  co coss   2 cos   1  8co8 coss   8 coscos  1 1 Use de Moivre’s theorem to show that    sisin 5  co coss  sinn   cos sin  s sin  ,,    5  c   cos5  s i n 5  c o s     s i n n   1co1 coss  5co 5 coss  sin 10cos  sin 10cos  sin 5c 5cos os sin 1 1sin  cos  5cos  sin  10cos  sin  10cos  sin  5cos  sin  sin  sisin 5  5co5 coss  sisin  10co  10coss sin  sisi n    5,   10, 10,  1 When Equating real parts LESSON 2 where and are integers determined. SOLUTION When Equating Imaginary Parts P a g e | 42 Use de Moivre’s theorem to show  3t 3 t a n   t a n tan3  13tan    c   cos3  s i n 3  c o s     s i n n   cos  3co 3 coss  sin 3cos 3cos sin  sin LESSON 3 that (ii)  −√ 1   √3 – 1t√3tan − √13   √√3  2 By DeMoi3 vre’s Theorem 3−  cos  3 3cosos  sisinn 3co  3cosssisin    sin  11  √3−  −2 cos os 3si n  3 sin3 3  2 cos cos 3   cos3  cos  3 3 cos  sin   118 cos    sinn    1  1 8   sisin 3  3co3 coss  sisin  sisi n    18 sin3   – √3√3 tan3  cos3    3 coscoss 3s3incossss ssiinn  3co 3 coscoss si  n  cossin  3co  1    3      3  1  2 √ √ coscos   3cossi n   cos 1  −   t a n    an  t tan  √3  6   3t3 t13tan √3    2 cos6 s sinn6 By De Moivre’s Theorem   cos si n  √3  2cos6si n 6 −√ 6  cos106   sisinn10  2 1 0 si n  5 5 cos  1024cos     s i n n   3 3 BycosDeMoisivren’sTheor e m   12  √23    1024 4 4 cos12× sin12×   1coscos 34sin3 4  51512  512√12√3 Rewriting form in Modulus Argument SOLUTION Equating Real Parts: Equating Imaginary Parts LESSON 5 Express argument form. Hence, find . SOLUTION LESSON 4 (i) (ii) SOLUTION (i) Find the value of in the modulus in the form P a g e | 43 INTRODUCTION sin cos 1  cos  1  sisin  coscos    sisin    cos    sinnco. coscos s  sinsisin  ∴   1  ccos    sin   ccos    sinn  2c 2 c o s    1  ccos    sinn  cos    sinn  22 sisin  By De Moivre’s Theorem 1   coscos    sisin ,so , so that  c1os os   sisin  1     2co2 coss  and     2sin  cos    1   4 4. 1 6 6. 1 4 4..1  1   4 4 6 6  4  1    14  1  6   cos   sin,n ,     2co2 coss  2cos   2cos2cos 4  42cos2 ∴ 2cos 2cos2  6 16cos 1 2cos2cos 4  8 coscos 2  6 cos   8 coscos 4  4 cos 2 3 3 Expressions for powers of and in terms of sines and cosines of multiples of can be derived using the following results If then LESSON 1 Express cosines of multiples of . SOLUTION If in terms of  s i n    1   3 3. 1 3 3..1  1   3 3  3  1    1  3  1   cos   sin,n ,    2sin ∴ 22sin  2sin 3 3 322sin 8sin  1 2sin336sin sin    4 sin3  4 sin LESSON 2 multiples of . Express in terms of sines of SOLUTION If From Maclaurin’s Theor em cos   1  2!2!  4!4!  6!6!  ⋯     sin    3!3!  5!5! ⋯ ⋯ !  !  !  !  c o s  si n   1 1    ⋯ coscos  sisi n    1 1   !  !  !  ! ⋯ ⋯      1            11  1       Then This series the expansion of appears to be similar to i.e. Looking at the powers of P a g e | 44     111  1        Now let’s try the expans ion     11    2!2!  3!3!  4!4!  5!5!  ⋯            11    2!2!  3!3!  4!4!  5!5!  ⋯   11  2  4  ⋯     3!3!  5!5! ⋯ ⋯   cos    sisin        cos    sinn      11+  √− |11| 1 1 1  √2 arg   tan−1  4 √+2− √  1     √3      arg  || 1 1 1  √2 4   ||  −√31 11 2 arg  t tan √3   6 ||    √22   5 arg√2 arg  a arg   4    6  12   2  Grouping Imaginary and Real terms NB: If    ∗  − then        We will be using the notation LESSON 1 If the point in the complex plane corresponds to the complex number , find the locus of in each of the following situations.  |||223|  4 | 3 3 |  2 | |  3   0 ,0, 0 ,,   ∴ |    0 0 0|  3 | |  9     3 (a) (b) (c) SOLUTION (a) The distance between the point and the point representing the complex number is 3 CARTESIAN FORM LESSON 1 Express the following complex numbers in the form . (a) i.e. a circle with centre at (0, 0) and radius 3 (b) SOLUTION (a) (b) Let and (b) | 2 2|  4 2,0 2,2, 0 ,,       ||  2|2 4 0|  4 | 2 22 | 4 4  2 2    16 Circle with centre and radius 4 The distance between the point and the point representing the complex number is 4 CARTESIAN FORM P a g e | 45  7 77   11227 0   2    4    , 00 Circle with centre (c) | 3 3 |  2 3,3,1   3,  3, 1  ,,  || 3 33 1 1|| 2 2  3 3   1 1  2 Circle with centre and radius 2 The distance between the point and the point representing the complex number is 2. CARTESIAN FORM LESSON 2 Determine the Cartesian equation of the locus of points satisfying the following conditions. (a) (b) 2−|−  3√|3 || + 2|  3|  || 22||   33 33|||||| 4444324 3 24 3636    3 33 8 824 2412 36360 0    4 4  40,0, 4 −+−  √3 || 1 1|  √1|3| √232||    2| | 1 11 | √33|222|    3 1  2 222 21 1 14   1 2  123 12 3   14 11 11  0 SOLUTION (a) CARTESIAN FORM Circle with centre (b) and radius 2 and radius √ . ,,  LESSON 3 Sketch the locus of the point representing the complex number , given that . Write down the Cartesian equation of the locus. |  3|  |  2  5| | 3|  |  2  5| |   3 3|  |    25 |  25 ,,       0, 0 , 3 ,,  ⊥ 2,2,55 |    0  3|  |    25 25| |  3 3| | 2 2   55|    3 3   2 2   5 5     6 6  9   4 4  4   10  10 2 25 1616 4 4  20  0 4  5 5  0 arargg 3 332   SOLUTION Rewriting The distance between the point , representing the complex number , and the point is equal to the distance between and the point . Therefore, we are finding the bisector of . LESSON 4 Describe and sketch the locus of the points satisfying the following conditions. (a)  (b) CARTESIAN FORM SOLUTION (a) arg 3 3   P a g e | 46 arg    30 3 0   4 arg− 3 3   4 tan  3 3 4  3 3 t3an>43 1 3,3, 0 ; This is the half line starting at positive real axis. (b) SOLUTION  | 2|  1 |  0 2 2|  1 Circle with centre , not including (3, 0), making an angle of | 2| ≤ 1 LESSON 6 Shade on an Argand diagram the region in which . with the  arargg32 32 32  3 ar 2g2 3 3  2 2  3  3 23  t√a3n333√3√3   √3– 2  3√3  > 33,3,2  ; The half line starting at , exclusive, which makes an angle of with the positive 0,0, 2 and radius 1. LESSON 7 (a) Sketch on one Argand diagram: (i) the locus of points satisfying |  |  | 2 2| arg    4 |  | ≤ | 2 2| –  ≤ arargg   ≤  |||  |022||  |    20 2 0|   0 ,0, 1 ar2g,2, 0     arg–    0  0,0, 14   (ii) the locus of points satisfying real axis. (b) Shade on your diagram the region in which and SOLUTION (a) (i)   22    1     21  11  11 1  2,2, 1 LESSON 5 where Describe and sketch the locus of SOLUTION Using vectors This is the line passing through the point and parallel to the vector , i.e  This is the perpendicular bisector of the line segment joining the points and (ii) Half line starting at making an angle of axis. , excluding, with the positive - P a g e | 47 (b) (ii) LESSON 8 arg   s sin−    LESSON 9 (a) Sketch on an Argand diagram the locus of points satisfying the equation |  6 6|  3 | 6 6|  3 | | 1 <  ≤ 1 (b) It is given that satisfies the equation . (i) Write down the greatest possible value of . (ii) Find the greatest possible value of , giving your answer in the form , where . SOLUTION (a) Circle with centre arg  0,0, 6 (a) On the same Argand diagram, sketch the loci of points satisfying: (i) (ii) |arg3 3 33|  5  (b) (i) From your sketch, explain why there is only one complex number satisfying both equations. (ii) Verify that this complex number is 74 – SOLUTION and radius 3 3,3,3,3,110   (a) (i) Circle with centre (ii) Half line, starting at making an angle of real axis. (b) (i) 9 is the largest possible value of | | and radius 5 , exclusive, with the positive . (b) (i) There is only one complex number satisfying both equations since there is only one point of intersection due to the P a g e | 48 73 ||743 | 4 3 3 |  4434 | 33 5 ar garg744 444 3    tan− 44   34 half-line which starts within the circle. (ii) If is the point of intersection it must satisfy both conditions. SEQUENCES P a g e | 49 At the end of this section, students should be able to: {}   1. define the concept of a sequence of terms as a function from the positive integers to the real numbers; 2. write a specific term from the formula for the th term, or from a recurrence relation; 3. describe the behaviour of convergent and divergent sequences, through simple examples; 4. apply mathematical induction to establish properties properties of sequences. P a g e | 50 SEQUENCES INTRODUCTION A sequence is a list of numbers which obey a particular pattern. Each number in the sequence is called a term of the sequence. These are usually denoted where is the first term, is the second term and is the term. In some cases the sequence can be defined by a formula an expression for the term.  , , , … , −,     –    +4+ 1   11 ++     +   441 11 1  3   4423 11 11  71111   4445 11 11  15191159 3, 7,++11, 15, 19, ….   1122 1 11  23   33 2 1  42   44 3 1  53   55 4 1  64  3 54 55 6 2,2, 2, 3, 4, 5, …  1 1   21  12   21  14   21  81   2  16 (d) LESSON 1 Write down the first 5 terms of the following sequences: (a) (b) (c) (d) SOLUTION (a) (b) (c) 1 1211 3211 1 2, 41, 18,++16, +32, …   11 1 12 1  12 2   11 2 3 1  3 3   11 3 4 1  4 4   11 4 5 1  5 5 1  112  35  14  56 2 ,  3, 4 ,  5, 6, …    5, 8, 11, 14, …. 2, …  ,  ,  ,  , , … 1,1,1, , ,, , , , , …, … × ,  ×, × , × , × , …. 2,2, , ,  , … LESSON 2 For each of the following sequences determine an expression for the term, . (a) (b) 8, (c) (d) (e) (f) (g) 6, 4, 2, 0,    SOLUTION 3  3  2 2 2   10  2    + (a) Consecutive terms differ by 3 therefore we try . To create the right formula we add 2 i.e. (b) Consecutive terms differ by therefore we try . To create the correct expression we need to add 10 i.e. (c) The numerators are the natural numbers and the denominators are two more than the numerator i.e.  (d) Ignoring the signs, each numerator is 1 and the denominators are the natural numbers . Since the signs alternate between positive and P a g e | 51 11++   11++ +       1  1    ++  1 1   + negative, starting with positive, we use . Therefore (e) Each numerator is 1 and the denominators are powers of 2 i.e. (f) Each numerator is 1 and the first number of the denominator is and the second is . Therefore (g) The numerators are the natural numbers but they begin with 2, i.e. and the denominators are the square numbers. Therefore The sequence above diverges since it does not converge to any specific value. A sequence can be classified as being convergent, divergent, oscillating or periodic. Convergent sequences as the name suggests converge to a definite limit. →lim    This oscillating sequence above is divergent. The sequence above is convergent because it is tending to a value. This divergent sequence is PERIODIC as it consists of a set of values which are constantly repeated. The repeating pattern of the sequence consists of three values therefore the sequence is said to have a period of 3. The sequence above is OSCILLATING and converges. Divergent sequences are sequences which are not convergent. P a g e | 52 LESSON 1 Determine which of the following functions is convergent or divergent. If the sequence is convergent, determine the limit of the sequence. (a) (b) (c) (d)   +  − −   √+   lim  3 →  →lim  31 1  →lim   1  →lim 1 3 1 3 lim   →  →lim  77  →lim 7 1   →lim 1  7 0 lim √−+12 →   1 2  →lim √3  4  →lim 434 44 1    4 4  4   1  →lim  3  4  SOLUTION (a)   4  1 4  →lim  3  4  4 1 4     →lim 3  4 l→im    →lim 1 By L’Hopital  0→lim 3 DOES NOT EXIST Not convergent (d) Convergent and converges to 0. LESSON 1 following A sequence is given by the   4 +    3 converges and it converges to 3 (b) is convergent and it converges to 0. (c) Write down the first four terms of the sequence. SOLUTION   4   +   3 3  7   +   3 3  1010   +   3 3  1313   1,1,  +    33 1, 1 ∈1ℤ+ 1,1,  ∈ ℤ+   1   1 1  1  1   1 LESSON 2 defined by A sequence of positive integers is Prove by induction that SOLUTION When , P a g e | 53        1 11 1 +   1 1  1 1 +    33  1    11 1 133  1      13 1 3     2 2   1   2 2  1  1   1 1 1 1 +     1 1  1,  ∈ ℤ+   5  +  2++11 2 2   2+ 1 1   2 1 1   5  +      2 +1 1 +  2 1 1 ++  2 2++  2 +12  22+2   1  2 1 1 +  Therefore is true Assume true for Now, Therefore, is true when Mathematical Induction LESSON 3 and is true. Hence, by A sequence is defined by . Prove by induction that . SOLUTION Therefore Assume is true is true for Now, Therefore is ture whenever is true. Hence by mathematical oinduction   2+ 1 1 . P a g e | 54 SERIES At the end of this section, students should be able to: 1. 2. 3. 4. 5. 6. 7. 8. Ʃ use the summation notation; define a series, as the sum of the terms of a sequence; identify the th term of a series, in the summation notation; define the th partial sum as the sum of the first terms of the sequence, that is,       =  ; apply mathematical induction to establish properties of series; find the sum to infinity of a convergent series; apply the method of differences to appropriate series, series, and find their sums; use the Maclaurin theorem for the expansion of series; 9. use the Taylor theorem for the expansion of series . P a g e | 55 SERIES   2 =  2 INTRODUCTION , , , , … ,          ⋯ Given the sequence corresponding series is , the                         ⋯  is the 11++ partial sum where: the first partial sum the second partial sum are powers of 2 i.e. partial sum LESSON 1 Write each of the following series using sigma notation. (a) (b) (c) (d) (e) (f) (g) 581114⋯   86420 86420 2  2  ⋯           ⋯        ⋯ 1    ⋯ 1     ⋯  ⋯ ⋯ ×2   ×  ×  ⋯× ×    33  2  3  2 = 2 2 102 102 = 1  ++  1   1   1 =    1 = 2− (v) Each numerator is 1 and the denominators the third partial sum the  (iv) Ignoring the signs, each numerator is 1 and the denominators are the natural numbers . Since the signs alternate between positive and negative, starting with positive, we use . Therefore   1 1  1 =   11  1 1   1  1  =  (vi) Each numerator is 1 and the first number of the denominator is and the second is . Therefore (vii) The numerators are the natural numbers but they begin with 2, i.e. and the denominators are the square numbers. Therefore SOLUTION (i) Consecutive terms differ by 3 therefore we try . To create the right formula we add 2 i.e. (ii) Consecutive terms differ by therefore we try . To create the correct expression we need to add 10 i.e.  (iii) The numerators are the natural numbers and the denominators are two more than the numerator i.e. + The following standard results can be used to find the sum of various series.      1          1 ,     1  1 2 2   1 , 2 6 = = =   4   1 ∑∑== 1 1 LESSON 1 Find each of the following sums (a) (b) SOLUTION (a) ∑==  1 1 P a g e | 56    =     =   =  4 (b)  406 4 1 124 1 1  42 4  1 ∑=    =  =    16164 471161 1 61  94 9 1 1 ∑∑====11221 1 ∑== 1 1 1 1  =  1 1  =   = 1  611112222116   1 1226  1  6  22 3 3616  22 3 36 156  22 65 1 1 ∑=  26 2      =  2 2  =  2=   4  1 1 2 2 6   1122  1 LESSON 2 Express each of the following in a factorized form. (a) (b) SOLUTION (a) (b)  31 111334 41121 141222111   1 133 3123  8  4  12 LESSON 1 that Prove by mathematical induction    1     =   1 1  12   1 1 33  2     1     : =   1 1  12   1 1 33  2 : 11 1 1  121 11 1 131 2 2 0 0          1     : =   1 1  12   1 1 3 2 2 +   1       +: =    1  12   1  1 1 1 1 3   1  2  121  1 1 2 233  5  121  1 1 2 23 5 5      1 1 for all positive integers . SOLUTION Therefore, Assume Now, is true. is true for +  term +  121  1 13 2 2   1 1 11 1 1  121  1 1 1 133  2  1212121 1  121  1 1 1 13 2 2 12 1 P a g e | 57  121  1 133    2  12  12  121  1 133 11  11 10 10  121  1 1 2 233  5 +    1 1  1       1     1 3 3   2 12 =     1 1   =   {={  1  } Therefore is true whenever is true. Hence by mathematical induction   :1: 1: 111112 2111 21211    1    =   1 1  3   1 1   22 ∑==  1 1  ∑==  ∑==  (iii) for all positive integers . If , then LESSON 1 LESSON 2 (i) Express in partial fractions. (ii) Use the method of differences to show that (i) Show that  1 1 2 2   1 1  1 ≡ 33 1 1 (iii)Write (iii) Write down the limit to which (ii) Hence use the method of differences to find an expression for   1 1 = ∑∑===  1 ∑1= . 1111222 211111 1  33 1 111 2 2     ∑= 1: 1 1  121311120211  1   2:3: 233445 23 231234    4: 456 4 435  converges as (iv) Find (iii) Show that you can obtain the same expression for using the standard results for and (ii) tends to infinity. giving your answer to 3 significant figures. SOLUTION SOLUTION (i)  16  1 12 1 1  2 3 1 1  16  1 122  1  6  1 1  16  1 12  1  3  16  1 122  4  3  1 1 2 2 4    1 1 2 2  4 3 2 1 =  1 1 2 2  2   1 1   2 2  4 =  1 1 2 2   4   =  1 1 2 2 (i)   +  + 4+ +++ + 1  1 2 2   2 2       1  1   0 44  02 12 2     1 4  11  1112 12 When When P a g e | 58 3  3   2 42  222  222  1 1  4  2 3  1 ++1+1+2 2   1 1  2 2 ∑= + 4   2 3 1     11:1 2 22  3  111  2 2 1 2 3 2 3   2: 22  33  141    3: 32  43  51    4: 4  5  6      1: 2 12 1 3 3  111 1  : 4     1 1   2 2 =  1 1 2 2  21  32  22   11 1   13 1   21 2 When (ii)  32   12 1   21 2  → ∞4   3 2 1 =  1 1 2 2  2    1    2 3 2  1 2 1  1 1 2 3 2   ∑= +++++    + +  =  141 2 2  =  141 2 2  0.3210410012  10021   32  4912  4921  (iii) As (iv) ARITHMETIC ARITHME TIC PROGRESSIONS PROGRESSIONS , , , … −, , …    −                2 2      1 1 for every  > 1 INTRODUCTION A sequence is called an or there exists a constant , called the such that if That is Therefore, LESSON 1 Find the common difference for each of the following arithmetic progressions. 3, 5, 7,2 9,7,11, … … 2, 5, 8, 11, … 3,3, 5,5, 53 5 37,7, …2 8,8, 3,3,222,23 , 55 5 2,2 , 5,511, 88,8, 3  11,… 11, 15, … is an arithmetic   −   4  1 −  4 1 1  1  44 5 5   −  44  1  44  55 4  4 (a) (b) 8, (c) 3, , SOLUTION (a) (b) (c) NB: Any pair of consecutive terms can be used. LESSON 2 Prove that the sequence 3, 7, progression. SOLUTION We need to show that Therefore, is a constant. P a g e | 59   5 2 2 LESSON 3 The sum, , of the first terms of a sequence is given by . Show that the sequence is an arithmetic progression with common difference 10.     −  5 2 2   1 15 1 1 2 2  5 2  2   1 15  5  2  5 2  2   1 155  7  5 2  2  55 7  7  55 7 7  5 2  2  55 1 12  7  5 2 2  5 12 12  7  1010 7 7  −  1010  1 1  7  1010  10  7  10 17     −  1010 7 7  1010 17 17  10 SOLUTION 502  1111  120120 502 21 11 13535 70 3636 25050  3636  12   30602 LESSON 2 The last term of an arithmetic progression of 20 terms is 295 and the common difference is 4. Calculate the sum of the progression. SOLUTION   20,20,   295, 4 201 201 141295   219 202 2219219  2020  14   5140 We need to determine LESSON 3 The sum of the first 6 terms of an arithmetic progression is 54.75 and the sum of the next 6 terms is 63.75. Find the common difference and the first term. SOLUTION 622254.756 1 1  54.75 6 15  15  54. 54 . 7 5    5,   9,   13   54. 75 7 5  63. 75 7 5  118. 11 8. 5  12   55101 1 011 14 1222 66  121 121118.118.5  118.5  66 2  41 6 15  15  54. 54 . 7 5  12 66  66  118. 11 8. , , , … ,       ⋯ ⋯  ×2: 5    3066  109.118.10119.8.55  30   12 12 66 36  9   2 22   1 1 1   8.4 5 250   120 LESSON 4 If the first three terms of an arithmetic progression are 5, 9, and 13, what is the value of the 10 th term? SOLUTION (1) The common difference, , is 4 (2) Solving (1) and (2) simultaneously Sum Formulae for Finite Arithmetic Sequence If is a finite arithmetic sequence, then the corresponding series is called a finite arithmetic series. The sum of the first terms of the series, which we denote , would be stated as LESSON 1 Find the sum of the even numbers from 50 to 120 inclusive. SOLUTION (1) P a g e | 60 GEOMETRIC PROGRESSIONS , , , … , , … INTRODUCTION A sequence is called a  or there exists a nonzero constant , called the such that if        −     ,  > 1    27, … is a Geometric Progression.    33−, −−  3−,   3,   3 −  33−  3 −  33−−− −−− − 3 Therefore, or LESSON 1 Prove that the sequence 1, 3, 9, SOLUTION We need to show that is a constant. LESSON 2 The first and fourth terms of a geometric progression are 6 and 20.25 respectively. Determine the 8 th term of the progression. SOLUTION   6  20.25 √20.3.6327553.1.3575   63  6561  2 64 Since length cannot be negative Sum of a Geometric Progression The sum of the first terms of a G.P is given by LESSON 1 The fourth term of a geometric progression is 6 and the seventh term is . Calculate (i) the common ratio, (ii) the first term, (iii) the sum of the first eleven terms. SOLUTION (i)     486      2 486  8 8 8 6 6   34−−  −3   4 1 22221 1   512. 25      1         51  −1  5 11− 4 5  5 13  5 45 1−3 4  5 413  5 41−4 3 5 44 313− 114 3 13 (ii) (iii) LESSON 2 LESSON 3 The lengths of the sides of a triangle are in geometric progression and the longest side has a length of 36 cm. Given that the perimeter is 76 cm, find the length of the shortest side. SOLUTION Let longest side be and shortest side be .    36   76  3636  363636367676 369 93693610 40 0 3 52 53 2 2  0 3    3636 23  16   1   , 1         ><1 1 or    111 1 , 48  Given that , find and prove that this sequence is a Geometric Progression. SOLUTION P a g e | 61 5 6 13−− −  556 131− −  65 31− 1 36 3 (ii) 1  12  14  18  161  ⋯ If we think about it we should realise that the sum appears to be 2. Since the sum appears to tend towards a specific number as it goes on indefinitely we refer to this series as a CONVERGENT series. The sum of this series can be given using the formula   1  , 1<1< 1       1 1 12  2 therefore and , Thus we see that our intuitive answer is indeed correct. LESSON 1 The first and fourth terms of a geometric progression are 500 and 32 respectively. Find (i) the values of second and third terms (ii) the sum to infinity of the progression   500  3232 500  32  250032  1258 5   500500 2   200 5 SOLUTION (i)  LESSON 2 The first term of a geometric progression is and the common ratio is . Given that and that the sum to infinity is 4, calculate the third term. SOLUTION Sum to Infinity What would be the sum of the infinite series For our series above we have  2  8085005000 5   −500 1 2500 25 3    12  121  44  41 1212 4  116 4   121212 1   3 4   31  34   16 LESSON 3 The first term of a geometric series is 120. The sum to infinity of the series is 480. Given that the sum of the first terms is greater than 300, determine the smallest possible value of . SOLUTION    1 120  4804804801 1  120 1 1 3  14 4     1    > 3001     P a g e | 62 1201334> 300 143  12011 4  > 300 4  3 4801   > 300 4  3 1343 > 58 43 < 8 3 ln 43 < ln 83 ln43< ln 8  > llnn 834 ln 34 is negative  > 43.4  1 = 2 1 <  < 1  1  1  1  1  1  ⋯ = 12 2 4 8 16   412  12 11<  < 1   1 2 12  1 LESSON 4 Determine whether the geometric series is convergent. If it converges, determine its sum. SOLUTION We need to show that . Since , the series converges. P a g e | 63 MACLAURIN’S MACLAURIN’S SERIES INTRODUCTION               0   0  2!2!  0  3!3!  0  ⋯  ! 0 ⋯ ⋯ must be differentiable  0    0  Us– e Maclaurin’s Theorcosem to find coscos0.0.2    coscos   0  coscos0  1  s sin 0 sisi n0 0     cos cos    0  co coss 0  1   sin   0 sin0 0      coscos     0  coscos 0  1     sin    0  sisi n0  0    c cos   0  co coss0 1 1   0  0 2!2!  0  3!3!  0 ⋯ ⋯  !  0  ⋯    1  0  112!2!  0 3!3!  1 4!4!  0 5!5!   11 6!6!  11  2!2!  4!4!  6!6! ⋯ ⋯          0. 0 . 2 0. 0 . 2 0. 0 . 2 0.0.2  1  2  24  720  0.98 must exist at The derivatives of must exist at Only within specific values of is the series valid. LESSON 1 the first four non zero terms for determine an approximation for SOLUTION , hence . 1   cos LESSON 2  Find the Maclaurin expansion for up to and including the term in . SOLUTION   11   cos  11 2    cos   11 2   11  2!2! ..  using result Ques tion 1  11  2!2! 2 2     ⋯ ⋯  11  2  12    Fi n d t h e Macl a ur i n ’ s s e r i e s f o r  tan  LESSON 3 up to . SOLUTIONhttp://sirhunte.teachable.com/courses /93027/lectures/2211764         0   0  2!2!  0  3!3!  0 ⋯ ⋯  !  0  ⋯    tan   sec      2tanse n sec   2se2secc   4sec tan   0  tan0  0 0  sec 0  1   0  2tann0 sec 0  0  2se2sec0c0 4sec 0tan 0 2       0    0 2!2! 2 2 3!3!  ⋯ P a g e | 64    23!3!   + Obtain the Maclaurin’s series expansion for 0.0.1 LESSON 4 A function is defined as . (a) up to and including the term in . (ii) Hence, estimate to four decimal places. SOLUTION    ++  0      2 + 0  2     4 +  0  4     8 +  0  8     16  0  16   0  0 2!2!  0  3!3!  0  ⋯  !  0 ⋯ ⋯          2  4 4 2!2! 8 84 3!3! 16  1624!4! 0.0.12 2 22 20.02.1  220.03.1 0.30.1  0.0.1  3.3201      −− ,′     13−−   13 0 1  −−   −−    22 13 1 3 33  6 13 1 3 (i) Valid for LESSON 6   l n 1    ln1   – ln +−  ln 3+−  3 (i) Use the Maclaurin series for and to obtain the first three non zero terms in the Maclaurin series for You are given that Find and Maclaurin series for (ii) Find the value of for which . Hence find an approximation to , giving your answer to three decimal places. SOLUTION (i) . . Hence obtain the as far as the term in . By considering the equivalent binomial expansion, give the set of values of for which the Maclaurin series is valid. SOLUTION . State the range of validity for this series. (ii) LESSON 5 0  6 −−    18 1813 1 3 33  545413 1 3−−  0  54 −−    96 9612 1 2 33  28828813 1 3−−  0  288     0  0 2!2!  0  3!3!  0 ⋯ ⋯  !  0  ⋯   1  6  36 4  288 6  16 1 6  9 48  48   1 < 3 < 1 →   <  <    ln11     1       1 1 0  ln11  0 0  1  0  1  01 1 10    1 2 20  1 020      1 6   0  661 60 P a g e | 65     1 24   0  241 240     0  0 2!2!  0  3!3!  0  ⋯  !  0 ⋯ ⋯   1      0    2!2!   2 3!3!  6 4!4!  2424 5!5! ⋯ ⋯           2  3  4  5 ⋯   ln1   0  ln1  0    1 1  10   1 1 0     1 1 10   1 10     1 2 20   1 20   6  0 6  6    1     1  0   24   024  24 1   1  0   0  0 2!2!  0  3!3!  0  ⋯  !  0 ⋯ ⋯           0    2!2!⋯2 3!3!  6 4!4!  2424 5!5!            2  3  4  5 ln 11    ln1    ln1      2  3 4 5 ⋯     2  3  4  5   2  23  25  1 <  < 1 +−  3 14 2 3 33   12 1 ln3  ln 11  212   1 2 1 2 1  221.0296  3 2  5 2 (ii) P a g e | 66 INTRODUCTION    Maclaurin’s Series is:      0  0  2!2!  0  3!3!  0 ⋯ ⋯  ! 0 ⋯ ⋯ For Taylor’s Series, we let               2!2!   3!3!  ⋯ ⋯  !  ⋯     1. 2. The function The function has to be infinitely differentiable has to be defined in a region near the value Furthermore, replacing with . we get         2!2!      3!3!    ⋯ ⋯  !    ⋯ ⋯    = !!    – ln 3 3 LESSON 1 Find the first three non terms of the Taylor expansion of SOLUTION zero . LESSON 2 + Find the first four non – zero  1   +   3 3−−    31 3 1  14     313 1   161     323  1  321      363  1   1283 terms for the Taylor expansion of . with centre   ln      ln 3 3 →   3   ln1  3  ln1 3    3  3     1  3   19       2!2!    3!3!   ⋯  !  ⋯ ln 3 3  ln3 n3  13   19 2!2!1                              2!2!  ln3 n3  13   181    3!3!    ⋯ ⋯  !    ⋯ ⋯ SOLUTION Let 1 3 3  14  161  1 1  321 2!2!1   1 1  1283 3!3!1   1 1 P a g e | 67 2  2 2 3 2 3 √ √ √ sin16  2  2√2 163 3 4  16     12 16  0.1920  14  161  1 1  641  1 1  2563  1 1 – sin LESSON 3 (i) Obtain the first four non zero terms of the Taylor Series expansion of in ascending powers of    . (ii) Hence, calculate an approximation to SOLUTION  sin            3!3!    ⋯ ⋯  2!2!!    4  sin    √2  4  sin4  2   coscos   4  cos4  √22    s sin  4   sisin 4   √22    cos        cos co s  cos    √2 4 2 sin  √22  √22 √241 √22 2!2!1  4       2 3!3! 4  √22  √22   4√2√42  4     12 4   (i) let (ii) sin  sin →    3   4  16  4   16 . P a g e | 68 BINOMIAL THEOREM At the end of this section, students should be able to:           ∈ ℚ 1. explain the meaning and use simple properties of 2. recognise that that is, !    , ∈ ℤ and , that is, , is the number of ways in which where objects may be chosen from distinct objects; 3. expand for ; 4. apply the Binomial Theorem to real-world problems, for example, in mathematics of finance, science. P a g e | 69 PASCAL’S TRIANGLE INTRODUCTION  For any positive integer :     1     111 1     11 2 2 1 1       11 3 3 3 3 1 1       11 4 4  6 6  3 3 1 1 !   1 1 2 2 3 3 … 321 8!  8×7× 8× 7×6×6×5×5×4×4×3×3× 2×1 2× 1 0!0! !! FACTORIALS For example, Is defined as 1. LESSON 1 Simplify SOLUTION ! 9!9!6!6!  9×8×7×6×5×4×3×2×1 6×5×4×3×2 6×5×4 9 ×3×2 × 8 × 7 ×1 504 !!   2 2! !   2 2!   1 1 2 2 3 3 … 321   2 2!   11 2 2!   2 2!  2 2! 1 1  1   2 2!    11 −−! − −!  72  13 13 24 24 35 35 …… 332211  72  1 1 2 2  72 LESSON 2 Simplify SOLUTION LESSON 3 SOLUTION Solve the equation . PASCAL’S TRIANGLE 1 1 1 1 1 1 2 1 3 4 3 6 1 4 1  3  3  2  7272  3  3  70  0  10 10 7 7  0   10   7 since LESSON 4 in invalid Show that 1 2 2!  31 31!   32 3! 1 2 2!  3131!   3 3332 2!  31 31!  33 33!  3131! 3331! 1   3   32 3! SOLUTION P a g e | 70 LESSON 5 +  +!! (a) Express  and in the form are integers. +   +! + +! , where (b) Hence find  1 =  2 2!! + + !!  + + !  + + ! SOLUTION (a)  11 11  212!!  2 2    2 2 1 1!   2 2!  1 1   2 2     1 2    1 2 1   1 1 1 1    2 2!!  1 1!  2 2!  1  1  1 =  2 2!!  =   11!  =  2 2!   1: 2!2!1  3!3!1     2: 3!3!1  4!4!1     3: 4!4!1  5!5!1   Equating coefficients of : Equating constants: (b)     1: !1   11 1!    :  11 1!   21 2!  1 1 1 =  2 2!!  2   2 2! P a g e | 71 THE BINOMIAL THEOREM INTRODUCTION We now look at an alternative to Pascal’s Triangle using!Factorials.     ! !!     For any positive integer :     = −  0   1 −  2 −  3 − ⋯ 1 1 −    32 3 2 LESSON 1 Determine the expansion . 32 3 25  5   5   5   5   5   0 3  1 3 22  2 322  3 3 22  43 22  5 22  243810 24243  581811080 22  101072 2727720440 24  240101009883232 531616   13232     2 3 3    23 2 3   5   0  105 2 23 3   1959552 2522523232243  13  1 32   13 1 3  16 1 6  9   2    64 2  65 2  66   60151541 16262    1      6   960  1  6  9…60  …60 1 12    172 6 12  12 6 0  54   72   540 0  469469 SOLUTION LESSON 2 Find the 6th term of the expansion SOLUTION The 6th term of the expansion begins with LESSON 3 . occurs when since the summation index . Find the coefficient of in the expansion of SOLUTION We only need the terms which will result in a term after multiplication. We isolate the multiplications which would create an term. . P a g e | 72 LESSON 4 Find the term independent of  in the expansion 3    . SOLUTION   86 33 121   28286399 64  16 The term independent of :  1    11    2!2! 1    1 13!3! 2 2  ⋯ ⋯ 1 <  < 1       1     1     1 1    2!2! 1    1 13!3! 2 2  ⋯ ⋯  1 <  < 1 →  <  <  12 1 2  12 1 23   11 2 23 3 1 3 3 3 1  1 2 22  2 2 121 22  2 2 12 2321 22  1  3  32 12 12 4  32 12  12 312 8  13 1 3  32   12   1 < 2 < 1  12 <  < 12 For any real number provided that LESSON 1 SOLUTION Find the binomial expansion of up to and including the term in . P a g e | 73 LESSON 2 Find the first three terms of the expansion of SOLUTION − . 423  423 2 3−−  421 421 32 −  424 2− 11  32 − − 3  21 21  2   3 1 3  221  11  2   1111  11 21  2   212 1  32   94  ⋯ ⋯   23 2 3  92  + ++ + +  +   + ++ +   9 1  25 2 5 + ++ +  +  +    99  2225 25 5   1     5      2    0  5 3  9   69  3  6 1  2 5 5 1   25 +331  −−  33331 11 11 1111  11 211   1111  1112  12 3121   33 3 3  3 3 3 LESSON 3 (i) Express in the form where (ii) Hence, or otherwise, find the expansion of including the term in . (iii)Find (iii) Find the range of values of SOLUTION (i) Equating constants: Equationg coefficients of : (ii) and are integers. as a power series in ascending order up to and for which the series expansion of P a g e | 74 625  6625 2 55−− −  621 21  2  −  62− 11 −52    31 31  52  5   1 5 1 5  31 31  511 252   1251111  11 21 2   1111  1112  12 321 2    313 115 2  754   3758    33  2   4   8  31    256  933 3 363 3 33513  3  152   754   3758   2   4   8  1 <  < 1 1 <   < 1 →   < 2<  2 5 <  < 5 √1   √82 √1    11    11  14   14 14 1211   14 14 114 23211   11  14   323   1287   82  8 1  1  181 1 81  811 811  811    1  81 11  81 (iii) Valid for LESSON 4 and Use the expansion of SOLUTION This works for small . to setimate to four decimal places. P a g e | 75  1  31 31  81   1 1 3 1 7 1  331  4 81  32 81  128 81  31.1.003  3.009  16  186 6 8 6 LESSON 5 (a) Find the binomial expansion of √144 1 6 6  14 11 4664   1  66    86 8 6  8 1    8 11  68    4141  618    4141  4118 418    42 4 2    4     81441446  121122   12 6 24 3 √144≅ 4223  14 23 ≅ 479 (b) Find the binomial expansion of (a) (b) (c)  . up to and including the term in (c) Hence, find an estimate for the the value of SOLUTION    up to and inclusing the term in  in the form where and . are integers. P a g e | 76 ROOTS OF EQUATIONS At the end of this section, students should be able to:   0 1. test for the existence of a root of where f is continuous using the Intermediate Value Theorem; 2. use interval bisection to find an approximation for a root in a given interval; 3. use linear interpolation to find an approximation for a root in a given interval; 4. explain, in geometrical terms, the working of the Newton-Raphson method; 5. use the Newton-Raphson method to find successive approximations to the roots of , where is differentiable; differentiable; 6. use a given iteration to determine a root of an equation to a specified degree of accuracy. 0    P a g e | 77 ,,    0    2 2  1 1  1 2 21 1 1  1  11 2  2 2 22 2 2 1  1  1 ×× 2  1 ,  THE INTERMEDIATE VALUE THEOREM If in is a continuous function on the closed interval such that . LESSON 1 and the product  < 0 then there exists  Use the Intermediate Value Theorem to show that has a root between 1 and 2. SOLUTION is a polynomial and therefore continuous on the By the Intermediate Value Theorem there must be some root between 1 and 2. 1 3 4 4 1  0   3 4 4  1 0  1  1  31 4 41 1 1  6  interval  ∈ 1,1, 2 such that   0 1,1, 2 . Therefore there is a LESSON 2 Use the Intermediate Value Theorem to verify that there is a root of the equation between 0 and 1. 3 4 4  SOLUTION Let 0,0, 1 0 ×× 1  6  ∈ 0,0, 1   0 is a polynomial and therefore continuous on the interval By the Intermediate Value Theorem there must be some root between 0 and 1. . such that . . Therefore there is a P a g e | 78 DETERMINING THE ROOTS OF AN EQUATION  3 3  1  LESSON 1 Given that has a root between 2 and 3, find this root to 1 decimal place using the bisection method. LESSON 1 Use Linear Interpolation, twice over, to determine the root of the equation in the interval .  3 3  1   3 3  1   3 3    1  0 SOLUTION SOLUTION  3 3  1 1   3 3   1 0    3 3  1 2  2 3 32 2 2  1  33 3  3 3 33 3 3 1  2 –      2.2.5  2.5 3 3 2.2.5 2.2. 5  1  1.1. 625625 – 2.2.755  2.75 3 32.2.755  2.2.7575  1  0.140625 –      2.2.87575  2.875 3 3 2.2.87575 2.875  1  0.8418 2.75 –      2.2.8125125  2.8125 3 3 2.2.8125125 2 2.8125  1  0.33 2,2, 3 Let Mid point of interval is 2.5 Since there is a sign change between 2.5 and 3 the root occurs within this interval. We therefore now bisect this interval. Mid point is 2.75 Due to sign changes the root must be in the interval 2.75 and 3. Mid point is 2.875 Due to sign changes the root is between 2.875 Mid and point is 2.8125    3 3    1 2  2 3 32 2 2  1  3 3  3 3 33 3  311  2 23  33 2 2 2 2 2  33   2  4  9  3  5 1313   5  2.6 Let Using similar triangles Therefore root lies between 2.75 and 2.8125 and since to 1 decimal place both limits are the same the root is approximately 2.8.   13 13 13   5    5  3 3  5   135   1  1.1. 1 P a g e | 79 21.1  32.2. 6 2 2.2. 6  1.13   2  5.5.2  3.3.3  1.1.11 3.1  8.5   2.74 INTRODUCTION    44 10900900  0 – LESSON 1 The equation has exactly one real root, . Taking as the first approximation to , use the Newton Raphson method to find a second approximation, , to . Give your answer to four significant figures.         4 4  900900    3 2 2  4   10       49000   1010  1031010110024241011010090   1010  28440   9.859 2  5 5 3 3  13  0  – 1<<2 2 51.55  3 3  13  0 1,1, 2  5  3       2  5  3   1 3 12  2215 552513 33231311313 13 2929 33 1 ×× 2 < 0 1,1,2 2 5 5 3 3  13 1.56 1 10 3   1.1.5  1921.1.561.15.551.1.51010 31.1.351.1.53 13 13   1.1.5  63 SOLUTION Let LESSON 2 The equation has exactly one real root . (i) Show that lies in the interval . (ii) Using the Newton Raphson method with initial estimate to estimate the root of the equation in the interval , correct to 2 decimal places. SOLUTION (i) Let Since , by the Intermediate Value Theorem there is a root in the interval (ii) P a g e | 80   1.1984   1.1984984 21.1.198498461.1.1598498451.1.110984984101.13.198498431.1.19849843   13   1.1.114679840.0516 1.1467 21.1.1467467 5 51.1.1467467 3 31.1.1467467  13  10     6 1. 1 . 1 467 467  10 1. 1 . 1 467 467  3   1.1984984 0.001357 01357   1.1453     1.15 +  +   2 4 4 1  0   1.2 Since and correct to 2 decimal places are both equal to 1.15, . DERIVING AN ITERATIVE FORMULA EXAMPLE 1 Show that is an approximate iterative formula for finding the root of . Apply the iterative formula with initial approximation , to obtain an approximation of the root to 2 decimal places. SOLUTION 2 4 4 1  0 2  4  1   42 1   42 1 +  421 1   1   41.1.22 1 1  1.2386   41.1.23863862  1 1  1.2551   41.1.25515512  1 1  1.2621 1.26 Therefore, the approximation is decimal places. correct to 2 P a g e | 81 MATRICES At the end of this section, students should be able to: 1. 2. 3. 4. 5. 6.  ××    2,2, 3 reduce a system of linear equations to echelon form; ; row-reduce the augmented matrix of an system of linear equations, determine whether the system is consistent, and if so, how many solutions it has; find all solutions of a consistent system; invert a non-singular matrix; solve a system of linear equations, having a non-singular coefficient matrix, matrix, by using its inverse. 3×3 3×3 P a g e | 82 MATRICES  A matrix is a rectangular array of elements enclosed in brackets. A matrix is defined: number of rows  ×  Two matrices are equal if they contain the same corresponding elements.  The addition and subtraction of matrices is only possible if the matrices are of the same size. number of columns (this is the size/order of a matrix). A square matrix contains the same number of rows and columns. o  Matrix addition is commutative and associative. Two matrices can be multiplied if the number of columns in the first matrix is equal to the number of rows in the second. o  Matrix multiplication is not commutative but it is associative.  2 × 2 10 01 3 × 3   11 1 222 311   210 111 101 − The identity matrix for LESSON 1 (i) find If matrices is and for matrices it is 100 010 001 and (ii) deduce SOLUTION (i) (ii)   11 1 222 131 210 111 101                         1 2 2  2 1   1  1 0 1 1  1  2 1   1  1 1 1 1 2  2 0  1  1 1  11122222211113300 111111122211 13 1311 11111222200 13 1311 4 0 0  00 40 04 −4  14    2 1 1 −   10 11 01 Therefore . P a g e | 83 × (i)          ℎ    ℎ       ℎ |||  0 |  ||| |   |||  0  |||   × | |  ∈ ℝ |        |||  ||| | | | |   0 – det−      311 121 213   – INTRODUCTION The determinant of is . o If the matrix zeros then contains a row with ALL . o For square matrices o If o . contains two identical rows or columns then . Interchanging two rows or columns of a matrix changes the sign of the determinant. o For , where is a matrix. o If a row or a column of a matrix is multiplied by a constant, , then the determinant of the matrix is multiplied by . o Adding a multiple of one row to another does not affect the determinant. o o 11 21 | 323| 13121 13111311 11312          2 1 1  1 2  11335 1 12 2 211 | | ≠ 0  –  2 31 14  20 3 22 310 143   0 2 10 43  32 4312 10  0 26393 32334  2810122  0 9 3232  9 36 61 105 925 SOLUTION (ii) Since LESSON 3 , is non singular. Find the value of for which the matrix is singular. SOLUTION LESSON 4 Determine by factoring. A square matrix is singular if SOLUTION otherwise it is non be obtained by factorizing the matrix. This is done singular. o LESSON 2 The matrix is given by . (i) Find the value of the determinant of (ii) State, giving a reason, whether or non singular. . is singular The determinant of a matrix can by factoring out the common factor from each row or column. 3 213 621 1055 1 6 3 35523 21 01  factoring out 3 from  factoring out 5 from  To find the determinant it is easier to use the elements of because it contains a zero.  1515 3 232 12121 16  1515377  1313  120      31 3 112 221            | ||  5  ,  –     6 4 4 1 9           3 1 6 3 5 5 9   4 26 81     5   7 9 3 1 6   54 26    2 2 3  3 2 3  3 2  |||  1 1 213 213 1   51122  11212 1 19 |||  4 69 38 2575 38 1575 69  4436590 90 2 241 41  187 87 55   16 52 46 4 2 1   16 52 46 57 9 6 38   567 6788 249 5×4  320 The transpose of a matrix is created by interchanging the rows and columns. The LESSON 5 The matrix . transpose of a matrix is denoted . o (i) Show that (ii) Matrix and . is changed to form new matrices . Write down the determinant of EACH o A square matrix is symmetric if o A square matrix is called skew symmetric is . . of the new matrices, giving a reason for your For example, if answers in EACH case. (a) Matrix is formed by interchanging column 1 and column 2 of matrix and then interchanging row 1 and row 2 of the resulting matrix. (b) Column 1 of matrix then LESSON 6 by and is formed by adding column 2 to column 1 of matrix . The . other columns remain unchanged. (c) Matrix The matrices is formed by multiplying each element of matrix Calculate by 4. SOLUTION (a) the determinant of (i) (b) SOLUTION (ii) (a) interchanging columns will change the (a) sign of the determinant to 5 and then interchanging the rows will change the sign of the determinant to (b) The determinant remains as . (b) since adding a multiple of a row does not affect the determinant. (c) Since each element is multiplied by 4 the value of the determinant is and are given                    2×2 ±1                  | |   |  , ∆ |||             If then , are called the minors of and , and respectively. A minor of an element is obtained by deleting the row and column containing that element and finding the determinant of the matrix which remains. The cofactors of a matrix are determined by multiplying the minor by in the following order Step 1: Find the determinant of the matrix Determinant of , det , The elements of any row or column can their corresponding cofactors can be used to determine |||                                  (        )  the determinant. i.e. using the first row and its corresponding cofactors. Step 2: Write the matrix, say Step 3: Write the matrix, the adjoint of ( adj) , of cofactors. . This matrix is called       −      ||  adj Step 4: Use the relation to find − 104 111 132   104 111 132 |||  11 111 324111 13  13155 4 422  11111 312 1041 321 1041 111     111 12  141 12  141 11  (5 1123 4  0 3   0 1)   3 263 31    5124 363 231  5 3 2 1 −   3 124 36 31  . LESSON 7 SOLUTION Let Find the inverse of   3×1 1 0 3   73 52 01 – − 10    128  |||  1 52 01  073 01373 52  1215 3 311  520 013 173 301 731 520    02 31 31 13 13 02 ( 55 0 7 117 0 7 5 )  156 821 25    75 86 2115 11 5 26 515 −  2 71 82 215  −  −   1−5 6 15 10   2 71 82 215  128  1521 2188  1 2 1751710101011010066121282281211212215        5 8  12 226  113 LESSON 8 and are are related by the equation is non Find (a) (b) , when SOLUTION (a) (b) singular. matrices and , where LESSON 9 Solve the equations 4 554519129  5 5    2020    , 1 5 5  2  41 55 41   2019  1 −15 30 45 −   150  28 5 06 2156 −     1501 158 306 4516  192   1 23005 0 2 5 20    150 450450    23  3 SOLUTION Step 1: Write the system in the form where and are matrices Step 2: Find Step 3: Multiply both sides of equation by This equation is said to be consistent since it has a solution.  → 104 111 132 100 100 001 1 1 1 1 0 0  4  4 → 00 13 36  04 10 01 –  → 10 11 1310 10 00 0 13 0 6 24 10 11 0    → 00 31 36  04 10 01 3 × 3 1    3 1 0 2 1 1 0  3 →    0 1 3 0 1 0  00 10 1 1 1 00 0 23 14 31 01 10 1   3  → 00 10 31 043 11 013 000 1 1 0 1 0 2 4 2 1  3  3 →   0 1 0   4 1 – 0 0 1 53 1  32 1 1 0 0 3  2  2 → (00 10 0144 21 131) 3 3    → 10   5  3 2 1 −     ℎ  0 ℎ  3 124 36 33 The following elementary row operations can be performed on a matrix. - The interchanging of rows - The multiplication of a row by a non zero scalar. - The adding of the multiple of a row to another row. These operations can convert a matrix (or any other matrix) to row echelon form or This can be done by 1. Beginning with the left most column and use row operations to make the first element in this column a 1 and the elements below it zeros. 2. Ignore the row and the column with the 1 created from step 1 and repeat step 1 on 100 ℎ  → 100 10  the remaining matrix. 3. Repeat this process until the desired matrix is obtained. 1 0 11 13 4 12 LESSON 10 SOLUTION Find the inverse of A system of equations can either be consistent or inconsistent. A consistent system can have  a unique solution  infinitely many solutions. An inconsistent system has no solution. . The use of row reduction greatly assists in To obtain the inverse of a matrix −|  |   |    104 111 132100 010 001 we can use row operations to convert the augmented matrix to . determining the consistency of a system of equations. An upper triangle matrix unique solution. 10 1  0 0 1 indicates a 100 10 00 100 10 0  3  333 4   2 2  6 Matrices of the form Since the system has a unique solution it is said to infinitely many solutions. Matrices of the form solution. LESSON 11 be consistent. indicate no A system of three equations is (i) Write the augmented matrix for the system. (ii) Use row reduction to solve the system of equations. 1 0 11 1333 41 11 12 36  04 11 3236  3 1 1 1  → 04 11 32 36   4  4 → 100 131 136 363  1 1 1 3  → 00 31 36 63     → 001 310 362 603  1 0 2 0  3  3 → 00 10 3333 13  → 010 100 321 103  :  3 13  3  3 3011  3   2 2  3 SOLUTION (i) (ii) From From : : : From :  2 211  20 indicate LESSON 12 Determine the general solutions of the system of equations. 22222  343 2  4  2  8 11 22 12 34  2 4 21 82 2 3    → 02 44 32 78  1 2 2 3  2  2 → 00 48 36147  14  → 10 212 234374  0 81 62 142 3  8  8 → 00 10 034 740  0  0  0  0 SOLUTION The row of zeros indicates that , therefore we can use a parameter for one of the variables and express the other variables in terms of this parameter. Furthermore, the row of zeros indicates that the system has infinitely many solutions.     343   774   43  47  4 4 Let From row 2: From row 1:  2 23 2 7 3   23 4 7 4  2  3   2 1 2 1 2  3   2 2  1    1          1    0   1  0 Therefore, if we say that and So then . would be one solution for the system. LESSON 13 Discuss the solutions of the given  4  8 3 33 410 2  2  2   13 2 132 121 410  4 1  1  1 Interchange  and  → 3 2 32 21 10  1 1 1 4  3 2 3  →   0 6 4  2 0 0 0 22 8  04  0  0    8  44  4  08  0  0   8 8  0 equations when (i) (ii) SOLUTION (i) When , we have This result indicates that when , the system of equations is inconsistent and has no solution. (ii) When , we have that This result indicates that when  8 , the system of equations has infinitely many solutions.   66  44  2222   113  23      11  2 44    31  13     4   3  3 Let From : From : DIFFERENTIAL DIFFERENT IAL EQUA EQUATIONS TIONS At the end of this section, students should be able to:                     0   a, ,, ∈ ℝ  1. solve first order linear differential equations using an integrating factor, given that is a real constant or a function of , and is a function; 2. solve first order linear differential equations given boundary conditions; 3. solve second order ordinary differential equations with constant coefficients of the form , where and is: (a) a polynomial, (b) an exponential function, (c) a trigonometric function; and the complementary function may consist of (a) real and distinct root, (b) 2 equal roots, (c) 2 complex roots; 4. solve second order ordinary differential equation given boundary conditions; 5. use substitution to reduce a second order ordinary differential equation to a suitable form. DIFFERENTIAL EQUATIONS A differential equation is an equation which contains derivatives of a function or functions. For a first order differential equation the highest derivative is the first derivative. For a second order differential equation the highest derivative is the second derivative. sin       where The solution of this type of equation can be achieved by separating the variables and integrating both sides of the equation with respect These solutions are called general solutions of the differential equation because the value of the constant is unknown. to the relative variables. LESSON 3 LESSON 1 SOLUTION   5 3 Solve the differential equations   53 3    55 3 3     55 3 3  2  53  3     103   6     coscos  tan    cost o s t a n  1tan   coscos      ln    cossin    coscos   lnsisin sisin+  sisin    LESSON 2 SOLUTION Solve the differential equation Find the particular solution of the    csc     4,4,    1  cscsc;c1 ;   3 when   4 csc1     −  csc       sisin     −   coscos   1 1  cos1 31  4   1  2  14 1 →  4   coscos     4 differential equation when SOLUTION . INTRODUCTION      ∫ Linear differential equations of the form can be solved by multiplying throughout by the Integrating Factor, LESSON 4 . Solve the following differential   3 3  7        3   7 .   ∫3      .   ∫−   −     − −  3 3−  7−               −  3 3−    7−     27   equations SOLUTION Step 1: Write the DE in the form Step 2: Calculate the I.F using Step 3: Multiply each term in the equation by I.F L.H.S resembles the product rule where I.F is actually Step 4: Integrate both sides of the equation w.r.t.    72     cotcot  2 coscos    2      cotcot  2 coscos    cotcot  cos  cotcot   sin   lnsin .   sicoscosn  sin s sin.n . sin .   2sincn cos  sin   cos cos   2c2 cosss sin   si nsinsin csc  +  +                        1 1     2     2  sin2  sin2 2  1  1   s1in cs csc  LESSON 5 Solve the differential equation given that when . SOLUTION when LESSON 6 Determine the particular solution of the differential equation −   4 4     3  0   4 4  −  4  4  − given that SOLUTION when . I.F  ∫    4 4   −    4 4        −     3,   0 32        − 2 2 When      0 andand  d       0 ,,   are first and second order homogenous equations where and are constants. The solution of these equations is called the complementary function (C.F). LESSON 7 5 2 2 1  5 2  1  5 2      5  ln  25       +        SOLUTION 5  2  2  0 Solve the differential equation In general, the solution (complementary function)      0   − of a first order differential equation is of the form is Auxiliary Quadratic Equation Given the equation                 0 If the auxiliary quadratic equation has a repeated root , , The complementary function is determined by the is the general solution of the differential equation. roots of the nature of the roots of the quadratic auxiliary equation which is       0  5 5   4  0  5 5  4  4  0 51,15,4 4  0              LESSON 8 Solve the equation SOLUTION Auxiliary equation If the auxiliary quadratic equation has real and distinct roots, and then  2 2  5  5  0  2 2  5 5  0 212± 25  0   cos  cos22   sin 22  ±    cos  cos  s sin  LESSON 10 SOLUTION Auxiliary equation If the auxiliary quadratic equation has complex roots of the form , is the general solution of the differential equation. is the general solution of the differential equation. 4  4 4     0 4  4 4     0 4 41 4  1  0   2twicee        LESSON 9 Solve the equation SOLUTION Auxiliary equation Solve the equation Non – – –               ,  ≠ 0 homogeneous first order and second order differential equation are of the form The particular integral is any solution of these types of differential equations.  is a Polynomial  5 5   6                   →    →   0 0665 5 5 66    6 11 6 55 6 5 6  0   36       LESSON 11 SOLUTION Solve the equation It seems sensible to think that the solution is of the form Substituting into original equation So is a solution of the given equation, but it cannot be the complete solution since it does not contain any arbitrary constant. However, it must be part of the complete solution, and is called a particular integral (P.I). The remainder of the solution can be found by considering the  5 5   6  0              16   365 simpler differential equation  4 4  3 2 2  1 4±24  0   −    SOLUTION Auxiliary equation C.F: Particular Integral:       →   22   →   2 24 4 44 4  22 4 3 3  22221 1 1 4  3 →    341 4  2 →   2 5 2  4  1  →    8         −     34   12   58  Substituting into original equation P.I is General solution: is a Trigonometric Function whose solution is LESSON 13 Thus the complete solution is differential equation which is the combination of the complementary SOLUTION function and the particular integral. LESSON 12  4  4  3 2 2  1 Solve the differential equation Find the complete solution of the 4  5 5     cos  sisi n  4  5 5     cos  s sin 44 51 1511 10 0   sisi n    cos  Auxiliary equation Trial Solution   14 , 1         cos cos  s  sinn   sin   coscos     coscos    sisin  cos csos sisinsinncoscos5sisisnin c 4co  cos  44 5  5  coscoscosos sisi44n5 5   sin 333 5 55  11    1741    17    174 coscos   171 sin    174 coscos   171 sisin        C.F is Substituting into original equation Particular Integral is The complete solution is LESSON 14 (i) Solve the D.E  4  4   3  65sin2   3,   7 when   0 (ii) Hence, find the particular solution for which SOLUTION Auxiliary equation 41,4 33  0   −   −   sisi n    cos   sisi n 2   coscos 2  22 coscos 2  2sin2   44sisin 2  4cos2   334sisisinn22 4co cos2 4cos s 2  4 2  c o s 2  2si  2si n 2  cos2  6565 sisin 2 C.F is Trial Solution Substituting into original equation 465si4 8 n82 33 sin2  44 8 8  33 cos2 88  8 si65n2  88   coscos 2  65sisi65 n 2 8    0  →   1si& n 2 8 8 coscos 2   −   − sisi n 2  8 cos 2   7   0,   3,  −  −− sisi−n 2  8 cos 2  3 2 2 coscos 2  16sin 2    3 33      8 sisi n0 8co  8coss0     11  3 77   3  332 2co  2coss0  16si16sin0  3  3  9 Particular Integral is General solution is When 3 3 119 →   10 &   21   21− 10  10− s sin2  8 cos 2   2 2  10   10   26   5  11    0  2222±1010240411010   21   1 ± 3    −cos  cos33   sin 33 Particular Solution is is an Exponential Function LESSON 15 Solve the differential equation given that and when your answer in the form . SOLUTION Auxiliary equation: Complementary function: Particular Integral: . Give 53  cos  cos00   sin 0  2   11;11;   0   −cos  cos33−  sin 33  2 33 sisin 3  3cos3 3cos3 cos 3si    11111  cos0 0   s i n 0    3si n 03cos0 03cos0  2     1 3 1  1 3 3  2  2 12 43    −3 cos 3  4 sisin 3  2  2 2  3  3  22−  → 0  → ∞   3  0  32 3231 10 0   1,1, 3−      −   −        −− −      LESSON 16 Solve the differential equation given that as and that when . SOLUTION Auxiliary equation: Trial Solution       Both the trial and the original have the same form. Coefficients differ.          2 2 10 10  26    26  13   22     −cos  cos33   sin 33  2   5;   0 General solution: Particular solution: C.F: In this case, the particular integral would be of the form but since this is already included in the complementary function we have to use . Trial Solution Both the trial and the original have the same form. Coefficients differ.   −   −  − −  −  2    4 2 2−− 2−−  2−   − 3−  2− 4 12  2    12 −   −    12 −  →→ ∞0 − →→∞0   3   0      → 0:0:   0   3,3,   0   0   −  12 −  12 − 3 5  12   12 0 2   52 −  12 − LESSON 17 (i) Show that by using the substitution may be written in the form Particular Integral: (ii) Find the general solution of General Solution: as and that Therefore we have From general solution ( Particular solution: and hence find the general solution of when : ):  2  2    2 2    2 2    2  2   differential equation    SOLUTION (i)       ×    1 ×   2  2    1  1 1   2     2 2    2 2  0  2            (ii) Auxiliary equation: Complementary function: Particular Integral Let , the ∴   2      2 2  22   21 1 2 12  2  0 14                     1 →   1 1    12   14 1  4 24 2  1   4 24 2  1  ≠0    2 233  1  3 33 2 2  18     6 6   9  18   2 233  1  3 33 2 2  18       Particular Integral: Complete solution: LESSON 18 It is given that the differential equation (a) Show that the substitution this differential equation into , satisfies transforms (b) Hence find the general solution of the differential equation giving your answer in the form SOLUTION (a) .                   2       2 23 1 1  3 333  2  18   6 6  2 2  9 9 6 6  1818   2 2   6     99  1818  6 6  9  9  1818  6 6 9  0   3,3    −                0 0 66  9    18 6 9  9 9 9  18 9  18  2 6  9  0 9  124  3   2  43 (b) Auxiliary equation Complementary function: Particular Integral Let    −   2 2     −   2 2  43    −   2 34 General solution: (b) LESSON 19   ,  > 0,0,  > 0    2     4  2  2         88 1 1  12  12  12  4 4   3  3    88 1 1  12  12  12           ×    12 − →   2   2      2    (a) Given that function of , show that: and is a (i) (ii) (b) Hence show that the substitution transforms the differential equation into (c) Hence find the general solution of the differential equation giving your answer in the form SOLUTION (a) (i) Now, (ii)   −  2 2        −      2 2 2     2  4 4    88 1 1  12  12  12  8 8  1  1 12   12 4  2 2  8 8  −2 12   12 4   2  16  2  12  12  1212 4  16  16  12  12  1212  4 4  3  3  3  4 4 3  0   1,1, 3               0 0 44  3    3 44 3  3 3  3  33 3  3  1 44 3  3  0 4  3  0 3  4 . (c) Auxiliary equation Complementary function:   43     43                ,    4          3 General Solution: Since LESSON 20 (a) A pond is initially empty and is then filled     5√41 5 5 gradually with water. After depth of the water, minutes, the metres, satisfies the differential equation Solve this differential equation to find  terms of .  in (b) Another pond is gradually filling with water,  after minutes, the surface of the water forms a circle of radius  metres. The rate of change of the radius is inversely proportional to the area of the surface of the water. (i)   Write down a differential equation, in the variables and and a constant of proportionality, which represents how the radius of the surface of the water is changing with time. (You are not required to solve your differential equation.) (ii) When the radius of the pond is 1 metre, the radius is increasing at a rate of 4.5 metres per second. Find the radius of the pond when the radius is increasing at a rate of 0.5 metres per second. SOLUTION (a)   √+ + + 1√45   51 1   45 4 5−   15 1  −−  45 4551  15 1  −−   2 25 √4  5   15 1 1      0,   0 25 4  50   15 1 1 0   12   1 1 5 √4  5 5 1 1 5 11   √4  5  2  2 1   4  5  52  211 5  52  211  4 4  15 52  211   45    9  1,   4.5 29  1 2  9  2   29    When (b) (i) (ii) When   12  29 39 LESSON 2 The number of bacteria in a liquid culture is observed to grow at a rate proportional to the number of cells present. At the beginning of the experiment there are 10,000 cells and after three hours there are 500,000. How many will there be after one day of growth if this unlimited growth continues? What is the of the bacteria? SOLUTION     1    ln1   +           0,         50050 000 10 000  ln50 1ln350 ≅ 1.304   103 000. 22  .. ln2 ln1.2304≅ 0.532 hourhourss 1.304 Let represent the number of bacteria present at time . Then the rate of change is where is the constant of proportionality where When Also, Therefore, (initial population) Using the information from the problem we now determine Doubling time refers to the amount of time for the bacteria to double in number from its original number.

pure-mathematics-notes

Related documents

Products

Support

pure-mathematics-notes

Related documents

Add this document to collection(s)

Add this document to saved

Suggest us how to improve StudyLib