2009 Q5 (a) - Weebly

2009 Q5 (a) - Weebly

Revision Index Page 2 Page 9 Page 16 Page 20 Functions and Graphs Rules of Indices Rules of Surds Co ordinate geometry recap 2009 Q5 (a) f(x) = 5x 12 and =a f(a) = 5a 12

a = 5a 12 4a = 12 a=3 f(a) (b)(i) f(x) = 5x 4 g(x) = 3x + 1 Domain = 0 x 3 x = 0, 1, 2, 3 f(1) = 5(0) 4 = -4 (0,-4) f(3) = 5(3) 4 = 11 (3,11) g(1) = 3(0) + 1 = 1

(0, 1) g(3) = 3(3) + 1 = 10 (ii) P.O.I = (2.5,8.5) (c) f(x) = 2x + x 15 -4 x 3 Finding points f(-4) = 2(-4) + (-4) 15 = 13 (-4, 13) f(-3) = 2(-3) + (-3) 15 = 0 (-3, 0) f(-2) = 2(-2) + (-2) 15 = -9 (-3, -9) f(-1) = 2(-1) + (-1) 15 = -14(-1, 14) f(0) = 2(0) + (0) 15 = -15(0, -15) f(1) = 2(1) + (1) 15 = -12 (1, -12) f(2) = 2(2) + (2) 15 = -5 (2, -5) f(3) = 2(3) + (3) 15 = 6 (3, 6)

(-4, 13) (-3, 0) (-3, -9) (-1, 14) (0, -15) (1, -12) (2, -5) (3, 6) (ii) Minimum value of f(x) (iii)Range of values of x for which f(x) 0 Q5 2008 P1

(i) Fnd the value of b and c (ii) Solve f(x) = -6 2007 P1 (a) Express 72km/hr in metres per second 1000m = 1km 72km/hr = 72,000m/hr 60mins in 1hour 3600seconds in 1hour 72,000 3,600 = 20m/s (ii) Snowman with mass of 12kg Melts at a rate of 0.2% of its mass per minute What is its mass after 3mins

Mass at start of 1st min: Less 0.2% Mass after 1min 12kg 0.024kg 11.976kg Mass at start of 2nd min: Less 0.2% Mass after 2mins 11.976 0.23952kg 11.952kg Mass at start of 3rd min: Less 0.2%

Mass after 3mins 11.952 0.23904kg 11.928kg Mass after 3min = 11.93kg (c) Indices Rules of Indices an x am = an + m an/am = an m Rule 2: (4) = 4 Example:

4 x 4 = 4 5 5 = 5 Example: Express 9 in the form 3n 9 = 3 9 = (3) 9 = 3 Exercise Express the following in 2n: 8 x 4 (2) x (2) 2 x 2

= 2 Express the following in 5n: 25 125 (5) (5) 5 5 = 5 Rule 3 a = a Example: 16 =4 Similarly: a = a

Eg. 27 =3 a = a Example: 16 =4 Similarly: a = a Eg. 27 =3 2006 3(a) paper 1 (i)

(64) (64) 8 = 512 (ii) (64) (64) 4 = 16 2005 3(a) paper 1 in the form 2k = 16 = (2) = 2

= 2 k=1 we want to change everything to 2 64 2 x (2) = 64 =8 2 x (2) = 2 2 x 2 2 x 2 2 2 = 2 (2)

= 2 = 2 Using Calculator 2 x 8 64 x 4 32 x 4 8 x 16 128 128 =1 = 2 2003 Q2(b) + 8 + 16 + 64 + 4 = 68

0.272479 + 62.41 + 4.0398 66.72228 = 66.72 3x3 3 x 3 3 3 = 3 Surds Rules of Surds: ab = a x b Example: 5 x 3 15 Simplifying Surds:

Simplify the following surd 45 15 x 3 ? 5 x 9 = 35 Express in their simplest form 32 16 x 2 42 18 75 9 x 2 32

25 x 3 53 Hint: 3(3 2) 36 52(45) 2010 20 4 x 5 25 Also: 43 + 63 = 103

6(53 - 36) - 23(53 - 36) 518 - 336 109 - 618 518 - 618 3(6) 10(3) -18 48 -(9)(2) 48 -32 - 48 218 49 - 350 - 2100 218 4(3) - 350 2(10) 218 12 - 350 20 2(9)(2) 12 3(25)(2) 20 2(3)(2) 12 3(5)(2) 20 62 - 152 32 32 92 Coordinate Geometry (Paper2 Q2) What you can be asked

Basic Formulae: Distance Midpoint Slope Years they come up 2011,2010, 2008,2006 2010,2009,2008,2007 2011,2007, 2005 Important Formula: Equation of a line - we need a point - and the slope 2011,2008,2007,2006, 2005

Other Basics: 2010 Getting slope of a line from the equation Showing that a certain point is on a line 2009,2008,2007, 2005 More complex questions Years Parallel/perpendicular Lines - as soon as you see this word you think SLOPE 2011,2010, 2009,2007,20 06,2005(b)(iv) Crossing the x-axis:

- As soon as you see this you think when crossing the x axis, y = 0 Vica versa with y-axis Point of intersection - Very easy with a graph - use simultaneous equations otherwise 2011,2008,20 07 2010,2007, 2006,2005 Ctd. Years they come up Symmetry: - axial - central - transformation

2011, 2004, 2003 Showing Lines on a graph: 2009,2008 Other: - Finding the area of a triangle on the graph 2010, 2009,2008 2005Q2 (a) a(1,4) and b(-2,-1) (x1,y1) (i) Slope of ab

(x2,y2) (ii) Equation of ab y y1 = m(x x1) y 4 = (x 1) 3(y 4) = 5(x 1) 3y 12 = 5x 5 3y 5x = 7 (b) L is the line 3x 4y + 7 = 0 Contains(-1,h) M is the line 4x + 3y -24 = 0 and contains point (k,0) (i) Find h and k 3x 4y + 7 = 0 3(-1) 4(h) + 7 = 0 -3 4h + 7 = 0 -4h = -4

h=1 (i) Find h and k 4x + 3y -24 = 0 4(k) + 3(0) -24 = 0 4k 24 = 0 4k = 24 k=6 (b) L is the line 3x 4y + 7 = 0 Contains(-1,h) M is the line 4x + 3y -24 = 0 and contains point (k,0) (ii) Point of intersection of L and M is r 3x 4y = -7 x=3 3x 4y = -7 4x + 3y = 24 3(3) 4y = -7

9 4y = -7 9x 12y = -21 y=4 16x + 12y = 96 25x = 75 P.O.I = (3,4) x=3 On a graph p(-1,1) q(6,0) r(4,3) Prove its a Right Angle If its a right angle the lines are perpendicular i.e. their slopes multiplied give you -1 Slope of L: 3x 4y + 7 = 0

-b/a = -(-4)/3 m= Slope of M: 4x + 3y 24 = 0 -b/a = -3/4 = m = - x (-) = -1 2006(b) p(-1,2) r(3,4) (i)Midpoint = Midpoint = (1,3) (ii) Slope = Slope = 2/4 m= (iii) Find equation of perpendicular bisector How will we find the

bisector? Midpoint! Wait! We already have it! Midpoint = (1,3) A perpendicular line Slope is inverse Slope of pr = Slope of L = -2/1 or Slope = -2 Equation = y y1 =m(x x1) y 3 = -2(x 1) y + 2x = 10 (iv) Equation of K is x 2y = 0 Point of intersection L:

y + 2x = 5 K: -2y + x = 0 2y + 4x = 10 -2y + x = 0 5x = 10 x=2 y + 2x = 5 y+8=5 y=1 Point of Intersection= (2,1) Trigonometry 2006 Paper 2 Q5 (a) Construct the angle A such that Tan A = Tan = opposite/adjacent We construct a right angled triangle where the

opposite side is 3 and the adjacent is 4 3 A 4 2006 P2 Q5(b) a C b 21.7 Area of a triangle = abSinC

a and b = 9 C: 21.7 + 21.7 = 43.4 180 43.4 = 136.6 Area = abSinC = (9)(9)Sin136.6 = 27.83 (c) f 38 d 65 80m

(i) Find ef Sin Rule

e w 51m 65 Sin65 = w/51 Width = 51Sin65 = 46.22 =46metres 2007 Q5 (a) If SinA = - find the 2 values for angle a where 0 A 360 90

All + Sin 180 0 Tan Cos 270 Sin-1 () = 30 A = 30 Sin30 = Sin150 = Sin210 = -

Sin330 = - A = 210 and 330 Values of A where CosA = - 0 A 360 Cos-1 () = 60 Cos60 = Cos300 = Cos120 = - Cos240 = - A = 120 and 240 Q5(b) a 9km 15km

(i) Find ac Pythagoras: 9 + 12 = ac ac = 225 (ii) Find CD Tan36.87 = 15/CD CD = 15/Tan36.87 CD = 20km ac = 15 12km 36.87 d c

(ii) Find AD 20 + 15 = AD AD = 25 Total Course = 25 + 9 + 12 + 20 = 66km (c) 10 h x 30 150 w 20 12m

For the Sine rule we need 2angles and a side 180 30 = 150 180 (150 + 20) = 10 Sine rule: 12 = x Sin10 Sin20 x = 12Sin20 Sin10 x = 23.64 (c) (ii) Find h

Sin30 = h/23.64 10 h 11.81 h = 23.64Sin30 h = 11.81m 23.64 (iii) Find w Tan30 = 11.81/w 30 150 w 20 12m

w = 11.81/Tan30 w = 20.45m

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