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+ 2 = 0 = Z = 6z²7z+2=0 (2z - 1)(3z - 2) = 0 2-√√3 12/13 x = + Sketch the graph of y=4x-x²-1 y=-(x² - 4x) -1 y=-((x-2)²-4)-1 y=-(x-2)² +3 www.mathsbox.org.uk n x³ = "√x¹ = (√x)" NIFT vertex at (-a, b) line of symmetry as equation x = -a y ^ 6- 4- 2+ The conjugate of the denominator 2-√√3 is 2+ √√3 so that z = (2-√3)(2+√3) = 2²-√√3² = 1 NIM x = + Line of symmetry x = 2 Vertex (2,3) 6 8 INIM X Quadratic formula (and the DISCRIMINANT) -b+√b²-4ac 2a x = The DISCRIMINANT b² - 4ac can be used to identify the number of roots d) SIMULTANEOUS EQUATIONS Solving by elimination b² - 4ac >0 there are 2 real distinct roots (graph crosses the x-axis twice) b²-4ac=0 there is a single repeated root (the x-axis is a tangent) b² - 4ac < 0 there are no real roots (the graph does not touch the x-axis) 3x - 2y = 19 x 3 2x-3y = 21 x 2 Solving by substitution for solving ax²+bx+c = 0 x+y=1 (y=1-x) x² + y² = 25 Solve x² + 4x-5 <0 9x - 6y=57 4x-6y=42 x² + 4x-5=0 (x - 1)(x + 5) = 0 x = 1 x = -5 f) POLYNOMIALS If you end up with a quadratic equation when solving simultaneously the discriminant can be used determine the relationship between the graphs If b² - 4ac > 0 the graphs intersect at 2 distinct points b²-4ac=0 the graphs intersect at 1 point (or tangent) b²-4ac <0 the graphs do not intersect Quadratic Inequality - always a good idea to sketch a graph plot the graph as a solid line or curve <> plot as a dotted/dashed line or curve x² + 4x-5 <0 5x - Oy=15 x=3 (9-2y = 19) e) INQUALITIES Linear Inequality - solve using the same method as solving a linear equation but remember to reverse the inequality if you multiply or divide by a negative number -5 <x< 1 which can be writte as {x:x>-5} n {x: x <1} x² + (1-x)² = 25 2x²2x24 = 0 2(x-4) (x + 3) = 0 x = 4 y = -3 If you are unsure of which area to shade pick a point in one of the regions and check the inequalities using the coordinates of the point Solve 4x² 25 ≥ 0 4x² 250 (2x - 5)(2x + 5) = 0 x = = x=-=-2 X 4x² 2520 y = -5 x = -3 y = 4 x=-=/orx²2/20 which can be written as {x:x≤ - }U (x:x> } XS www.mathsbox.org.uk -10- 20- ● ● ● A polynomial is an expression which can be written in the form ax + bxn-1 + cxn-2+ ... where a, b, c are constants and n is a positive integer. The order of the polynomial is the highest power of x in the polynomial Polynomials can be divided to give a Quotient and Remainder Divide x³x²+x+15 by x + 2 X +2 X³ x³ g) GRAPHS OF FUNCTIONS Sketching Graphs y=mx + c ● y = kx² x² -3x + X 2x - 4 = -x 3x = 4 x² + 2x² X = -3x² -3x² <x<4 +X -6x Factor Theorem - If (x-a) is a factor of f(x) then f(a) = 0 and is root of the equation f(x) = 0 Show that (x-3) is a factor of x³ - 19x +30=0 f(x) = x³ 19x + 30 f(3) = 3³ -19 x 3 + 20 = 0 f(3) = 0 so x-3 is a factor of f(x) 7x 7x Identify where the graph crossed the y-axis (x = 0) Identify where the graph crossed the x-axis (y = 0) Identify any asymptotes and plot with a dashed line y=kx³ Asymptotes at y == VALEE y is proportional to x² y is proportional to x² y is proportional to +7 + 15 + 15 + 14 1 Quotient Remainder www.mathsbox.org.uk x = 0 and y = 0 Modulus Graphs ● |x is the 'modulus of x' or the absolute value |2|=2 |-2|=2 To sketch the graph of y = f(x)| sketch y = f(x) and take any part of the graph which is below the x-axis and reflect it in the x-axis Solve |2x-4|<|x| 2x - 4 = x X = 4 V y = Asymptotes at x = 0 and y = 0 y is proportional to h) FUNCTIONS f(x)=√√x domain x ≥ 0 A function is a rule which generates exactly ONE OUTPUT for EVERY INPUT DOMAIN - defines the set of the values that can be 'put into' the function RANGE - defines the set of values 'output' by the function - make sure it is defined in terms of f(x) and not x f:xx² x ER means an input a is converted to a² where the input 'a' can be any real number Range f(x) > 0 ● ● INVERSE FUNCTION denoted by f¹(x) The domain of f¹(x) is the range of f(x) The range of f¹(x) is the domain of f(x) Using the same scale on the x and y axis the graphs of a function and it's inverse have reflection symmetry in the line y = x fg(x) = 4(x²-1) = 4x² - 4 i) TRANSFORMING GRAPHS COMPOSITE FUNCTIONS The function gf(x) is a composite function which tells you 'to do' f first and then use the output in g f(x) = 4x g(x)=x²-1 Reflection flection in the x-axis replace y with Reflection in the y-axis replace x with -x f(x) = Translation To find the equation of a graph after a translation of [] replace x by (x-a) and y by (y - b) y = f(x -a) + b The graph of y=x²-1 is translated by Stretch Stretch in the y-direction by scale factor a y = af(x) Stretch on the x-direction by scale factor 1 a y = f(ax) Reflection y = (-x)² -6(-x)+9 = x² + 6x + 9 y = -f(x) y = f(-x) The graph of y = x² is translated by [3] and then reflected in the y axis. Find the equation of the resulting graph Translation y = (x - 3)² = x² -6x + 9 Find the equation of the resulting graph. (y + 2) = (x-3)² - 1 y=x² - 6x + 6 y = z²2 x+2 x = 3³ -2 f-¹(x) = -2 Combining Transformations Take care with the order in which the transformations are carried out. x+2 gf(x) = (4x)²-1 = 16x² - 1 www.mathsbox.org.uk [_³] find f-¹(x) resulting graph Reflection y=(-x)² = x² The graph of y = x² is reflected in the y axis and then translated by [3]. Find the equation of the Translation y = (x - 3)² = x² - 6x +9 j) PARTIAL FRACTIONS Any proper algebraic fractions with a denominator that is a product of linear factors can be written as partial fractions Useful for integrating a rational function Useful for finding binomial approximations ● A B px+q + + (ax+b)(cx+d) (ex+f) ax+b cx+d ex+f = 2 COORDINATE GEOMETRY a) Graphs of linear functions Gradient = Positive gradient y = mx + c change in y change in x Express A B + x-2 x+3 A(x + 3) + B(x - 2) = 5 5 (x-2)(x+3) = Parallel and Perpendicular Lines y = M₁x + C₁ in the form + x-2 px+q (ax+b)(cx+d)² A(x+3)+B(x-2) (x+3)(x-2) Negative gradient Mid-point = 5 (x-2)(x+3) B x+3 y = m₂x + C2 If m₁ = m₂ then the lines are PARALLEL If m₁ x m₂ = -1 then the lines are PERPENDICULAR Gradient of y-2x = 7 is 2 (y = 2x + 7) Gradient of the perpendicular line = - Finding the equation of a line with gradient m through point (x₁, y₁) Use the equation (y - y₁) = m(x-x₁) If necessary rearrange to the required form (ax + by = c or y = mx + c) the line intercepts the y axis at (0, c) x = 2 5A = 5 x= -3 -5B = 5 A = 1 B = -1 = A ax+b 1 1 x-2 x+3 (2x - = -1) Equation of the line with gradient - passing through (4, -6) (y + 6) = -½(x-4) 2y + 12 = 4 x x + 2y = -8 Finding the mid-point of the line segment joining (a,b) and (c,d) = (atc, b+d) Find the equation of the line perpendicular to the line y-2x = 7 passing through point (4, -6) www.mathsbox.org.uk + Calculating the length of a line segment joining (a,b) and (c,d) Length = √√(ca)² + (d − b)² B cx+d + (cx+d)² b) Circles A circle with centre (0,0) and radius r has the equations x² + y² = r² A circle with centre (a,b) and radius r is given by (x - a)² + (y - b)² = r² Finding the centre and the radius (completing the square for x and y) Find the centre and radius of the circle x² + y² + 2x - 4y -4 = 0 x² + 2x + y²-4y-4 = 0 (x + 1)²-1+(y-2)²-4-4-0 (x + 1)² + (y-2)² = 3² Centre (-1,2) Radius = 3 The following circle properties might be useful Angle in a semi-circle is a right angle ● Finding the equation of a tangent to a circle at point (a,b) The gradient of the tangent at (a,b) is perpendicular to the gradient of the radius which meets the circumference at (a, b) 4-1 Gradient of radius= = Find equation of the tangent to the circle x² + y² - 2x - 2y - 23 = 0 at the point (5,4) (x - 1)² + (y-1)²-25= 0 Centre of the circle (1,1) -10 The perpendicular from the centre to a chord bisects the chord Equation of the tangent (y-4)=(x-5) b²-4ac > 0 10 Lines and circles Solving simultaneously to investigate the relationship between a line and a circle will result in a quadratic equation. Use the discriminant to determine the relationship between the line and the circle b²-4ac0 (tangent) 10 Gradient of tangent = 3 x 3y-12 = 20 - 4x 4x + 3y = 32 -10 The tangent to a circle is perpendicular to the radius -5 ++++++++>x b²-4ac <0 www.mathsbox.org.uk -10 c) Parametric Equations Two equations that separately define the x and y coordinates of a graph in terms of a third variable The third variable is called the parameter x To convert a pair of parametric equations to a cartesian equation you need to eliminate the parameter (you may need to use trig identities if the parametric equations involve trig functions) Find the cartesian equation of the curve given by the parametric equations given by x = cose y = sin20 y = sin20 y = 2sin cose 3. SEQUENCES AND SERIES a) Binomial Expansion of (1+x)" |x|<1 n E Q (1+x)” =1+nx+ ● b) Sequences ● ● Expansion of (a+b)" (a + b)" = a + nan-¹b + ● -2 2-² (1-3x)² = y² = 4sin²0cos²0 = 4(1- cos²0) cos²0 y² = 4(1-x²)x² Use the binomial expansion to write down the first four terms of = 2−² (1 + −2 × (− 3x) + −²x23³ (−³x)² +· 27 1x2 27 = (1 + 3x + +²/7x³) n(n-1) n(n-1)(n-2) 1x2 1x2x3 neZ+ -x² + 3 27 = 1 + ³x + ²7/x² +² n(n-1) an-2b² + 1x2 n(n-1)(n-2) an-3³ 1x2x3 x 26 x (3x)³ = 145152 (x³) -x³ An increasing sequence is one where Un+1 > Un for all n An decreasing sequence is one where Un+1 < Un for all n A sequence may converge to a limit Lun+1 = Find the coefficient of the x³ term in the expansion of (2 + 3x)⁹ (3x)³ must have 26 as part of the coefficient (3+6 = 9) 9x8x7 1x2x3 The sequence defined by Un+1 = 0.2un + 2 0.8L 2 L = 2.5 L = 0.2L + 2 (2-3x)² -2x-3x-4 1x2x3 Find the first 3 terms of a sequence defined by Un+1 = 2Un + 1 U₁2 U₂ = 2x2+1 = 5 U3 = 2 x 5 + 1 = 11 An inductive definition defines a sequence by giving the first term and a rule to find the next term(s) Un+1 = f(un) U₁ = a f(un) as n → ∞0 + nxn-1 + xn c) Sigma Notation - sum of 6 (r² + 1) = (1²+1) + (2²+1) + (3²+1) + (4²+1) + (5²+1) + (6²+1) r=1 = 2 + 5+10 +17 +26+37 = 97 (-²/x)³ + nabn-1 + bn www.mathsbox.org.uk Un+1 = Un = L U₁3 converges to a limit L. Find L A periodic sequence repeats itself over a fixed interval Unta un for all n for a constant a which is the period of the sequence U₁ = 2 Staring with the 1st term r = 1 Ending with the 6th term r = 6 d) Arithmetic sequences and series Each term is found by adding a fixed constant (common difference d) to the previous term The first term is a giving the sequence a, a +d, a + 2d, a + 3d...... where un = a + (n − 1)d The sum of the first n terms can be found using: Sn (2a + (n − 1)d) = or Sn (a + 1) where / is the last term ● ● e) Geometric sequence and series Each term is found by multiplying the previous term by a fixed constant (common ratio r) The first term is a giving the sequence a, ar, ar², ar³, arª... S∞ = ₁ |r|<1 ● ● ● 4. TRIGONOMETRY MAKE SURE YOU KNOW AND CAN USE THE FOLLOWING FROM GCSE Sin Cos The sum of the first n terms can be found using a(n-1) Sn = a(1-rn) 1-r or Sn = Area = ● a b SinA SinB Sinc 0 0 1 Tan 0 = a² = b²+ c²2bcCosA FIGHINK |~ √3 1 3 absinc √3 √2 e - PANSIN √2 √3 2 2-8² 20 W NEN WW or √3 a) Radians 2π radians = 360° EIN - ÷0 ÷0 0 1 0 sine 0 cose 1- SinA a T When is small show that (1–0²) + 2-8² π radians = 180° You MUST work in radians if you are integrating or differentiating trig functions For an angle at the centre of a sector of 0 radians Arc Length = re Area of the sector = ½r²0 b) Small angle approximations (0 in radians) 1 0 0² 2 SinB Sinc b с cose sine y = sin x y = cos x tane 0 can be written as 2-8² 20 www.mathsbox.org.uk b B a y = tan x y TAWA c) Inverse Functions (sin ¹x, cos¹¹x, tan¹ x) By definition a function must be one-to-one which leads to restricted domains for the inverse trig functions y = sin’x (arcsin x) Domain: -1 ≤ x ≤ 1 П 2 secx = o ● EIN d) Reciprocal Trig Functions and identities (derived from sin²x + cos²x = 1) cos x 1 + tan²x = sec²x y = cos ¹x (arccos x) Domain -1 ≤ x ≤ 1 πT -1 cosec x== sin 2A = 2 sinAcosA cos 2A = cos²A - sin²A = 2cos²A - 1 = 1 - 2sin²A 2 tan A 1-tan² A Tan 2A = 1 sin x e) Double angle and addition formulae The addition formulae are given the formula booklet Make sure you can use these to derive : DOUBLE ANGLE FORMULAE 0 Solve for 0° < 0 < 360° the equation sec40 tan¹0 = 2 sec²x = 1 + tan²x sec¹x = 1+ 2tan²x + tan¹x cot x = 1 + cot²x = cosec²x 1 tan x 1 + 2tan²x + tan^x - tan²x = 2 2tan²x = 1 tan x = + cos x - Useful to solve equations cos 2A often used to integrate trig functions involving sin²x or cos²x y = tan¹¹x (arctan x) Domain XE R sinx EXPRESSING IN THE FORM rsin(0+ a) and rcos (0 + a) Useful in solving equations asin 0 + bcos 0 = 0 Useful in finding minimum/maximum values of acose +bsine and sin(A + B) = sinA cosB + cosAsinB cos(A + B) = cosA cosB + sinAsinB tanA+ tang 1 F tanAtanB www.mathsbox.org.uk tan(A + B) = x = 35.3°, 145°, 215°,325° 3 s.f. EIN asine + bcose Find the maximum value of the expression 2sinx + 3cosx and the value of x where this occurs (x< 180°) 2sinx + 3cosx = Rsin(x + a) (Rsin x cos a + Rcos x sin α) rcos a = 2 rsin a = 3 3 R = √2² +32 tana = 2 = √13 a = 56.3° 2sinx + 3cosx = √13 sin(x + 56.3°) Max value = √13 occurs when sin(x + 56.3°) = 1 X = 33.7° 5 LOGARITHMS AND EXPONENTIALS ● ● A function of the form y = ax is an exponential function The graph of y = ax is positive for all values of x and passes through (0,1) Logarithms - rules to learn loga a = 1 A logarithm is the inverse of an exponential function y = ax x = loga y logam + loga n = loga mn y = Ax" y = Abx loga 1 = 0 Write the following in the form alog 2 where a is an integer 3log 2 + 2log 4-log16 Method 1: log 8 + log 16 - log 4 = log (³×16) = Method 2: 3log 2 + 4log 2-2log 2 = 5log 2 loga ax = x logam-loga n = loga (™) An equation of the form ax = b can be solved by taking logs of both sides log y = log (Ax") = log 32 = 5log 2 a) MODELLING CURVES Exponential relationships can be changed to a linear form y = mx + c allowing the constants m and c to be 'estimated' from a graph of plotted data log a = 3 a = 10³ log y = n log x + log A y = mx + C log y = log (Ab*) log y = x log b + log A y = mx + c V and x are connected by the equation V = axb The equation is reduced to linear form by taking logs log V = b log x + log a (y = mx + c) (log V plotted against log x) From the graph b = 2 y = a* aloga x = x kloga m = loga mk log V 14 40- 9- www.mathsbox.org.uk y = ax Plot log y against log x. n is the gradient of the line and log A is the y axis intercept Plot log y against x. log b is the gradient of the line and log A is the y axis intercept log x Gradient = 2 Intercept = 3 6 b) The exponential function y = ex Exponential Growth y = ex The inverse of y = ex is the natural logarithm denoted by In x The rate of growth/decay to find the 'rate of change' you need to differentiate to find the gradient LEARN THIS y = Aekx DIFFERENTIATION ● ● LEARN THESE y = xn dy dx dy dx y = sin kx = Akekx =nxn-1 y = ekx dy = kekx dy dx Exponential Decay y = ex The gradient is denoted by dy if y is given as a function of x The gradient is denoted by f'(x) is the function is given as f(x) dx = kcos kx The number of bacteria P in a culture is modelled by P = 600 + 5e0.2t where t is the time in hours from the start of the experiment. Calculate the rate of growth after 5 hours 3e⁰.2t P = 600 + 15e0.2t P dt t = 5 dp dt QUOTIENT RULE for differentiating y = y = axn = 3e0.2x5 = 8.2 bacteria per hour y = lnx y = cos kx PRODUCT RULE for differentiating y = f(x)g(x) f(x) dy g(x) dx PARAMETRIC EQUATIONS y = f(t) x = g(t) dy dx dy dx a) Methods of differentiation CHAIN RULE for differentiating y = fg(x) y = f(u) where u = g(x) = dy dx dy dx = dy 1 dx x = = -ksin kx = Solve 2ex-2 = 6 leaving your answer in exact form ex-2 = 3 In(ex-2) = In 3 x - 2 = ln 3 x = ln 3+2 = naxn-1 dy dt + dx dt f'(x)g(x)-f(x)g'(x) [g(x)]² www.mathsbox.org.uk dy du X du dx f'(x)g(x) + f(x)g'(x) dx y = a y = akx dy dx y = tan kx dx = 0 = (klna) akx dy dx = ksec²kx IMPLICIT DIFFERENTIATION- take care as you may need to use the product rule too (xy², xy, ysinx) d[f(y)] d[f(y)] dy dx X dy dx b) Stationary (Turning) Points dy dx ● The points where = 0 are stationary points (turning points/points of inflection) of a graph The nature of the turning points can be found by: Maximum point 1분 ● >0 Maximum if dy d²y dx² Convex curve : <0 dy = 6x² - 6x = 0 at a stationary point dx <0 6x(x - 1) = 0 Turning points at (0, 18) and (1,17) Concave curve : c) Using Differentiation Find and determine the nature of the stationary points of the curvey = 2x³ - 3x² + 18 dy dx Points of inflection occur when = 0 (f"(x) = 0) dx² d²y dx² Minimum Point dy dx <0 Minimum if > 0 for all values of x in the 'convex section of the curve' dx² 82x Gradient of tangent at (2,12)=8-4 = 4 Gradient of the normal = -¼ d²y dx² d²y = 12x6 x = 0 ² <0 (0,18) is a maximum x = 1 >0 (1,17) is a minimum. d². dx² <0 for all values of x in the 'concave section of the curve' dx² wwwww.mmmatfisbox.org.ak d²y ...but = 0 could also indicate a min or max point dx² Find the equation of the normal to the curve y = 8x - x² at the point (2,12) dy dx (y - 12) = -4 (x-2) 4y + x = 50 >0 dy dx Tangents and Normals The gradient of a curve at a given point = gradient of the tangent to the curve at that point The gradient of the normal is perpendicular to the gradient of the tangent that point >0 d) Differentiation from first principles 7 0.5- (x,f(x)), a) Methods of Integration (x+h,f(x+h)) 0.2 Sekx dx = 1ekx + c f sin kx dx = -cos kx + c h Sf'(x) [f(x)]¹ =-(1-x²)+c = INTEGRATION Integration is the reverse of differentiation LEARN THESE x²+1 fx" dx = + c (c is the constant of integration) n+1 0.4 3 Look out for integrals of the form f(x) dx = In f(x)| + c f(x) + C As h approaches zero the gradient of the chord gets closer to being the gradient of the tangent at the point f'(x) = lim (f(x+h)-f(x)) h→0 [f(x)]¹+¹+c Find from first principles the derivative of x³ - 2x + 3 f'(x) = lim (f(x+h)-f(x)) = lim ((x+h)³-2(x+ h→0 = lim h→0 = lim h→0 INTEGRATION BY SUBSTITUTION Transforming a complex integral into a simpler integral using 'u = ' and integrating with respect to u fx√1-x² dx Let u = 1- x² = -2x so dx = - du dx du -2x [x√1-x² dx = fx√u dux -2x =-²/fu² du −2(x+h)+3−(x³−2x+3)) (x³+3x²h+3xh²+h³-2x-2h +3-x³ + 2x-3) -3-x³ + 2x-3)) (3x²h+3xh²+h³-2h h = lim(3x² + 3xh+h² - 2) h→0 = 3x² - 2 S=dx = lnx + c 5 cos kx dx = sin kx +c k h www.mathsbox.org.uk If it is a definite integral it is often easier to calculate the limits in terms of u and substitute these in after integrating Look for integrals of the form Seax+bdx f cos(ax + b) dx settbe ·dx INTEGRATION BY PARTS dv Sudx = uv-fvudx Take care in defining u and dv Sxe²x dx f xln dx U = x u = lnx dx = 1 dv dx dv dx = x ₁/dt = [4lnt]³ = 4ln3 4ln1 = 4ln 3 f lnx dx du == dx = (8-4) (0-0) = 4 1 PARAMETRIC INTEGRATION To find the area under a curve defined parametrically use area = Remember that the limits of the integral must be in terms of t A curve is defined parametrically by x = t - 1 y = included by the line x= 2, the x-axis and the y-axis. x=2 t=3 x = 0 t = 1 dx dt x For an area below the x-axis the integral will result in a negative value u = ln x V = x f lnx dx = xlnx - fx dx = xlnx − x +C lim y¡dx = i=1 = Sydx dt =Calculate the area of the region b) AREA UNDER A CURVE The area under a graph can be approximated using rectangle of height y and width dx. The limit as the number of rectangles increases is equal to the definite integral n b - fydx a dv dx Calculate the area under the graph y = 4x - x³ between x = 0 and x = 2 ²4x-: - x³ dx = 1 www.mathsbox.org.uk c) AREA BETWEEN 2 CURVES If no limits are given you need to identify the x coordinates of the points where the curve intersect Determine which function is 'above' the other [f(x) = g(x)]dx g(x) f(x) X₁ y X₂ d) SOLUTION OF DIFFERENTIAL EQUATIONS Separating the variables If you are given the coordinates of a point on the curve a particular solution can be found if not a general solution is needed Find the general solution for the differential equation dy y = xy² + 3x dx 8 NUMERICAL METHODS a) CHANGE OF SIGN - locating a root For an equations f(x) = 0, if f(x₁) and f(x₂) have opposite signs and f(x) is a continuous function between X₁ and X₂ then a root of the equation lies in the interval X₁< X < X2 First we will write it in the form x = f(x) x³ + 12 = 12x + 1 = x dy yax= x(y² + 3) b) STAIRCASE and COWBEB DIAGRAMS If an iterative formula (recurrence relation) of the form Xn+1=f(xn) converges to a limit, the value of the limit is the x-coordinate of the point of intersection of the graphs y = f(x) and y=x The limit is the solution of the equation f(x) = x 12 S2dy = fx dx y²+3 In|y² +31 = x² + c A staircase or cobweb diagram based on the graphs y = f(x) and y = x shows the convergence e.g Solve the equation x³-12x + 12 =0 Plotting the graphs y = + 1 and y = x X 12 the solution is the point of intersection of the two graphs. The solution of x ³-12x + 12 =0 is 1.8 1.6- 1.4 1.2 1 0.8 0.61 We can confirm that there is a point of intersection between x = 1 and x = 2 by a change of sign the values are substituted. x = 1.1157 0.4 0.2 x³ Substituting x = 2 into y = + 1 gives y = 1.66... (shown on the diagram) 12 Substituting x = 1.66.... y = 1.38... Repeating this the values converge to 1.1157 c) NEWTON-RAPHSON iteration f(x) = 0 Xn+1 = Xn- The equation e-2x -0.5x = 0 has a root close to 0.5. Using 0.5 as the first approximation use the Newton-Raphson you find the next approximation X₁ = 0.5 f(0.5)=e¹ -0.25 f'(x) = -2e 2x0.5 f'(0.5) = -2e¹-0.5 e-¹-0.25 X₂ = 0.5- X2 = 0.595 -2e-1-0.5 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 X f(xn) fi(xn) www.mathsbox.org.uk Use you ANS button on your calculator Limitations of the Newton-Raphson method As the method uses the tangent to the curve, if the starting value is a stationary point or close to a stationary point (min, max or inflection) the method does not work d) APPROXIMATING THE AREA UNDER A CURVE TRAPEZIUM RULE - given in the formula book but make sure you know how to use it! The trapezium rule gives an approximation of the area under a graph An easy way to calculate the y values is to use the TABLE function on a calculator- make sure you list the values in the formula (or a table) to show your method ● The rule will underestimate the area when the curve is concave The rule will overestimate the area when the curve is convex ● 9 UPPER and LOWER bounds - Area estimated using the area of rectangles For the function shown below if the left hand 'heights' are used the total area is a Lower Bound - the rectangles calculated using the right hand heights the area results in the Upper Bound b-a Soy dx ≈h[(yo+Yn) + 2(y₁ + y₂+... Yn-1)] where h = n a) NOTATION Lower Bound a = 2 63.4 VECTORS A vector has two properties magnitude (size) and direction 3 4 6 5 Vectors can be written as = (²³) a = 3i + 4j where i and j perpendicular unit vectors (magnitude 1) Magnitude-direction form (5, 53.1°) also known as polar form The direction is the angle the vector makes with the positive x axis Express the vector p = 3i-6j in polar form 3 |p|= 3² + (-6)² = 3√5 p = (3√5, 63.4°) 1 Upper Bound www.mathsbox.org.uk 2 3 4 53.1° 3 5 4 The Magnitude of vector a is denoted by lal and can be found using Pythagoras |a| = √3² +4² A Unit Vector is a vector which has magnitude 1 A position vector is a vector that starts at the origin (it has a fixed position) (²) b) ARITHMETIC WITH VECTORS Multiplying by a scalar (number) - (²) 3i+2j ² (²) = (6) 6i+4j a = a = 2a = 2 Addition of vectors = (₁³) a and 2a are parallel vectors Multiplying by -1 reverses the direction of the b) a + b = OA = b = = (²3) + (²³) =(5) 2i + 4j a b a+b A and B have the coordinates (1,5) and (-2,4). a) Write down the position vectors of A and B OA = (²) OB= (²) a 2a Subtraction of vectors a = (²3) b = (²) a-b= -a Write down the vector of the line segment joining A to B AB= -0A + OB or OB-OA AB = (¯²) - (¹²) = (-³) This is really a +-b (²3)-(²³) =(2₂¹) B www.mathsbox.org.uk -(29) 7 a Show that A(3,1,2) B(7,4,5) and C(19,13,14) AB = 4i +3j + 3k BC = 12i + 9j + 9k BC= 3AB AB and Collinear - vectors in 2D and 3D can be used to show that 3 or more points are collinear (lie on a straight line) a-b are parallel vectors sharing a common point B and are therefore collinear 10 PROOF Notation → ← If x=3 then x² = 9 x= 3 x² = 9 x = 3 is a condition for x² =9 x = 3 x² = 9 is not true as x could = - 3 x+1=3 ⇒ x = 2 a) Proof by deduction - statement proved using known mathematical principles Useful expressions: 2n (an even number) 2n + 1 (an odd number) Prove that the difference between the squares of any consecutive even numbers is a multiple of 4 Consecutive even numbers 2n, 2n + 2 (2n + 2)²-(2n)² 4n² + 8n +4-4n² =8n + 4 = 4(2n +1) a multiple of 4 b) Proof by exhaustion - showing that a statement is true for every possible case or value Prove that (n + 2)³ ≥ 3n-1 for neN, n<4 We need to show it is true for 1,2 and 3 n=1 272 1 n = 2 642 3 n = 3 12529 True for all possible values hence proof that the statement is true by exhaustion c) Disproof by counter example - finding an example that shows the statement is false. Find a counter example for the statement '2n + 4 is a multiple of 4' n = 2 4+4=8 a multiple of 4 n=3 6+4 = 10 NOT a multiple of 4 d) Proof by contradiction - assume first that the statement is not true and then show that this is not possible Prove that for all integers n, if n³ + 5 is odd then n is even Assume that n³ + 5 is odd and n is odd Let n³ + 5 = 2k + 1 and let n = 2m +1 (k and m integers) 2k + 1 = (2m + 1)³ + 5 2k + 1 = 8m³ + 12m² + 6m + 6 2k 8m³ + 12m² + 6m + 5 2k = 2(4m³ + 6m² + 3m) + 5 k = (4m³ + 6m² +3m) + = 2 4m³ + 6m² + 3m has an integer value leaving k as a non-integer value (contradicting assumptions) Our initial assumption that when that n³ + 5 is odd then n is odd is false so n must be even www.mathsbox.org.uk