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  • 1. y = cos xe.g. y = (x + 3)2 (x 1)21 Copyright itute.com 20060 2 Free download & print from www.itute.com30 1 Do not reproduce by other means 1 Mathematical Methods 3,4 e.g. y = (x + 2 )3 (x 1) y = tan x Summary sheets Distance between two points 20 1 0 2 = (x2 x1 )2 + ( y2 y1 )2x +x y +y Mid-point = 1 2 , 1 2 e.g. y = (x + 2)4 2 2 Modulus functions Parallel lines, m1 = m2 x, x 0 y= x = Perpendicular lines, x , x < 0120m1m2 = 1orm2 = Transformations of y = f (x )m1Examples of power functions: (1) Vertical dilation (dilation away from the Graphs of polynomial functions in x-axis, dilation parallel to the y-axis) by 1 factorised form: y = x 1 y = factor k. y = kf (x ) Quadratics e.g. y = (x + 1)(x 3)x (2) Horizontal dilation (dilation away from 0 the y-axis, dilation parallel to the x-axis) by 1 factor . y = f (nx ) n10 3 (3) Reflection in the x-axis. y = f (x )y = x 2 (4) Reflection in the y-axis. y = f ( x ) e.g. y = (x 3)21 (5) Vertical translation (translation parallely = 2 to the y-axis) by c units. x y = f (x ) c , + up, down. 0031 (6) Horizontal translation (translation Cubics e.g. y = 3(x + 1)(x 1)(x 2 )y= x2 parallel to the x-axis) by b units.y = f (x b ) , + left, right.(y = x ) *Always carry out translations last in 0 sketching graphs. Example 1 Sketch y = 2(x 1) + 2Exponential functions: -1 0 1 22y = a x where a = 2, e,10 01 2 e.g. y = (x + 1)2 (x 1) 10x ex2x 1 0 1Example 2 Sketch y = 2 1 x . Rewrite as y = 2 (x 1) .1 2asymptotic 0 e.g. y = (x + 1) 30 1Logarithmic functions: Relations and functions:10 y = log a x where a = 2, e,10A relation is a set of ordered pairs (points). If no two ordered pairs have the same first 2xelement, then the relation is a function. Quartics e.g. y = (x + 3)(x + 1)(x 1)(x 2 )ex *Use the vertical line test to determine whether a relation is a function.10x*Use the horizontal line test to determine3 1 0 1 2 01 whether a function is one-to-one or many-to-asymptotic one. *The inverse of a relation is given by its reflection in the line y = x . *The inverse of a one-to-one function is a e.g. y = (x + 3)2 (x 1)(x 2 )Trigonometric functions: function and is denoted by f 1 . The inversey = sin x of a many-to-one function is not a function1 and therefore cannot be called inverse 3 0 1 2 02 function, and f 1 cannot be used to denote the inverse.1

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Factorisation of polynomials: Quadratic formula: Index laws: ( ) (1) Check for common factors first.2 Solutions of ax + bx + c = 0 aream n (2) Difference of two squares, a ma n = a m+ n , = amn , am = a mn an( ) 3 = (x 3)(x + 3)2 2 b b 4ac e.g. x 4 9 = x 222 2x= .11(ab )n = a nbn , = an , am = ( 3 )(x + 3 )(x + 3) 2a= x 2 nm Graphs of transformed trig. functions aa (3) Trinomials, by trial and error,1 1 e.g. y = 2 cos 3 x + 1 , rewrite a 0 = 1, a 2 =na e.g. 2 x 2 x 1 = (2 x + 1)(x 1)2= a,a n (4) Difference of two cubes, e.g. ()Logarithm laws: equation as y = 2 cos 3 x + 1 . x3 y 3 = (x y ) x 2 + xy + y 2 6 alog a + log b = log ab, log a log b = log3b (5) Sum of two cubes, e.g. 8 + a =The graph is obtained by reflecting it in the (23 + a3 = (2 + a ) 4 2a + a 2 )x-axis, dilating it vertically so that its amplitude becomes 2, dilating it horizontallylog ab = b log a, log 1 b = log b, log a a = 1 (6) Grouping two and two, 2 e.g. x3 + 3x 2 + 3 x + 1 = x3 + 1 + 3 x 2 + 3 x( ) () so that its period becomes3, translating log1 = 0, log 0 = undef , log(neg ) = undef ( = (x + 1) x x + 1 + 3 x(x + 1) 2 ) upwards by 1 and right by.Change of base:= (x + 1)(x) 6 log b x 2 x + 1 + 3x log a x = ,log b a = (x + 1)(x+ 2 x + 1 = (x + 1)3 2 ) 3log e 7 (7) Grouping three and one,e.g. log 2 7 == 2.8 .log e 2 e.g. x 2 2 x y 2 + 1 5 0Exponential equations: ( )= x 2 2 x + 1 y 2 = (x 1)2 y 2166e.g. 2e3 x = 5, e3 x = 2.5 , 3x = loge 2.5 ,= (x 1 y )(x 1 + y )1 (8) Completing the square, e.g.x=loge 2.531 1 22Solving trig. equations x2 + x 1 = x2 + x + 1e.g. 2e 2 x 3e x 2 = 0 ,2 2 32 2 e.g. Solve sin 2 x = 2, 0 x 2 .( ) 3(e ) 2 = 0 ,2 ex 2 x = x2 +x+ 1 4 5 4 =x +1 2 2 5 0 2 x 4 , 2 2(2e + 1)(e 2) = 0 , since 2exxx+1 0 , 2x = , , + 2 ,+ 2 x x e 2 = 0 , e = 2 , x = loge 2 . 3 3 3 31 5 x + 1 + 5 = x + 7 4 2 2 2 2 x = , ,, .Equations involving log: 6 3 6 3e.g. loge (1 2 x ) + 1 = 0 , loge (1 2 x ) = 1 , (9) Factor theorem,xx e.g. sin = 3 cos , 0 x 2 .1 1 e.g. P(x ) = x3 3x 2 + 3 x 1 221 2 x = e 1 , 2 x = 1 e 1 , x =1 .2 e P( 1) = ( 1)3 3( 1)2 + 3( 1) 1 0sin xe.g. log10 (x 1) = 1 log10 (2 x 1)x2 = 3 , tan x = 3 , P(1) = 13 3(1)2 + 3(1) 1 = 00 , xlog10 (x 1) + log10 (2 x 1) = 122 (x 1) is a factor.cos 2log10 (x 1)(2 x 1) = 1 , (x 1)(2 x 1) = 10 , Long division: x 2 = , x = . 2 x 2 3x 9 = 0 , (2 x + 3)(x 3) = 0 , x2 2x + 12 3 3 3 x 1)x3 3 x 2 + 3 x 1 x = , 3 . 3 is the only solution because2( x3 x 2 ) Exact values for trig. functions:x= 3 makes the log equation undefined. 2 x 2 + 3xxo x sin x cos x tan x2 ( 2x2 + 2 x ) 030 0 /6 01/2 1 3/2 0 1/3 Equation of inverse:x 1Interchange x and y in the equation to obtain45 /4 1/21/21 (x 1)60 /3 3/2 1/2 3the equation of the inverse. If possibleexpress y in terms of x. 090 /2 1 0undef ( P(x ) = (x 1) x 2 2 x + 1 = (x 1)3 )1202/3 3/21/23e.g. y = 2(x 1)2 + 1 , x = 2( y 1)2 + 1 , x 1 1353/4 1/2 1/2 12( y 1)2 = x 1 , ( y 1)2 =, Remainder theorem:1505/61/2 3/2 1/32 e.g. when P (x ) = x3 3 x 2 + 3 x 1 is180 01 0x 1 2107/6 1/2 3/21/3 y= +1 . divided by x + 2 , the remainder is2 2255/41/2 1/21P ( 2) = ( 2)3 3( 2)2 + 3( 2) 1 = 112404/33/21/2 3e.g. y = 2 +4, x = 2+4 , When it is divided by 2 x 3 , the remainderx 1 y 1 2703/21 0undef3 13005/33/2 1/2322 is P = .x4= , y 1 = ,2 83157/41/21/2 1y 1x4 330 11/61/23/21/3 2y=+1 . 360 20 10x4 3. e.g. y = 2e x 1 + 1 , x = 2e y 1 + 1 , Differentiation rules: The approx. change in a function isThe product rule: For the multiplication of= f (a + h ) f (a ) = hf (a ) , 1 x 2e y 1 = 1 x , e y 1 =,two functions, y = u (x )v(x ) , e.g.e.g. find the approx. change in cos x when x 2y = x 2 sin 2 x , let u = x 2 , v = sin 2 x , 1 x 1 x changes from to 1.6. Let f (x ) = cos x ,y 1 = loge , y = log e +1. dy dudv 2 2 2 =v +u dx dxdxthen f (x ) = sin x . Let a =, then e.g. y = loge (1 2 x ) 1 ,( ) = (sin 2 x )(2 x ) + x 2 (2 cos 2 x ) 2 x = loge (1 2 y ) 1 , = 2 x(sin 2 x + x cos 2 x )f (a ) = sin = 1 and h = 1.6 = 0.0322The quotient rule: For the division of loge (1 2 y ) = (x + 1) , 1 2 y = e ( x +1)u (x ) log e xChange in cos x = hf (a ) = 0.03 1 = 0.03functions, y =, e.g. y =,12( 2 y = 1 e ( x +1) , y = 1 e ( x +1) .)v(x )xRate of change:dydxis the rate of change ofdu dv The binomial theorem:vu dx dydxdx = y with respect to x. v =, velocity is the e.g. Expand (2 x 1)4dx v2dt = 4C0 (2 x )4 ( 1)0 + 4C1 (2 x )3 ( 1)1 rate of change of position x with respect to (x ) 1 (loge x )(1) dv + 4C2 (2 x )2 ( 1)2 + 4C3 (2 x )1 ( 1)3 x 1 log e x time t. a = , acceleration a is the rate of = 2 =. dtxx2 + 4C4 (2 x )0 ( 1)4 = ......The chain rule: For composite functions, change of velocity v with respect to t. e.g. Find the coefficient of x2 in they = f (u ( x) ) , e.g. y = e cos x . Average rate of change: Given y = f (x ) , expansion of (2 x 3)5 . dy dy duwhen x = a , y = f (a ) , when x = b ,Let u = cos x , y = eu , = The required term is 5C3 (2 x )2 ( 3)3 dx du dxy = f (b ) , the average rate of change of y( ) = 10 4 x 2 ( 27 ) = 1080 x 2 .( )( sin x) = e= eu cos xsin x .with respect to x = y= f (b ) f (a ) . the coefficient of x2 is 1080.dy xbaFinding stationary points: Let= 0 anddx Differentiation rules: solve for x and then y, the coordinates of the Deducing the graph of gradient functiondy from the graph of a functiony = f (x ) = f ' (x ) stationary point. f(x)dxNature of stationary point at x = a : ax nanx n 1Local LocalInflection max.min. point a(x + c )nan(x + c )n 1x0 0 , (< 0) a(bx + c )n abn(bx + c )n 1 dxdx dx x=adydy dy a sin x a cos x=0=0 =0odxdx dx a sin (x + c )a cos(x + c ) x>adydy dyf(x) a sin (bx + c ) ab cos(bx + c )0 > 0 , (< 0)dxdx dxo o a cos x a sin x Equation of tangent and normal at x = a : a cos(x + c ) a sin (x + c ) 1) Find the y coordinate if it is not given. 0 o xdy a cos(bx + c ) ab sin (bx + c ) 2) Gradient of tangent mT = at x = a .dx a tan x a sec 2 x3) Use y y1 = mT (x x1 ) to find equationDeducing the graph of function from the a tan (x + c )a sec (x + c )2 of tangent.graph of anti-derivative functiona tan (bx + c )1 ab sec 2 (bx + c ) 4) Find gradient of normal mN = . f(x)dx+ c mTx x aeae 5) Use y y1 = m N (x x1 ) to find equationx+c aeae x + c of the normal.Linear approximation:0 x ae bx + c abe bx + c To find the approx. value of a function, use a log e x a f (a + h ) f (a ) + hf (a ) , e.g. find the xapprox. value of 25.1 . Let f (x ) = x , o a log e bxa1 xthen f (x ) = . Let a = 25 and h = 0.1 ,f(x) 2 x o o a log e (x + c )athen f (a + h ) = 25.1 , f (a ) = 25 = 5 , x+c 0 ox 1 a log e b(x + c ) a f (a ) == 0.1 . 2 25 x+c 25.1 5 + 0.1 0.1 = 5.01 a log e (bx + c ) ab bx + c 4. Anti-differentiation (indefinite integrals):Estimate area by left (or right) rectanglesGraphics calculator :Pr ( X = a ) = binompdf (n, p, a )f (x ) f (x)dx Left RightPr ( X a ) = binomcdf (n, p, a ) ax n for n 1a n +1 xa b a b Pr ( X < a ) = binomcdf (n, p, a 1)n +1 Area between two curves: Pr ( X a ) = 1 binomcdf (n, p, a 1) a (x + c )n , n 1 a (x + c )n +1y = g (x ) Pr ( X > a ) = 1 binomcdf (n, p, a )n +1y = f (x )Pr (a X b ) = binomcdf (n, p, b ) a (bx + c )n , n 1a(bx + c )n +1 a0b binomcdf (n, p, a 1)(n + 1)bProbability density functions f (x ) for aa log e x , x > 0 x [a, b] . yy = f (x ) xa log e ( x ) , x < 0 Firstly find the x-coordinates of the aa log e (x + c ) intersecting points, a, b, then evaluateb acbx x+c aalog e (bx + c ) A= [ f (x) g (x)]dx . Always the functionaFor f (x ) to be a probability density above mi