Core 4 Module Revision Sheet

1 Core 4 Module Revision SheetThe C4 exam is 1 hour 30 minutes long and is in two A(36 marks) 5 7 short questions worth at most 8 marks B(36 marks) 2 questions worth about 18 marks are allowed a graphics you go into the exam make sure you are fully aware of thecontents of the formula bookletyou receive. Also be sure not to panic; it is not uncommon to get stuck on a question (I vebeen there!). Just continue with what you can do and return atthe end to the question(s)you have found hard. If you have time check all your work, especially the first question youattempted.. always an area prone to M SI cannot stress enough that with A2 modules such as C3 and C4 success is en-tirely through practice.

2 The topics and questions are harder and longer andonly diligent practice will let you gain top grades. This Revision Sheet willtherefore be of only limited have been warned!!!1. Algebra Review binomial expansion from C1 for (x+y)nfor integern. Notice that it is valid foranyxandyand that the expansion hasn+ 1 terms. The general binomial expansion is given by(1 +x)n= 1 +nx+n(n 1)2!x2+n(n 1)(n 2)3!x3+ and is valid for anyn(fractional or negative) but 1< x <1 ( |x|<1). Notice alsoit must start with a 1 in the brackets. For example expand (4 x) 1/2.(4 x) 1/2=(4(1 x4)) 1/2=12(1 x4) 1/2=12[1 +( 12)( x4)+( 12)( 32)2!( x4)2+( 12)( 32)( 52)3!( x4)3+ ]=12[1 +x8+3x2128+15x33072+ ].

3 It is only valid for|x/4|<1 |x|<4. Partial fractions is effectively the reverse of combining together two algebraic example1x+ 1+1x+ 2 Algebraic FractionsPartial Fractions 2x+ 3(x+ 1)(x+ 2).To review algebraic fractions quickly; you must be able to manipulate algebraic fractionsfluently. The rules for fractions MultiplyingDividingab cd=ad bcbd,ab cd=acbd,ab cd=ab dc= for example simplify the following:aa+ 1+1a 1=a(a 1) + 1(a+ 1)(a+ 1)(a 1)=a2+ 1a2 1. The general principle for partial fractions is that we need to split an algebraic fractioninto two (or more) fractions. So we create anidentitythat must be true for allxand thenfind certain missing constants. For example we expandx 1(x 4)(x+ 1)toAx 4+Bx+ 1 x 1(x 4)(x+ 1) Ax 4+Bx+ need to discoverAandB.

4 Multiplying through by (x 4)(x+ 1) and cancelling wefindx 1 A(x+ 1) +B(x 4).We can either multiply out and equate coefficients1or choose values ofxto work out theconstantsAandB. I prefer the latter. Here we would letx= 1 to discoverB= 2/5and letx= 4 to discoverA= 3/5. Thereforex 1(x 4)(x+ 1) 35(x 4)+25(x+ 1). The overall methods you need to know are:px+q(ax+b)(cx+d) Aax+b+Bcx+d,px2+qx+r(ax+b)(cx2+d) Aax+b+Bx+Ccx2+d,andpx2+qx+r(ax+b)(cx+d)2 Aax+b+Bcx+d+C(cx+d) Trigonometry By definitionsecx 1cosx,cosecx 1sinx,cotx remember these by the third letter (underlined). You must be able to produce graphsof these in both radians and degrees (P184). By dividing sin2x+ cos2x 1 by sin2xand cos2xwe can derive 1 + cot2x cosec2xand tan2x+ 1 sec2xrespectively.

5 These allow us to solve a whole family of equationsthat we couldn t solve before. With any equation with a combination of trigonometricfunctions (and if one of them is squared) we should be using these relationships to helpus. It is important to note that is you must be careful when you seethings like 2 tanxsinx=tanx. It issotempting to divide by tanxto yield 2 sinx= 1. But you must bringeverything to one side and factorise; tanx(2 sinx 1) = 0. The full solutions can thenfound by solving 2 sinx 1 = 0 andtanx= 0. [It is completely analogous tox2=x. Ifwe divide byxwe findx= 1, but we know this has missed the solutionx= 0. Howeverwhen we factorise we findx(x 1) = 0 and both solutions are found.]

6 ]1 Here we would findx 1 Ax+A+Bx 4B (A+B)x+(A 4B). ThereforeA+B= 1 andA 4B= solve toA= 3/5 andB= 2 Parametric Equations Curves are usually given in the formy=f(x). However we have seen in C1 that this isnot always the best form; for example a circle of centre (a, b) and radiusris usually givenin the form (x a)2+ (y b)2=r2. A curve can also be expressed by aparameter(usuallytor ). In this systemy=f(t) andx=f(t). Therefore instead ofybeing linkeddirectlywithx, they are linkedindirectlythrought. Usuallytcan vary over all the real numbers, but it can also be restricted( >0 or 06 62 ). Graphs of more exotic relationships can be obtained inparametrics (see P225). To graph the parametric you just vary the parameter and plotthe resulting (x, y) points.

7 You must be able to convert any reasonable parametric systemto a normalxyrelationshipby eliminating the parameter. If it contains merely a polynomial relationship then itshould be easy enough. For example eliminatetfromy= (t+ 1)2,x=t 1. From thexrelationshipt=x+ 1, soy= ((x+ 1) + 1)2=x2+ 4x+ 4. In harder examples you must use algebraic trickery; for exampley=t1 t,x=t1+t. Fromthe second we discoverx=t1 +t x+xt=t t=x1 x,so by substitutiony=x1 x1 (x1 x)=x1 x1 x1 x (x1 x)=x1 2x. If the relationship involves sin s and cos s then you will need to be more cunning. Bearin mind that most of these will be circles or ellipses so we will probably be dealing withx2 s andy2 s. Always think about the ways we know to get rid of trig functions likesin2 + cos2 = 1 and the other trig identities from the previous section.

8 For example eliminatetfromy= 1 + 3 sin ,x= 4 + 3 cos . We find(y 1)2= 9 sin2 and (x 4)2= 9 cos2 .Adding and using that 9 sin2 + 9 cos2 = 9 we find (x 4)2+ (y 1)2= 9. So it is acircle. This is a specific example of a general type you must learn:x=a+rcos y=b+rsin (x a)2+ (x b)2= similar relationship for the circle centre (0,0) is trivial from the above if we leta=b= 0. Calculus is possible with a parametric relationship. We know from the chain rule thatdydx=dydt dtdx. A simple extension of this shows thatdydx=dy/ example finddy/dxforx=e2t+ 1,y= 3e4t;dydx=dy/dtdx/dt=12e4t2e2t=6(e2t)2e2 t= The questions asked on this can be rather algebraic! Insteadof giving you a specific pointon the curve they get you to consider a general point where theparameter is some value(sayp).

9 They then get you to consider the gradient of the tangent (or normal) at thatpoint. You may then need to substitute this intoy y1=m(x x1). There are twodetailed examples on P238/9. Please try to do these by yourself; they are not easy. Obviously turning points are found by puttingdy/dx= 0 (when aren t they?!). Oncethe value of the parameter is found, put it back into the original relationships to find thecartesian coordinates. To discover their nature you must goback to basics and considerpoints either side; don t try to do double Further Integration ba y2dxis the volume of revolution of the curveyrotated about thex-axis betweenx=aandx=b. All that is needed for you to do is calculatey2in terms ofxfromy.

10 (For volumes of revolution around the they-axis switch thexand theyand use qp x2dybetweeny=pandy=q.) You must be on the lookout for integrals of the form 1(x 1)(x 2)dx. Use partial fractionsto split the terms andthenintegrate. 1(x 1)(x 2)dx= 1x 2 1x 1dx= 1x 2dx 1x 1dx= ln(x 2x 1)+ must be very fluent with using partial fractions to simplify terms. The area underanycurve can beapproximatedby the Trapezium Rule. The governingformula is given by (and contained in the formula booklet youwill have in the exam) bay dx 12h[y0+yn+ 2(y1+y2+ +yn 1)],wherehis the width of each trapezium,y0andynare the end heights andy1+y2+ +yn 1are the internal VectorsTwo Dimensional Vectors The vector(34)can be written 3i+ 4jand represents a vector going 3 right and 4 up.