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# Core 2 revision notes

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### Core 2 revision notes

1. 1. CORE 2 Summary Notes1 Indices· Rules of Indicesxa´ xb= xa + bxa¸ xb= xa – b(xa)b= xabSolve 32x + 1– 10 ´ 3x+ 3 = 03 ´ (3x)2– 10 ´ 3x+ 3 = 0Let y = 3x3y2– 10y + 3 = 0(3y – 1)(y – 3) = 0y =1gives 3x=13 3x = – 1y = 3 gives 3x= 3 x = 1Solve 9x=12732x= 3– 32x = – 3x = –32· x0= 1·x– a=1ax· xn= n1x· x n n m n mmx)x( ==2 Graphs and TransformationsASYMPTOTESStraight lines that are approached by a graph which never actually meets them.yx1 2 3 4 5– 1– 2– 3– 4– 512345– 1– 2– 3– 4– 5Horizontal Asymptotey = 1Vertical Asymptotex = 2y =1x – 2+ 1TRANSLATION - to find the equation of a graph after a translation of ú you replace xby (x-a) and y by (y - b)ûùêëéab1yx1 2 3 4 5– 1– 2– 3– 4– 512345– 1– 2– 3– 4– 5e.g. The graph of y = x2-1 is translated through ê ú . Write down the equation ofûùëé 3- 2the graph formed.(y + 2) = (x-3)2– 1y = x2– 6x + 6y - b = f(x-a)ory = f(x-a) +by = x2– 6x + 6y = x2-1www.mathsbox.org.uk
2. 2. y = -f(x)REFLECTINGReflection in the x-axis, replace y with –yReflection in the y-axis, replace x with – xSTRETCHINGStretch of factor k in the x direction replace x by1xkStretch of factor k in the y direction replace y by1yky = f(-x)y = f(1xk)e.g. Describe a stretch that will transform y = x2+ x -1 to thegraph of y = 4x2+ 2x -14x2+ 2x -1 = (2x)2+ (2x) -1So x has been replaced by 2x.Stretch of scale factor ½ in the x directiony = kf(x)3 Sequences and Series 1· A sequence can be defined by the nth term such as un = n2+1u1 = 12+1 u2 = 22+1 u3 = 32+1 u4 = 42+1= 2 = 5 =10 =17· An INDUCTIVE definition defines a sequence by giving the first term and a rule to findthe next terms.Un+1 = 2un + 1 u1 =3 u2 =7 u3 =15· Some sequences get closer and closer to a value called the LIMIT – these are knownas CONVERGING sequencese.g The sequence defined by Un+1 = 0.2un + 2 u1 =3 converges to a limit lFind the value of l.l must satisfy the equation I = 0.2l + 20.8l = 2l = 2 ÷ 0.8= 2.5ARITHMETIC SEQUENCEEach term is found by adding a fixed number (COMMON DIFFERENCE) to theprevious one.u1 = a u2 = u1+ d u3 = u2 + da is the first term ,d is the common difference the sequence isa, a + d , a + 2d, a + 3d,……….Un = a + (n - 1)dSUM of the first n terms of an AP (Arithmetic progression)Sn =21n ( 2a + (n-1)d ) or21n(a + l) where l is the last term· The sum of the first n positive integers is12n(n + 1)· SIGMA notation Si = 05i3= 03+ 13+ 23+ 33+ 43+ 532www.mathsbox.org.uk
3. 3. Show that Sr = 120(3n – 1) = 610= 2 + 5 + 8 + .········59a = 2 d = 3S20 =202(2 ´ 2 + (20 – 1) ´ 3)= 10 ´ 61= 6104 Geometric Sequences· A geometric sequence is one where each term is found by MULTIPLYING theprevious term by a fixed number (COMMON RATIO)· The nth term of a geometric sequence a, ar, ar2, ar3, …is arn-1a is the first term r is the common ratio· The sum of a geometric sequence a + ar + ar2+ ar3….+ arn-1is a geometricseriesr 11)=a(rn-Sn-· If -1 < r < 1 then the sum to infinity is1 ra-Find the sum to inifinty of the series 1 +23+49+827Geometric Series first term = 1 common ratio =23Sum to infinity =1= 331-25 Binomial Expansion· The number of ways of arranging n objects of which r are one type and (n-r) areanother is given byr!(n r)!n!rn=-÷÷øöççèæwhere n! = n(n-1)(n-2)….x 3 x 2 x 1·(1 + x)n= 1 +æèçn ö1 ø÷ x +èæçn ö2 ø÷ x2+æèçn ö3 ø÷ x3+ .········.xn(a + b)n= an+èæçn ö1 ø÷ an – 1b1+æèçn ö2 ø÷ an – 2b2+ .··········.æèçnn – 1 øö 1÷ a bn – 1+ bnFind the coefficient of x3in the expansion of (2+3x)9n = 9 , r = 3 ççæ9(3x)3(2)63÷÷øöèNote 3 + 6 = 9 (n)= 84 × 27 x3× 64 = 145152 x33www.mathsbox.org.uk
4. 4. 6 Trigonometry· Exact Values – LEARN45 30 60sin 45 =12cos 45 =12tan 45 = 111√245sin 30 =12cos 30 =32tan 30 =131 12 2sin 60 =32cos 60 =12tan 60 = 33060√3COSINE RULEa bcABCba2= b2+ c2– 2bc cos A2= a2+ c2– 2ac cos Bc2= a2+ b2– 2ab cos ASINE RULEasinA=bsinB=csinCAREA of a TRIANGLE1ab sinC =12 2bc sinA =12ac sinBe.g. Find the area of triangle PQRFinding angle P82= 92+ 122– 2 ´ 9 ´ 12 ´ cos Pcos P =92+ 122– 822 ´ 9 ´ 12= 0·745P = 41·8·····.Area =12´ 9 ´ 12 ´ sin 41·8·····.= 35·99991 36 cm2== 36 cm29cm12 cm8cmRP QTry to avoid rounding until you reach your finalanswer180 = p radians 90 =p2radians 60 =p3radians45 =p4radians 30 =p6radiansRADIANS and ARCS360 º = 2π radians4www.mathsbox.org.uk
5. 5. LENGTH OF AN ARC  l = rθlAREA of SECTOR = ½ r2θrr7 Trigonometryy = sin θsin (­θ) = ­ sin θ  sin (180 – θ) = sin θyx90 180 270 360– 90– 180– 2701– 1sin (π – θ) = sin θ  (in radians)y = cos θcos (­θ) = cos θ  cos (180 – θ) = ­ cos θyx90 180 270 360– 90– 180– 2701– 1cos ( π – θ) = ­ cos θ (in radians)y = tan θ yx90 180 270 360– 90– 180– 270510– 5– 10tan x =sinxcos xTan (­θ) = ­ tan θsin2x + cosx2= 1LEARN THIS IDENTITYSOLVING EQUATIONSe.g. Solve the equation 2sin2x = 3 cos x for 0 < x < 2 π22 sin2x = 3 cosx2(1 – cos2x) = 3 cos xcos2x + 3cosx – 2 = 0(2cos x – 1)(cos x + 2) = 0cos x = -2 has no solution5www.mathsbox.org.uk
6. 6. cos x = -½ x =p3or 2p -3p=5p3PROVING IDENTITIESShow that (sin x + cos x )2= 1 + 2sinx cos xLHS: (sin x + cos x)2= (sin2x + 2 sinx cos x + cos2x)= sin2x + cos2x + 2 sinx cos x= 1 + 2 sin x cos xLHS = RHSTRANSFORMATIONSStretch in the y direction scale factor a  y = a sin θ6y = sin x y = sin (x+30)yx90 180 270 360– 90– 18012– 1– 2AStretch in the θ direction scale factor1by = cos  bθy = sin x y = 4 sin xyx90 180 270 360– 9080– 112345– 1– 2– 3– 4y = cos θ  y = cos 2θyx90 180 270 360– 90– 18012– 1Translation êéy = sin (θ + c)úû- cùë 0www.mathsbox.org.uk
7. 7. Translation êéú  y = cosθ + dû0ùëdy = cos θ  y = cosθ +2yx90 180 270 360– 90– 18012345– 1A8 Exponentials and Logarithms· A function of the form y = axis an exponential function· The graph of y = axis positive for all valuesof x and passes through the point (0,1)· A Logarithm is the inverse of anexponential functiony = axÞ x = loga ye.g log2 32 = 5 because 25= 32LEARN THE FOLLOWINGloga a = 1 loga 1 = 0loga ax= x alog x= xLaws of Logarithmsloga m + loga n = loga mnloga m – loga n = logaæ mèçn øö÷kloga m = loga mkSolve logx 4 + logx 16 = 3logx 64 = 3x3= 64x = 4y = 3– xy = 4xy = 2xyx1 2 3– 1– 2– 3246810121416E.g7www.mathsbox.org.uk
8. 8. · An equation of the form ax=b can be solved by taking logs of both sides9 Differentiation and Integration· If y = xnthendy= nxdxn – 1·ò xndx =xn + 1n + 1+ c for all values of n except n = – 1e.gò1 + x2x=ò x–12 + x32 = 2x12 +25x52 + cTRAPEZIUM RULEThe trapezium rule gives an approximation of the area under a graph.If the gap between the ordinates is h thenArea = ½ h (end ordinates + twice ‘sum of interior ordinates’)Use the trapezium rule 4 strips to estimate the area under the graph of y = 1 + x2from x = 0 to x = 2x y0 10.5  √1.251  √21.5  √3.25yx1 2 3– 112– 1Area =½ x 0.5 ( 1 + √3.25 + 2(√1.25 + √2))= 1.966917= 1.97 (3sf)8www.mathsbox.org.uk