A sequence is a list of numbers in a specific order. A series is a sum of the terms of a sequence.


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1 Core Module Revision Sheet The C exm is hour 30 minutes long nd is in two sections. Section A (36 mrks) 8 0 short questions worth no more thn 5 mrks ech. Section B (36 mrks) 3 questions worth mrks ech. You re llowed grphics clcultor. Before you go into the exm mke sureyou re fully wre of the contents of theformul booklet you receive. Also be sure not to pnic; it is not uncommon to get stuck on question (I ve been there!). Just continue with wht you cn do nd return t the end to the question(s) you hve found hrd. If you hve time check ll your work, especilly the first question you ttempted...lwys n re prone to error. J MS. Sequences & Series A sequence is list of numbers in specific order. A series is sum of the terms of sequence. A periodic sequence repets fter some fixed number of terms (i.e. k+p = k for some fixed vlue of p). For exmple 4,,6,7,4,,6,7... is periodic. Anoscilltingsequenceoscilltesboutsomemiddlevlue. Therefore3,4,5,4,3,4,5,4,3... is oscillting. Sequences re sometimes defined inductively. For exmple the sequence n+ = n + 3 with = 0 defines the sequence 0,3,6,9... We know tht this is n rithmetic sequence which cn lso be defined deductively by n = 0+3(n ). Arithmetic Sequences An rithmetic sequence increses or decreses by constnt mount. The letter lwys denotes the first term nd d is the difference between the terms (negtive for decresing sequence!). The nth term is denoted n nd stisfies the importnt reltionship n = +(n )d. For exmple if told the third term of sequence is 0 nd the seventh term is 34 then we cn use the bove eqution to find the nd d. 0 = +(3 )d 34 = +(7 )d 4d = 4 d = 6 =. The sum of the n terms of n rithmetic sequence is given by S = n (First+Lst) = n (+(n )d). For exmple the sum of the first 0 terms of sequence is 30 nd the first term is 4. Wht is the difference? S = n (+(n )d) 30 = 0 (8+(0 )d) d =. J.M.Stone
2 Geometric Sequences A geometric sequence is one where the terms re multiplied by constnt mount. For exmple,,4,8,6,...,[ n ] is geometric sequence with = nd r =. The nth term is given by n = r n. So for the bove exmple the 0th term is 0 = 9 = The sum of n terms of geometric sequence is given by S = rn r. For exmple sum the first 0 terms of 4,,,,...,[4 n ]. This is given by S = 4 ( ) 0 = If the rtio (r) is between nd (i.e. < r < ) then there exists sum to infinity given by S = r. Therefore S for the bove exmple is S = 4 = 8. We cn see tht the sum to 0 terms is very close to S.. Differentition Differentition llows us to clculte the grdient function dy. This tells us how the dx grdient on the originl function y chnges with x. The rules re tht; y = constnt dy dx = 0 y = x dy dx = y = x n dy dx = nxn For exmple y = 4x 4 3x +x 5 dy dx = 6x3 6x+. Turning points re where the grdient of the curve is zero. They re either mxim, minim or points of inflection. To find the turning points of curve we must find dy/dx nd then set dy/dx = 0 nd solve for x. To determine the nture of turning point we must consider the sign of the grdient either side of the turning point. Present this in tble. In the exmple of y = x +x+3 we find dy/dx = x+ so we solve 0 = x+ to give the turning point when x = : x x < x > dy/dx negtive 0 positive minimum We cn lso use the second derivtive to determine the nture of turning point. This is found by differentiting the function twice; y = x 3 +3x x+4 dy dx = 6x +6x d y dx = x+6. You then evlute the second derivtive with the x vlue t the turning point nd look t its sign. If it is positive it is minimum, if it is negtive it is mximum. If it is zero then it is probbly point of inflection, but you need to do the bove nlysis either side of the turning point. J.M.Stone
3 3. Integrtion Know tht integrtion is the reverse of differentition. Tht is if dy dx = f(x) then y = f(x)dx. For exmple if dy dx = 3x3 then y = 3x 3 dx = 3 4 x4 +c. The generl rule is therefore x n dx = xn+ n+ +c. ydx is n indefinite integrl becuse there re no limits on the integrl sign. When evluting these integrls never forget n rbitrry constnt dded on t the end. For exmple 6x dx = x 3 +c. b ydx is definite integrl nd is the re between the curve nd the xxis from x = to x = b. Ares under the xxis re negtive. (For res between the curve nd the yxis switch the x nd the y nd use q p xdy between y = p nd y = q.) To find the re between two curves between x = nd x = b evlute b (top bottom)dx. The re under ny curve cn be pproximted by the Trpezium Rule. The governing formul is given by (nd contined in the formul booklet you will hve in the exm) b ydx h[y 0 +y n +(y +y + +y n )], where h is the width of ech trpezium, y 0 nd y n re the end heights nd y +y + + y n re the internl heights. 4. Trigonometry The sine rule sttes for ny tringle sina = sinb b = sinc. c The cos rule sttes tht = b +c bccosa. Prctice both sine nd cos rules on pge 93. Wedefinetnθ sinθ cosθ. Thisidentity isvery usefulinsolvingequtions likesinθ cosθ = 0 which yields tnθ =. The solutions of this in the rnge 0 θ 360 re θ = 63.4 nd θ = 43.4 to one deciml plce. Know the following (or better yet, lern couple nd be ble to quickly derive the rest from your knowledge of the trigonometric functions): θ sinθ cosθ tnθ undefined Never, never, never, never! It s n esy mrk on 7 mrk pper; don t lose it! 3 J.M.Stone
4 Be ble to sketch sinθ, cosθ nd tnθ in both degrees nd rdins. By considering right ngled tringle (or point on the unit circle) we cn derive the importnt result sin θ+cos θ. This is useful in solving certin trigonometric equtions. Worked exmple; solve = cos θ +sinθ for 0 θ 360. = cos θ+sinθ = ( sin θ)+sinθ get rid of cos θ, 0 = sin θ +sinθ qudrtic in sinθ, 0 = sin θ sinθ fctorise s norml, 0 = (sinθ+)(sinθ ). So we just solve sinθ = nd sinθ =. Therefore θ = 0 or θ = 330 or θ = 90. By considering hlf of generl prllelogrm we cn show tht the re of ny tringle is given by A = bsinc. There re (by definition) π rdins in circle. So 360 = π. To convert from degrees π to rdins we use the conversion fctor of 80. For exmple to convert 45 to rdins we clculte 45 π 80 = π 80 4 rd. From rdins to degrees we use its reciprocl π. When using rdins the formule for rc length nd re of sector of circle become simpler. They re s = rθ nd A = r θ. Extending the results from C we find tht given y = f(x) then: Function Grph Shpe f(x) Norml Grph f(x) Grph stretched by fctor of wy from the xxis i.e. every vlue of f(x) in the originl grph is multiplied by f(x) Grph squeezed by fctor of towrds the yxis 3f(4x) Grph squeezed by fctor of 4 towrds the yxis followed by stretching by fctor of 3 wy from the xxis f(x)+6 Grph moved verticlly up 6 units f(x) 6 Grph moved verticlly down 6 units f(x+4) Grph moved 4 units to the left f(x 6) Grph moved 6 units to the right f(x 6)+9 Grph trnslted 6 units to the right nd 9 units up. This is trnsltion nd cn be expressed s ( 6 9 f(x) Grph reflected in the xxis f( x) Grph reflected in the yxis ) ( where chnge in x ) chnge in y 5. Logrithms & Exponentils Logrithms nd exponentils (powers) re the inverse functions of ech other. Therefore So if log = 5.4 then = log 0 0 x = x nd 0 log 0 x = x. On pge 35 there re lots of questions of the type simplify log 4. In these types of questions you must write the number you re logging s power of the bse of the logrithm. So in this cse log 4 = log 4 4 = log 44 =. 4 J.M.Stone
5 There re few rules tht you must lern bout logrithms. They re (for ll bses): log(b) = log+logb log = 0 log(/b) = log logb log = log( n ) = nlog log(/) = log log b = log cb log c. Whenweneedtosolvenequtionwheretheunknownisintheexponentsuchs5 x = 8 tke log 0 of both sides nd simplify: 5 x = 8 log 0 (5 x ) = log 0 8 (x )log 0 5 = log 0 8 x = log 08 log 0 5 x = ( log0 8 log x =.5 (3sf). There re two types of reltionships tht cn be modelled by logrithms. They re exponentils of the form y = b x nd polynomils of the form y = x b. When one of these reltionships is suggested in the exm, then tke logs of both sides nd rerrnge to compre with Y = mx +c. Therefore (A) y = b x (B) y = x b logy = log(b x ) logy = log(x b ) logy = (logb)x+log ) logy = blogx+log. So in (A) we plot x ginst logy nd find grdient of logb nd yxis intercept of log. In (B) we plot logx ginst logy nd find grdient of b nd yxis intercept of log. Do some exmples of these questions on (nd round) pge Further Differentition & Integrtion Nothing relly new in this chpter, just extending the two importnt results for differentition nd integrtion to frctionl indices.. If y = x n then dy dx = nxn.. If y = x n then x n dx = xn+ n+ +c. For exmple if y = 4x x 4 5 then dy dx = 5x x J.M.Stone
6 For exmple, given tht y = x+ x find the re under the curve from x = nd x =. x+ x dx = x +x dx [ ] = 3 x3 +x ( ) ( ) = =.05 (3sf). 6 J.M.Stone
x 2 + n(n 1)(n 2) x 3 +
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