GCSE Further Mathematics

THOMAS WHITHAM SIXTH FORMGCSE FurtherMathematicsRevision GuideS J CooperThe book contains a number of worked examples covering the topics neededin the Further Mathematics Specifications. This includes Calculus,Trigonometry, Geometry and the more advanced Algebra.

Page 1Algebra Indices for all rational exponents a m a n a m n , a m a n a m n , a m11a 2 a , a n n a , a n a b amn n b a , a mn , a 0 1 ,1 an , nan n am Example1ann a nm183 3 8 2Example 9 12 1923232 x x3 4 x3ExampleSimplify 2 x 1 3x 6 x 3 4 3 19 822 x 1 3x 6 x 3 2 42Example122118 x 23 9x 2 3 24x6x6xxQuadratic equationsax 2 bx c 0Examples (i) 4 x 2 9(ii) 4 x 2 9 x (iii) 4 x 2 9 x 2 13

Page 2(i)(ii)x2 943 x 2Example(iii)4x 2 9x 04x 2 9x 2 0x(4 x 9) 0(4 x 1)( x 2) 0 x 0 or x 94 x 2 or x Solve (i) 2 x 2 x 5 0 using the method of CTS(ii) 5x 2 7 x 1 0 using the formula.(i) 2 x 2 x 5 0x2 x 5 2 22x 1 5 1 x 2 4 2 4 22 x 14 2 52 161 x 14 2 1641x 14 4116x 14 4116x 1.35, 1.85 Simultaneous equationsExampleSolve simultaneously 2 x 3 y 8, y x 2 x 2Here we substitute for y from the second equation into the first 2 x 3 y 8 2 x 3 x 2 x 2 8 3x 2 x 2 0 3x 2 x 1 0 x 23 , 114

Page 3when x 23 , y 94 23 2 when x 1 , y 1 1 2 2289}Solutions,The geometrical interpretation here is that the straight line2 x 3 y 8 and the parabola y x 2 x 2 intersect at points 23 ,289 1,2 (1,2)0 Intersection points of graphs to ‘solve’ equations. There are manyequations which can not be solved analytically. Approximate rootsto equations can be found graphically if necessary.ExampleWhat straight line drawn on the same axes as the graphof y x 3 will give the real root of the equation x 3 x 3 0 ?x3 x 3 0 x3 3 x draw y 3 xAs can be seen from the sketch there isonly one real root .0

Page 4ExampleObtain the points of intersection of the circlelineand the𝑦𝑥()()(())UsingHence points are ()Expansions and factorisation –extensionsExample 2 x 1 x 2 x 3 2 x 3 2 x 2 6 x x2 x 3 2 x 3 3x 2 7 x 3Expanding

Page 5 x 3 9 x x x 2 9 x x 3 x 3 ExampleFactorisingThe remainder TheoremIf the polynomial p(x) be divided by ax b the remainder willbe p ba ExampleWhen p( x) 2 x 3 3x 5 is divided by 2 x 1 theremainder is p 12 14 32 5 154ExampleFind the remainder whendivided by.f 1 4 1 3 1 11 1 2 103 is2 remainder is 10The factor theorem Following on from the last itemp ba 0 ax b is a factor of p(x)ExampleShow that x 2 is a factor of 6 x 3 13x 2 x 2 andhence solve the equation 6 x 3 13x 2 x 2 0Let p( x) 6 x 3 13x 2 x 2p(2) 6 2 13 2 2 2 48 52 2 2 032 x 2 is a factor of p(x) p( x) x 2 6 x 2 x 1 by inspection x 2 3x 1 2 x 1 Solutions to the equation are x 2, 13 ,12

Page 6Geometry Gradient/ intercept form of a straight line Equationy mx cyGradient m ( tan )c 0 xDistance between two pointsGiven A x1 ,y1 B x2 ,y 2 thenAB 2 x2 x1 y 2 y1 2 Gradient of a line through two points A x1 ,2y1 and B x2 ,saym y 2 y1x 2 x1Equation of a line through (x’, y’) of gradient my y m x x y2

Page 7 Equation of a line through two pointsFind the gradient using m y 2 y1and use the formula as above.x 2 x1Parallel and perpendicular linesLet two lines have gradients m1 and m2 Lines parallel m1 m2Lines perpendicular m1m2 1ory1 and B x2 ,Mid-point of line joining A x1 ,m1 1m2y 2 coordinatesare 12 x1 x2 , 12 y1 y2 General form of a straight lineax by c 0. To find the gradient, rewrite in gradient/interceptform.

Page 8ExampleGiven points A 2, 3 and B 1, 1 find(a)distance AB(b)the coordinates of the mid-point M of AB(c)the gradient of AB(d)the equation of the line through C 5, 2 parallel to AB(a)AB 2 1 2 1 3 9 16 25(b)M 12 , 1 (c)Gradient AB (d)Point 5, 2 Gradient 432Equation2 AB 5 1 34 1 23y 2 43 x 5 3 y 6 4 x 203 y 4 x 26ExampleFind the gradient of the line 2 x 3 y 12 and theequation of a perpendicular line through the point 0, 4

Page 92 x 3 y 12 3 y 2 x 12 y 23 x 4 gradient 23Gradient of perpendicular Equation y 32 x 41 2332 y mx c The CircleAngles in semicircle is 90 Perpendicular to a chord from centre ofcircle bisects the chord. Centre, radius form of equation x a 2 y b 2 r 2Centre (a, b) radius rExampleCentre (2, -1) radius 3 equation x 2 y 1 922

Page 10ExampleCentre (1, 2) touchingEquation x 1 2 y 2 2 4(1, 2)0 General form of equation x a 2 y b 2 r 2Circle centre () with radiusTo find centre and radius, use the method of CTS to change intocentre/radius form.x 2 y 2 2x 3 y 3 0Examplex 2 y 2 2x 3y 3 0 x x22 2x y 2 3y 3 2x 1 y 3y 2 3 22 3 1 3 22 x 1 2 y 32 2 254 Centre 1, 32 radius 52

Page 11 TangentsAngle between tangent and radius drawnto point of contact is 90 Tangents drawn from extendedpointLine of symmetryExampleFind the equation of the tangent to the circlex 2 y 2 2 x 4 y 5 0 at the point P(2, 1)x 2 y 2 2x 4 y 5 0 x x 2 x 1 y 4 y 4 5 1 422 2x y 2 4 y 52 x 1 2 y 2 2 10 Centre at 1, 2 , radius 10

Page 12Gradient CP (-1, 2) gradient of tangent at P 30P(2, 1)Equation

Page 13Calculus Differentiation by ruleExamplesddx x dxd x 12 12x 12 12 xd 4 d44 x 1 4 x 2 2 dx x dxxd x d 1 2 x 1 dx 2 dx2d 10 0dxd3x 2 x 5 6 x 1dx Vocabulary and more notationdyis the derivative of y (with respect to x)dxdyis the differential coefficient of y (with respect to x).dxExampleExamplef ( x) x2 2x x2x 2x31 x 2 2x 2

Page 14 f ( x) 32 x 2 12 2 x1 321 32 x 2 1x32 31x 2x xThe gradient of a curve at any point is given by the value ofdyatdxthat point.ExampleFind the gradient at the point P(1, 5) on the graph ofy x 2 2 x 2 . Hence find the equation of the tangentat P.y x 2 2x 2ydy 2x 2dxP(1, 5) At P(1, 5) gradient 4Tangent at P0xy 5 4 x 1 y 4x 1 Stationary points on the graph of a function are points where thegradient is zero.

Page 15STATIONARY POINTSTURNING POINTS POINTS OF INFLEXION MAXIMUM POINT MINIMUM POINTTANGENT PASSING THROUGH THE CURVETo obtain coordinates of a SP. on the graph of y f (x)(i) Put f ( x) 0 and solve for x.(ii) If x a is a solution of (i) the SP will be a,f (a) .(iii) If f (a) 0 there will be a minimum point at x aIf f (a) 0 there will be a maximum point at x aIf f (a) 0 there could be max or min or inflexion so thesecond derivative rule fails. Investigate the gradient to theimmediate left and right of the stationary point. (see the and signs on the diagrams in the previous section).Example Find the stationary points on the graphs of(i)y x 2 2x 2(ii)y x 3 3x 2and sketch the graphs.

Page 16(i)Here we have a quadratic function, which will have a true maxor min.y x 2 2x 2ydy 2x 2dx SP at 2 x 2 0i.e.at x 1i.e.at 1, 1 2Check point (0, 2)0x(-1, 1)d2y 2 0dx 2y SP is a minimum.(ii)(-1, 4)y x 3 3x 22dy 3x 2 3dx(-2, 0)For SP 3x 2 3x 0 x2 1x 1 SPs at (1, 0) (-1, 4)d2y 6xdx 2(2, 4)0(1, 0)x

Page 17d2y 6 0 MinAt (1, 0)dx 2At (-1, 4)d2y 6 0 Maxdx 2Check points (0, 2) (2, 4) (-2, 0)Note that the turning points are Local Max and Local Mind2y 6xdx 2d2y 6 0 MinAt (1, 0)dx 2At (-1, 4)d2y 6 0 Maxdx 2Check points (0, 2) (2, 4) (-2, 0)Note that the turning points are Local Max and Local Min

Page 18Trigonometry Trig ratios for 30 , 60 , 45 30 22 3145 60 111sin 30 cos 60 sin 60 cos 30 tan 45 132tan 60 3 tan 30 sin 45 cos 45 1213Trig ratios for all angles NB the CAST DIAGRAMFor the sign of a trig ratioAll positive in first quadrantSine (only) in second quadrantEtc SATC12

Page 19ExampleWithout using a calculator find(i) cos 150 (i)(ii) tan 210 (iii) sin 240 (ii)S(iii)SA ASA210150306030-240CTTcos 150 cos 30 Ctan 210 tan 30 32CT sin 240 sin 60 1332Trig of Scalene trianglesSine ruleBAab sin A sin B c sin C abCGiven AAS use it to find a second sideGiven SSA use it to find a second angle (but take care to choose theangle size appropriately –it could be acute or obtuse).

Page 20Cosine rulea 2 b 2 c 2 2bc cos ABcAcos A ab2 c2 a22bcBoth formulae with two morebsets.CGiven SAS use it to find the third sideGiven SSS use it to find an angle (no possible ambiguity here).ExampleTriangle PQR has PR 3cm, QR 7cm and QPˆ R 36 Find (i) QR using the cosine rule and then (ii) PQˆ R usingthe sine rule.Q(i) QR 2 9 49 42 cos 36 24.021.3R36P7QR 4.901. 4.90(ii)74.901. sin PQR sin 36sin PQR 7 sin 36 0.8394.4.901. PQR 57.086. or PQR 122.914.

Page 21It can’t be 57.08. since R would be 86.92. and would be the largestangle in the triangle, but R faces the smallest side so is the smallestangle. Hence PQR 122.91Area “ 12 absin C ” rule given SASArea of triangle 12 absin CGraphs of trig functions (all periodic) 1. Graph of y sin xy1Period 3600x-12. Graph of y cos xy1Period 3600-1x

Page 223. Graph of y tan xPeriod 180y0xVertical asymptotes Boundary values of trig ratiosS 1T - C 0T Verify these from graphsS T 0C -1S T 0C 1T S -1 T - C 0

Page 23 Two important trig identitiessin tan cos Examplesin 2 cos 2 1Given is obtuse and sin 178 find the values of cos and tan .sin 2 cos 2 1 cos 2 1 sin 2 S64 1 289 A 22528915 cos 17sin tan cos tan 8171517 T 815NB Learn how to rearrange the identities sin tan sin cos tan cos cos 2 1 sin 2 sin 2 1 cos 2 Trig equations Remember that from your calculator sin 1 , cos 1and tan 1 give the principal value (p.v.)C

Page 24ExampleSolve the equations(i) tan 1.5 for 0 360 (ii) sin 2 0.5 for 180 180 (iii) 2 cos 2 1 sin for 0 360 (iv) 2 sin 2 sin cos for 0 360 (v) sin 80 (i)3for 180 180 2AStan 1.5PV -56.30 124 , 304 TC(ii) sin 2 0.5 .first solve for 2 for 360 360 2 30, 150; 210, 330 15 , 75 ; 105 , 165SA PV 30 TC

Page 25(iii) (In this example, use cos 2 1 sin 2 ) 2 cos 2 1 sin 2 1 sin 2 1 sin 2 2 sin 1 sin 22 sin sin 1 02 sin 1 2 sin 1 0 sin 1 90 orASsin PV -30 12T 210 , 330 C 90 , 210 , 330 (iv) Don’t cancel out sin . Bring to LHS and factorise2 sin 2 sin cos 2 sin 2 sin cos 0sin 2 sin cos 0 sin 0or 0 , 180 2 sin cos sin 1 cos 2SPV 26.56 T12 27 , 207 tan 0 , 180 , 27 , 207 AC

Page 26(v) sin 80 32 80 60 ,solve first for 260 100 AS 240 PV 60 140 , 160 TExampleCSolve the following equations(i) cos x 0.3 for 0 x 360 , answers correct to 2d.p.(ii) tan(i)x 3 for 0 x 360 , answers in exact form2cos x 0.3 .ASx 75.5, 287.5PV 72.5T(ii)tanCx 3 solve first for 0 x 3602x 60 ,240 2 x 120 SAPV 60 TC

Page 27 Matrices1. Multiplying matricesIn general a b x ax by cdy cx dy A 2x2 matrix multiplied by a 2x1 gives a 2x1 matrixax bz a b w x aw by c d y z cw dy cx dz A 2 by 2 matrix multiplied by a 2 by 2 gives a 2 by 2 matrixExample If A (Work out(i)) and B ((i) AB (ii) AAAB ()( ((ii)))AA ( ())()Worked out by the sum-1x2 3x-1)Worked out by the sum2x3 1x1

Page 28ExampleFind the values of a and b when 2 3 a 5 35 b 2 From the first rowFrom the 2nd rowNow solve simultaneouslyX5X3And2. Multiplying a matrix by a number.ExampleGiven() work out(())Each element in the matrix is multiplied by the constant 3.

Page 293. Using matrices for transformationsExampleWhich transformation is equivalent to a reflection in the x-axis?yBMaking use of the1unit square reducesthe amount of workrequired.1AxIf we look at the movement of A(1, 0)And B(0, 1) when the transformation is about 1 1 A A 0 0 the origin then we can obtain the matrix 0 0 B B 1 1 A’B’ x 1 0 x y 0 1 y Hence Example 0 1 Which transformation is defined by the matrix 1 0 ?

Page 30Again drawing the unit square and looking at where A(1,0) and B(0, 1)moves to will help identify this matrix.y B1B’-1 1 0 A A 0 1 A1x 0 1 B B 1 0 -1 A’ A reflection in the line–4. Combinations of transformations. 0 1 M represents the transformation given by 1 0 1 0 N represents the transformation given by 0 1 a) Describe matrix M.b) Describe matrix N.c) Find the single transformation for the transformation MN andits description.

Page 31y B1a)B’-1 1 0 A A 0 1 A1x 0 1 B B 1 0 -1 A’Hence M represents a reflection inb)y B1A’-1 1 1 A A 0 0 A1xHence N represents a reflection in thec) 1 0 0 1 0 1 1 0 0 1 1 0 0 0 B B 1 1 axis

Page 32y B1 1 0 A A 0 1 B’A1-1x 0 1 B B 1 0 -1 A’Hence MN represents a rotation ofclockwise centre (0,0)

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Revision Guide S J Cooper The book contains a number of worked examples covering the topics needed in the Further Mathematics Specifications. This includes Calcul