ADDITIONAL MATHS OCR FSMQ LEVEL 3

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ЛЗ f(-2) = 0 f(-1) = 0 f(0) = -2 X f(1) = 0 f(2) 소 12 X = factorise X³ + 2x²-x-2 the result must = 0 to be (x-4)²-13 4 ± (6²-4ac) discrimanent: f(x)= x² - 4x + 3 x value = 7.6 = 0.39 Jy value 42 OV 2(1) = [-2] [-1] [1] use the 3 factors above: (x + 2) (x + 1) ( x −1 ) x³ + 2x²-x-2 + 2(1)(3) not a factor x = 3 not a b2- час хо factor of the vevtex - b ± Quadratic formula: √b² - 4ac 2a x = a factor f(x)=x²-4x+4 / 4 ± √4² +211) (4) 261) x = 2 NN 3 2 Equation ( line ) Equation (normal) 6² - 4ac = 0 (0,3) flx) % = y = mx + c = m ( x-x₁) y-y₁ = (4,-13) b²-4ac70 = x² - 4x + 5 4 ± √ 4²- 211) (5) 2413 x = - 4 no Solution eg. 3 Application of equations and Inequalities In one Simultane o u s : Substitution I elimination Inequalities 2x ≤ 4 X ≤ 4 An Inequality remains unchanged if any number IS added or Subtracted on both sides = eg. Make : An Inequality remains unchanged it any number DV operation is multiplied or divided on both sides Solving linear Inequalities: 7/2 An Inequality Will be reversed If both sides ave multiplied or divided by a negative value 0.9 x 3% 12% -4 3x ≤ x + 8 > < y I x²2x-3 < 0 (x + 1) (x-3) < 0 the 2 3 Cinside Solution < x < 3 = subject Solving quadratic Inequalities: + 3 0 4 eg 0.34 <-1.8 -0.3y < -1.8 -0.9% -y 2 5 y > y > - 2 variable 6 + 3% 2y + 6 < 5y + 12 2y < 5y + 6 -3y <6 divide by 6 + 3% O Solution x < -1 + -2 x²2x3 70 (x + 1)(x-3) 7 0 ↑ outside Đ 3 or operation U_" + - 1 x > 3 + 0 + } + 2 sequences and Un = Ui 2un + 3 201 U₂ = 2(1) + 3 5 eg. Unti = U13 14 = U16 = = U5 = 61 2012 2 (6) +3 13 29 2021 2022 = 125 = (1) The value 2024 recurrence relationship in modelling Anti where к IS recurrence ити term 1st term - The Initial Investment is denoted £A, - value on June 1st 2022 is £Az value on June 2nd 2023 IS £A3 ( U₁ = 1) 2025 A4+1 David Invests an amount of money on be paid anually at rate of 3% Will the relationship David's Initial Investment is £5000 (li) Find the value of his Investment on June 1ST 2025 i) K = 1.03 A₁ = 5000 Altı = 1.03 x 5000 = 5150 2023 A2+1 e 1.03 x 5150 = 5304.50 A3 + 1 = 1.03 x 5304.50 = 5463.64 1.03 x 5463.64 = 5627.55 eg. Uv = U₂ = = U 3 = = = = = I recurrence relationship) ↳ you must find out the term in front 20v-1 + 3 242-1 + 3 260) + 3 3 2013-1 +3 2 (3) + 3 9 204-1 +3 219) + 3 21 is given by the recurrence relationship Anti -KAn Constant, write down the value of K 2 1.03 (n₁ = 0) June 1st 2021. Compount interest (iii) Find an expression. An = 5000 x 1.03" (n-1) ↑ Power 0 year 1 = year 2 = 1 year 3 = 2 5 points, lines and Civeles P(3.1) Circle geometry Q (6,9) R QR = 5-1 4 PR = 63 3 = PQ = y = Ø B (2,6) 42 +3² 5 Circles with Centre (0,0), x² + y² = v² (4,-4) P(x, y) 4+2 2 Find the midpoint c 3 > 6-4 2 = > 1 19 = 0 (x₂ - X₁ )² + hypotenuse X1 + X2 2 midpoint уг-у, x2-x1 of AB: (4,5) Find the equation of the circle: (x-3)² + (y-1) ² 26 The equation of the civele c is x² + y² - 8x + 2y. Express in the form (x-^)² + (y-6) ² = √²: x² - 8x + y² + 2y - 19 = 0 → (x-4)² - 16 + (y + 1)² - 1 LX-4)² + y + 1)² = 36 (x-4)² + (y + 1)² = 6² " y · 19 = 0 = Civeles with centre (a, b) (x₂-x₁) ² + Find the distance AC (use mid Point): (4-3)² + (-4-1) ² √26 eq of Civele above: (уг-д,) г MUST BE IN THIS FORM! уі туг 2 (x-4)² + (y-5) ² Hence find gradient P(x, y) (9₂- y₁)² = √² x (8-4)²+(3+1) ² = 3² where 8, 3 lies: = 32 3236 lies Inside Graphs plotting linear & quadratic functions. Complete the square. y=(x-1)² To the power even t even odd odd + ✓ ✓ ^ of ✓ 1 unear Inequalities with 2 variables y > x + 3 X-3 X - exponential functions: y = 2% 2y = x + 4 y ≤ 1/₂ x + 2 2y-x ≤ 4 To test which side: -1+3 = 2 3> 2 V 2 ^ Never touches the neg y av 1 black ones y = 3x the the number, the closer it is to the line y = 2%. put a point Using Inequalities for problem solving: A boy decides to buy a white rabbits and b a) he wants at least 2 black ones b>2 b) he wants at least one white one c) his father says he can have no more than five altogether The ways he can buy his vabbits: 2⁰ = 1 21 = 2² = wwxxkiuly 3° = 1 31 Imove down 1 y=-2* reflect y=2-x 3 on x reflect on y mmmm, a a+b≤5 b≤-a+5 180 Trigonometric functions for angles of sine graph: - 360 -270 -180 -90 cosine graph: Tan graph: 90 ·360-270-180 - X 270 -1 LS -/90 7. -1 90 180 270 y ਨ x= cos o sine y = ਵ UJJ -360 -210-180 -90. 90 180 270/ 360 - dang 1 = 180 Size tane = 30 170 Sino cose x + h Cos ²e + Sin ² 0 = 1 360 / 1- dos 6 Sin ²0 1- inº@ = os 9 Positive 180 180-0 T (80+ 0 S " + 1 - Solve the equation: LOOK at the value if it IS }} plug the num- ber into the equations -360+0.739 2 90 270 2 cos ²0 Sino - \ = 0 2 (Cos²0) - Sine - 1 = 0 2 (1-sin²0) - sine - 1 = 0 2- 2sin²o- sine - ( = 0 0 = 2 sin²0 + sine - I -1 2 Sino sino 1 L2 Sin-1) (Sine +1) = 2 Sine = 1 Sino : 0 = If there is Sin2x = 0.1 2x A C (sin / +) -only A & S will work - 177 11 2 Cos²0 0 1/2 sin (1/2) 30 360-0 1) 30° 2) 180-30 = 150 Sin-¹ (0.1) 5.739 0 (+ / sin) -180-5.739 2 Negative -180-0 a number in front of the "x" - 180 ≤ x ≤ 180 -180 = sine -1 = 0 180+ 0 - 93 11 S :. 1. 30 2.150 3.270 Sin = -1 9 = - ·90 (sin 1-) · Only T & C will work = 180 +90 270 -270 for the values 90 divide answer 180-5.739 2 = A с 360-90 8711 -360-0 o all answers fit in the range 0 0° - 360° 5.739 2 = 3 " COS 20 = 0.4 20 = COS' (0.4) 66.422 66.422 = 2 (360 66.422) KEY sin 2x 00360 ( + I cos) 33 - go another (360+ 66.422) = 2 = 213 (720 2 = 147 : vound 66.422) = 2 = 327 but 33, 147, 213, 327 } only 2 answers } (4 answers) divide 2 to answer tan multiply 3 to answer Cos (X-30) add 30 to the answer values Application of triganometry of CAST 180 S T both still fit in the ranges 540 change T A с A C D 360/720 (10 Permutations and combinations Factorial different no. of arrangements eg. there are 7 children standing in a are there? 7 x 6 x 9 x 4 x 3 x 2 x 1 = 7! using each of these digits no more than once 3, 1, 7, 9, 8,5 i) How many different four digit numbers can be made ? 4 ili) How many digit First: second: 6 X 5 X ii) How many even three digit numbers can be made ? 1 3 digit 2 digit 1 digit even: 8 ↑ odd : 58 1 x 6 x 6x5 b third 1 4 5 x 4 X 3 third. 4 X numbers less than 4000 can be made - has to be smaller than 4 2 x 5 x 4 x 3 3 x second: 5 fourth: X 3 = 360 line iv) How many odd numbers greater than 500 000 can be made? 571915 odd 2 X X Ihow many different first 276 // I = 20 5 4 [5-1] todd 2 × 1 × 5 = arrangements 0! = 1 1! = 1 408 Permutation order matters Factorial notation = ед. I people. nl (n-v)! find first 3 5 7 x 5 x 4 x 3 x 2 x 1 (7-3) x 3 x 2 x 1 Combination. order doesn't matter Factorial notation = people are to be ท! v! (N-V)! 9x8x7x6 = chosen from 9 people. 4 x 3 x 2 X I 9 x 8 x 7 x 6 x 5*4*3*2*+ 6x4x 3 x 2 x 1 x 4x3x zx1 -Order of people/objects matters -"Hot Words" arrangements / orders/ rank (1st 2nd 3rd place) Examples - any type of race (w/ rank) - playlist/ order on a shelf - student gov't (P, VP, S, T) -question uses the "Hot Words" How do I use the calculator to help? First, type in TOTAL number of people/objects. Next, locate PRB button/menu, choose nPr option. Then, type in smaller number of the group. Finally, hit the equal sign. Key words First 3... 7x5x4 793/1 = 905 = 126 Permutations v. Combinations Key words. How many ways -Order of people/objects does not matter Examples -people sitting in a car/vehicle -handshakes or high fives with classmates/teammates - any type of race (no rank, top 10 get tshirts) - student senators How do I use the calculator to help? First, type in TOTAL number of people/objects. Next, locate PRB button/menu, choose nCr option. Then, type in smaller number of the group. Finally, hit the equal sign. 11) Binomial expansion Eg. Find the probability of 6 heads & 4 tails ↓ Probability 0 (at b) (at b)' = 0 (a + b) = (a+b) ³ (at b) 4 = [ 1 / ₂ ] ₁ = × [ 1½ ₂2 ] ^^ x 10C610C4 no. of Success 1 a² + 2ab + b² adds up to total power a³ + 3a²b + 3ab² + b³ a4 + 4 a³ b + no. of Success ba² b² + 4ab³ + b² decrease ач э аз э аг э аэ: 0 0 → b → b²b³b4 Increase 1004 (m426) 210 m 4 x 64 Binomial distribution P(x~ B) the number of ways of choosing V objects from a collection of n objects eg. Find the coefficient of m4 in the expanssion of (m+2) 10 In S 13440m4 ✓ Probability ↑ number of attempts different Combinations occuring Probability Idaffodil bulbs): (a+b) ⁰ (a+b)' (a+b)² (a+b)³ (a+b)" → (a+b)5 (a+b) 6 → 1 1 equal no. of each type of bulb 1 bCo X~ B = 120, 4/5) (4/15) 20 x (¹/5) ⁰ x 200 20 15 1 6 ↑ всі X~B ( 20, 4/5) [ 4/5] 16 × [1/5]" x 200 16 in the sample 1 5 1 4 A Bag contains tulip bulbs and daffodil bulbs. In the ratio 1:4 Selected at random. = 0.0115 X~B ( 20, 4/5) [415] ¹⁰ × [¹15] x 200 10 = 0.002 10 1 1 2 3 1 10 15 ↑ 662 = 0.218 1 3 6 4 the sample contains the 2 types of bulbs In the same ratio as " 1 10 5 20 15 6 ↑ 6C3 1 20 bulbos 1 6C4 6C5 1 the bag: ↑ are ьсь 12 Exponentials and logarithms y=5xy= 2x Exponential growtи: у= схак Li) log x³y √2 (ii) Logarithms: p = = E 2 loga logb- 3 loge loga² - logbloge s log at loge & юда x годећ log bes log x³ + logy - log √z 3 logx + logy - log2 z 3 logx + logy - = log2 (base must be the same) f(x) log f(x) = g(x) loga b = x -base (usually to) Log g(x) Exponential decay = y= cxa-kt eg. → log x. logx" 20 3% log3(20) X= 2.727 y=1te" = ax = b = x LAWS OF LOGARITHMS Lonly applies to same base) log x + logy = подху log2 + log 3 годь = n log x Solve: 3x = log 3x = log2x+1 x log 3 = (x+1) log 2 x log 3 = xlog₂ + log: 2 If no base = assume 10 2x+1 1. 2%= 8 3 году = годиц юдю - год 2 = log10² 2. 3% = 81 4 3. 4* = 54%-1 61 log5 (61) = 4x-1 4%-1 = 2.554 4x= 3.554 X = 0.889 = x log 3 - x log₂ = log2 xllog3 - log2) = год 2 X Log 2 log3-log2 put into calculator = 1.7095 0-5 x = -22 Log x² = n log x -|N = сод 5 2 log10 Reduction to linear form: Evariable y= ахи logy = logaxn & variable году юду Fg. 1. Take 0.5 году y = at" logarithm for both sides logy = log at " году log y = 0.5 y = 1.5t 0.5 10 = 1.5 +0.5 6.667 = t 0.5 logt 401648 t = = X logy = nlogt + log a (straight line → y=mx+c) loga + года a - = + + nlogt содхи людх one value log 6.667 = 0.5 logt and logs Log 6.667 0.5 logt 1.648 t 44.5 11 T y = grams 1 t = minutes 3. Assumming the relationship continues for at least the first hour, now long would there be log of the substance ? 1 Simplify first find the x & y values logy y = (variable) t у >logt год у →logt 1.5 2 nlogx mx ^ ^ Constant (gradient) Find "a" + года с Constant (Intercept) 4 9 14 19 24 29 3.00 4.50 5.61 6.54 7.35 8.08 0.602 0.954 1.15 1.28 1.38 1.46 0.477 0.653 0.749. 0.816 0.866 0.907 + (variable) a a = intercept loga = 0.18 0.18 10 = a 1.51 y= 1.51 t" n= -gradient 0.5 = logy = 0.5 log x + log 1.51 y = 1.51 +0.5 (y=at") Eg. 1. Take logarithm for both sides года содкас = logk + 2. Find E variable y = abx variable log y = logabx logy = loga + logb* году года a + x logb 0 = kat = года года = a = года logat logo = Floga + logk X y и к" unter cept 80 = a = and "a" gradient -0.025 10 - 0 025 0.944 logo ^ 2 1.5 1 0.5 t 9 log y + 4 (variable) ox 0 = = 80 logo 1.90 + 8 log b x m x constant (gradient) log O = + loga + logk сод loge = tlogo.944 + Log 80 0 80 (0.944)* + + 12 + log a с Variable 4 8 12 16 20 63 50 40 32 25 1.80 1.10 1.60 1.50 1.40 Constant (Intercept) 16 +t 20 half way Numerical methods. XO locating a root: 2% = % +4 I root Interval 2 < x < 3) f(x) = 2x-x-4 f (2) - 2 f(3) = 1 f (5) Smallest largest f (5.8) Interval bisection: f(x) = 2*11 - 5% - 4 f (5.4) = 4.22 15.1 = 5 %2 change of sign root Interative sequence: Converge: Inwards no. keep getting smaller Decimal search: detailed Interval bisection use decimal search to find a root with interval "0.001" f(x)= x3 % -1 f(1) = -1 f(2) 1 %1 f(1.1) = f(1.2) = f (1.3) = f (1.4) = →>>>> negative Positive (interval Disection interval of 0.1 → 5 < x < 5.8) 0.89 1.324 < x < 1.325 11 0.76 -0.61 0.344 point X₁1%₂ f(x)= diverge outward log 10 x2 + 5 I root Interval x = 2.3 1 a.p) f( 5.2) = -0.24> neg no. keep getting bigger (2.25) = 0.896 f (2.35). = -0.1514 fl1.31) = f(1.32) = f (1.33) = -0.062 -0.02 0.023 Xnti x₁ = X = 0 X 2 -4 X3 12 X4 = 5.2 < x < 5.3 F15.3) 1.90 ↓ pos (Xn)² - 4 (ans) ² - 4 140 f(1.321) = -0.016 f(1.322) = -0.012 f(1.323) 7.31 x 66³ f(1.324) = -3.06 x00⁰ fl1.325) = 1.20 x 10-³ } Xo = 2 2 Put 2" In cal. diverge Gradient of tangent р има роис owea unded L curve: 11 2 12 3 4 ить 2 3 4 5 6 (1.5,3%) (2,8) Central estimate = f(x+h)-f(x-h) 20 +2 = rectangles X2 +1 fi 3 18 trapezium с forward = f(x) + flxти) и I. Find width уг backward = flx) - flx-n) h 4 rectangles underestimate I start f(0) × 1 + fl1) x 1 вх-х2 = (4-0) = 4 1 Lof one rectangle) from 0). + f(2) x 1 6 rectangles + fl3) x 1 width (a+b)h =Lfl1) + f(2)) 1 formula = (yo+ 16 + 2 ly ₁ + y₂ + Ys + Yu + Y5 ) h) W smallest & largest 15 Integration dy dx dy dx = = X 1. 4x3 4 add one to - 12/2 add c 4 goes back & add 1 4x = first =1/12 223 eg. S (x ³²2 + 2 ) d x x= What multiply 2/2 = 4? 4÷1/1/2=8 : d dx 1 put x behind 2: - : 2x f'(x) differentiated = find f(x) (5x2 ) = \x y = x² + c y = 8x² + C - + 2 -2x = 2% + 2 - 1/2 + 2x + 2% + C S ( ²³ x ² - 1 x ² = ² + 1) dx = dy x > 0 ✓ Integrate A Curve with the equation y=f(x) passes through ²x² - 10x² dx 2. find an equation of the normal to the curve 25 = = (4) ³ - 2014) = + 4 + C C = 53 25 = m(4) + C m = - 1 2 dy du = (2x²¹) y = -x²6 x² 7 10x dx = 5x² + C ^ with respect to x C = = y x² - 1 equation = асс - velocity * * Velocity distance no limit x³ - 20x² + x + C y Intercept 14, 25) y = = x ²³ x + c the point 14, 25) = -1% + 53 + C 10 1 (1,0) y = 4x3 = 5 R % highest value Si [4x³] ax ^ lowest value (4,0) y = x (x-1)(x-2) X % [x4] = Integrate first · [ x* ] 5 = [ 5 ] - [14] limits 624 Add Limits The finite region R IS } bounded by the x axis and the curve with equation y = 27- 2x - 95% - 16/12 S₁ (21-2x - 9x² - 16x²²) dx 2 = ( 27 x 4 - ( 4 ) ² − 6 (4) ³ + 16 (4)¯¹ ) - ( 27(1) - (1) ² + 6(1) ¾ - 1611)¯') 12 = - [ 27x − x² − 6x² + 16 %¯¹ ] 4 -1 x(x-1)(x-2) = (x²-x)(x-2) 0-1 curve: 1-2 curve % 70 = = Só (x³ - 3x² + 2x) dx = [=x*- x³ + x² ] ! 1/4 x³ 2x²x² + 2x x³ 3x² + 2x S² (x³ - 3x² + 2x) = [ 4 x “ - x³ + x² ] } 24 4 total = 4 + 4 Area between curve & unes y ←3 3 y = x → % area between X(4-x) = x 4x - x² = x x² + x - 4x = 0 x14-x) & y = x Sol x² + x - 4x) dx = [ 3 x ² + 1 - 2 x ² ] } 0 9 area of red triangle. A = 3x3 2 4.5 black area = 9 -4.5 = 4.5