THOMAS WHITHAM SIXTH FORM Additional Mathematics Revision Guide S J Cooper The book contains a number of worked examples covering the topics needed in the Additional Mathematics Specifications. This includes Calculus, Trigonometry, Geometry and the more advanced Algebra. Plus the topics required for Mechanics and statistics. Algebra Indices for all rational exponents n a m a n a mn , a m a n a mn , a m a mn , a 0 1 , n n 1 1 1 1 a b a 2 a , a n n a , a n , a n , a n a n b a m m a n n a m n a 1 1 3 3 2 1 1 1 Example 8 8 2 Example 9 1 9 2 9 3 3 2 2 2 3 3 Example x 3 4 x 3 4 2 x 4 8 Example Simplify 2x1 3x2 6x3 2 1 18x 2 3 2x 1 3x2 6x3 9x 2 x 6x3 6x 4 x 2 2 Quadratic equations ax bx c 0 Examples (i) 4x2 9 (ii) 4x2 9x (iii) 4x2 9x 2 (i) (ii) (iii) 9 2 2 x 2 4x 9x 0 4x 9x 2 0 4 x(4x 9) 0 (4x 1)(x 2) 0 3 x 9 1 2 x 0 or x x 2 or x 4 4 Example Solve (i) 2x2 x 5 0 using the method of CTS (ii) 5x2 7x 1 0 using the formula. (i) x 5 x 2 2 2 2 2 x 1 5 1 x 2 2 4 2 4 1 2 5 1 x 4 2 16 1 2 41 x 4 16 1 41 x 4 16 1 41 x 4 16 x 1.35, 1.85 Simultaneous equations Example Solve simultaneously 2x 3y 8, y x2 x 2 Here we substitute for y from the second equation into the first 2x 3y 8 2x 3x2 x 2 8 3x2 x 2 0 2 3x 2x 1 0 x 3 , 1 2 4 2 28 Solutions, 2 28 when x 3 , y 9 3 2 9 x 3 , y 9 x 1, y 2 when x 1 , y 11 2 2 } The geometrical interpretation here is that the straight line and the 2 2 28 parabola y x x 2 intersect at points 3 , 9 1, 2 y 2 , 28 3 9 (1,2) 0 x Intersection points of graphs to ‘solve’ equations. There are many equations which can not be solved analytically. Approximate roots to equations can be found graphically if necessary. Example What straight line drawn on the same axes as the graph of y x3 will give the real root of the equation x3 x 3 0? y x3 x 3 0 x3 3 x draw y 3 x 0 x As can be seen from the sketch there is only one real root . Expansions and factorisation –extensions Example 2x 1x 2 x 3 2x3 2x 2 6x x 2 x 3 Expanding 2x3 3x 2 7x 3 Example x3 9x xx2 9 xx 3x 3 Factorising The remainder Theorem b If the polynomial p(x) be divided by ax b the remainder will be p a Example When p(x) 2x3 3x 5 is divided by 2x 1 the remainder is p 1 1 3 5 15 2 4 2 4 Example Find the remainder when 4푥3 − 3푥2 + 11푥 − 2 is divided by 푥 − 1. f 1 413 312 111 2 10 remainder is 10 The factor theorem Following on from the last item b p a 0 ax b is a factor of Example Show that x 2 is a factor of 6x3 13x2 x 2 and hence solve the equation 6x3 13x2 x 2 0 Let p(x) 6x3 13x2 x 2 p(2) 623 1322 2 2 48 52 2 2 0 x 2 is a factor of p(x) p(x) x 26x2 x 1 …by inspection x 23x 12x 1 1 1 Solutions to the equation are x 2, 3 , 2 n Expansion of a b where n is a positive integer. This has the same coefficients as the expansion of 1 xn . Each term will have degree n, powers of a descending from a n Example 1 x8 1 8x 28x2 ....... a b8 a8 8a7b 28a6b2 ....... n nr r The general term will be Cr a b Example Use the binomial theorem to expand 3 2x4 , simplifying each term of your expansion. 4 4 3 2 2 3 4 a b a 4a b 6a b 4ab b 3 2x4 34 433 2x 632 2x2 432x3 2x4 2 3 4 81 216x 216x 96x 16x Geometry Gradient/ intercept form of a straight line Equation y mx c y Gradient = m (= tan ) c 0 x Distance between two points Given A x1, y1 B x2 , y2 then 2 2 2 AB x2 x1 y2 y1 Gradient of a line through two points… Ax1, y1 and Bx2 , y2 say y2 y1 m x2 x1 Equation of a line through (x’, y’) of gradient m y y mx x Equation of a line through two points Find the gradient using and use the formula as above. Parallel and perpendicular lines Let two lines have gradients m1 and m2 Lines parallel m1 m2 1 Lines perpendicular m1m2 1 or m1 m2 Mid-point of line joining … and coordinates are 1 x x , 1 y y 2 1 2 2 1 2 General form of a straight line ax by c 0. To find the gradient, rewrite in gradient/intercept form. Example Given points A 2, 3 and B1, 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 C5, 2 parallel to AB (a) AB 2 1 22 1 32 9 16 25 AB = 5 1 (b) M 2 , 1 1 3 4 (c) Gradient AB = 1 2 3 4 (d) Point 5, 2 Gradient = 3 4 Equation y 2 3 x 5 3y 6 4x 20 3y 4x 26 Example Find the gradient of the line 2x 3y 12 and the equation of a perpendicular line through the point 0, 4 2 2 3y 2x 12 y 3 x 4 grad = 3 1 Gradient of perpendicular = 3 2 2 3 3 Equation y 2 x 4 y mx c The Circle Angles in semicircle is 90 Perpendicular to a chord from centre of circle bisects the chord. Centre, radius form of equation x a2 y b2 r 2 Centre (a, b) radius = r Example Centre (2, -1) radius 3 equation x 22 y 12 9 Example Centre (1, 2) touching 0x Equation x 12 y 22 4 (1, 2) 0 . General form of equation x a2 y b2 r 2 Circle centre 푎, 푏 with radius 푟 To find centre and radius, use the method of CTS to change into centre/radius form. Example x2 y 2 2x 3y 3 0 x 2 y 2 2x 3y 3 0 x 2 2x y 2 3y 3 2 2 3 2 3 2 x 2x 1 y 3y 2 3 1 2 2 3 2 25 x 1 y 2 4 3 5 Centre 1, 2 radius = 2 Tangents Angle between tangent and radius drawn to point of contact is 90 Tangents drawn from extended point Line of symmetry Example Find the equation of the tangent to the circle x2 y 2 2x 4y 5 0 at the point P(2, 1) x 2 y 2 2x 4y 5 0 x 2 2x y 2 4y 5 x 2 2x 1 y 2 4y 4 5 1 4 x 12 y 22 10 Centre at 1, 2 , radius 10 2 1 1 Gradient CP = (-1, 2) 1 2 3 gradient of tangent at P = 3 Equation P(2, 1) 0 y 1 3x 2 y 3x 5 Calculus Differentiation by rule Examples d d 1 1 1 x x 2 1 x 2 dy 2 y dx dx 2 x dx d 4 d 4 x n nx n1 4x 1 4x 2 dx x dx x 2 ax n anx n1 d x d 1 1 ax a 2 x dx 2 dx 2 a 0 d du dv dw 10 0 u v w dx dx dx dx d 2 3x x 5 6x 1 dx Vocabulary and more notation dy is the derivative of y (with respect to x) dx is the differential coefficient of y (with respect to x). Example 푦 = 푥3 − 4푥2 + 3푥 − 1 푑푦 2 = 3푥 − 8푥 + 3 푑푥 2 2 x 2 x 2 3 1 Example f (x) x 2 2x 2 x x x 1 3 1 3 2 1 2 3 2 1 3 1 f (x) 2 x 2 2x 2 x 3 x x 2 2 x x dy The gradient of a curve at any point is given by the value of at that point. dx Example Find the gradient at the point P(1, 5) on the graph of y x2 2x 2 . Hence find the equation of the tangent at P. 2 y y x 2x 2 dy 2x 2 dx P(1, 5) At P(1, 5) gradient = 4 Tangent at P 0 x y 5 4x 1 y 4x 1 Stationary points on the graph of a function are points where the gradient is zero. STATIONARY POINTS TURNING POINTS POINTS OF INFLEXION _ + + _ _ + + _ MAXIMUM POINT MINIMUM POINT TANGENT PASSING THROUGH THE CURVE To obtain coordinates of a SP.