1) The point A has coordinates (7, 4). The straight lines with equations x + 3y + 1 = 0 and 2x + 5y = 0 intersect at B. (a) (b) Find the gradient of AB. Hence show that AB is perpendicular to only one of these two lines. (a) B is the point of intersection of and Soln x + 3y + 1 = 0 2x + 5y =0 2x + 6y + 2 = 0 2x + 5y =0 2x + 6y + 2 = 0 2x 5y =0 y+2=0 ( 2 both sides ) y = 2 2x + 5y =0 2x + (5)(2) = 0 2x 10 = 0 ( + 10 both sides ) 2x = 10 ( 2 both sides ) x=5 B is the point (5, 2) ( 2) ( 1) Add A is the point (7, 4) B is the point (5, 2) (b) mAB = y2 y 1 y2 y 1 = 2 4 57 = 6 2 = 6 2 mAB = 3 A line perpendicular to AB has gradient 1 3 If perpendicular m1 m2 = 1 i.e. 3 1 = 1 3 x + 3y + 1 = 0 2x + 5y = 0 3y = x 1 5y = 2x y = 1 x 1 3 3 y = 2 x 5 m = 1 3 m = 2 5 AB is only perpendicular to x + 3y + 1 = 0 2) f(x) = x3 x2 5x 3 (a) (i) (ii) Show that (x + 1) is a factor of f(x). Hence or otherwise factorise f(x) fully. (b) One of the turning points of the graph of y = f(x) lies on the x-axis. Write down the coordinates of this turning point. Soln (a) (i) factor (x + 1) root x = 1 1 1 1 (a) (ii) f(x) = = = f(x) (b) = 1 1 2 5 2 3 3 3 0 x3 x2 5x 3 (x + 1)( x2 2x 3) (x + 1)(x + 1)(x 3) (x 3)(x + 1)2 Stationary points x-coordinates = x3 x2 5x 3 f’(x) = f’(x) = 3x2 2x 5 (3x 5)(x + 1) f(x) Set f’(x) = 0 differentiate y = f(x) and set equal to zero (3x 5)(x + 1) = 0 (3x 5) = 0 or (x + 1) = 0 3x = 5 x=5 3 or x = 1 y-coordinates f(x) = x3 x2 5x 3 x = 1 x=5 3 f(5/3) = (5/3)3 (5/3)2 5(5/3) 3 f(1) = (1)3 (1)2 5(1) 3 f(5/3) = 125 25 25 3 27 9 3 f(1) = 1 1 + 5 3 f(5/3) = 125 75 225 81 27 27 27 27 f(1) = 0 f(5/3) = 175 81 27 27 f(5/3) = 256 27 f( /3) = 9 /27 5 13 27 9 243 The point ( 5/3 , 913/27 ) is a stationary point The stationary/turning point ( 1 , 0 ) lies on the x-axis i.e. y = 0 The point ( 1 , 0 ) is a stationary point 3) Find all the values of x in the interval 0 x 2 for which tan2(x) = 3. Soln tan2(x) = 3 ( root both sides ) tan(x) = 3 = 3 (o) 1 (a) All four quadrants 0 60 3 2 0 30 6 (o) 1 (a) 3 tan = 3 = 3 3 1 tan ( ) = tan 2 = 3 3 3 tan ( + ) = tan 4 = 3 3 3 tan (2 ) = tan 5 = 3 3 3 4 solutions x = , 2 , 4 , 5 3 3 3 3 1800/ S A T C 0/3600/2 4) The diagram shows the graph of y = g(x). (a) Sketch the graph of y = g(x). (b) On the same diagram, sketch the graph of y = 3 g(x). y (b, 3) y = g(x) (0, 1) x O (a, 2) Soln y = g(x) (a, 2) (0, 1) (b, 3) y = g(x) ( reflect the graph of y = g(x) in the x-axis ) ( multiply the y-coordinates of y = g(x) by 1 ) (a, 2) (0, 1) (b, 3) y (a, 5) y = 3 g(x) (a, 2) (0, 2) O x y = 3 g(x) = g(x) + 3 ( add 3 to the y-coordinates of y = g(x) ) (a, 5) (0, 2) (b, 0) y = g(x) (b, 0) (0, 1) (b, 3) 5) A, B and C have coordinates (3, 4, 7), (1, 8, 3) and (0, 10, 1) respectively. (a) Show that A, B, and C are collinear. (b) Find the coordinates of D such that AD = 4AB. Soln (a) AB = b = 1 8 3 = = 1 + 3 a BC = c 3 0 4 10 7 1 = 10 8 37 13 = 4 2 4 2 2 2 2 AB = 2 BC 8 3 1 1 = 1 0+1 84 2 b 2 = 1 2 4 4 B BC = 1 AB 2 Parallell / same direction / common point B collinear 2 A 1 C (b) AD = 4 AB 4 AB = 2 4 AD 4 = d = x 4 y z 4 AB = 8 16 AD = 16 x+3 y4 z7 x+3=8 y 4 = 16 z 7 = 16 x=5 y = 20 z = 9 D(5, 20, 9) a 3 4 7 6) Given that y = 3sin(x) + cos(2x), find dy . dx Soln y = 3sin(x) + cos(2x) dy = 3cos(x) + ( sin(2x) ) (2) dx dy = 3cos(x) 2sin(2x) dx 7) Find Soln = = 2 2 2 4x + 1 dx 0 4x + 1 dx 0 (4x + 1)1/2 0 (4x + 1) 3 2 3/2 1 2 4 0 = (4x + 1) 3/2 21 3 2 4 0 = = 2 2 (4x + 1)3/2 12 (4x + 1)3/2 6 0 2 0 = ( (4(2) + 1) )3 6 = (9)3 (1)3 6 6 = 33 13 6 6 = 26 6 = 41/3 units2 = = 13 3 ( (4x + 1) )3 6 ( (4(0) + 1) )3 6 = 27 1 6 6 2 0 8) (a) Write x2 10x + 27 in the form (x + b)2 + c. (b) Hence show that the function g(x) = 1 x3 5x2 + 27x 2 3 is always increasing. Soln (a) = x2 10x + 27 2 (x 5)2 (x 5)2 + 2 (x 5)2 + 2 = x2 10x + 25 + 2 = x2 10x + 25 = x2 10x + 27 (b) g(x) = 1 x3 5x2 + 27x 2 3 g’(x) = x2 10x + 27 g’(x) = (x 5)2 + 2 (x 5)2 25 for all x (x 5)2 + 2 27 for all x g’(x) 0 for all x the function g(x) is always increasing g’(x) = (x 5)2 + 2 g’(x) is a parabola with a minimum turning point at (5, 2). There is no point of contact of g’(x) with the x-axis (i.e. no real roots). Therefore g’(x) 0 for all x. g’(x) is the gradient of g(x) and g’(x) 0 for all x. Therefore the function g(x) always has positive gradient and is always increasing. 9) Solve the equation log2(x + 1) 2log2(3) = 3 Soln log2(x + 1) 2log2(3) = 3 log2(x + 1) log2(3)2 = 3 log2(x + 1) log2(9) = 3 log2 x + 1 = 3 9 23 = x + 1 9 8=x+1 9 ( 9 both sides ) 72 = x + 1 ( 1 both sides ) 71 = x x = 71 10) In the diagram angle DEC = angle CEB = xo and angle CDE = angle BEA = 900. CD = 1 unit ; DE = 3 units. B C 1 By writing angle DEA in terms of xo, find the exact value of cos (DEA). x D o A xo E 3 Soln C Pythagoras 10 1 xo D sin xo = 1 10 cos (DEA) E 3 CE2 = = = CE = 32 + 1 2 9+1 10 10 cosxo = 3 10 = cos(2x + 90)0 = cos(2x)0cos900 sin(2x)0sin900 cos900 = 0 = sin900 = 1 0 sin(2x)0 sin(2x)0 = 2sinxocosxo = 2sinxocosxo = 2 = 3 5 1 3 10 10 = 6 10 10) The diagram shows a parabola passing through the points (0, 0), (1, 6) and (2, 0). (a) The equation of the parabola is of the form y = ax(x b). y (2, 0) O Find the values of a and b. (b) This parabola is the graph of y = f’(x). Given that f’(1) = 4, find the formula for f(x). (1, 6) Soln (a) roots of the parabola are x = 0 and x = 2 factors are x and (x 2) y = a(x)(x 2) y = ax(x 2) y = ax2 2ax the point (1, 6) lies on y = ax2 2ax ( x = 1, y = 6 ) 6 = a(12) 2a(1) 6 = a 2a 6 = a a=6 y = ax(x b) y = 6x(x 2) a = 6 and b = 2 x (b) y = f’(x) = 6x(x 2) f(x) = 6x(x 2) dx = 6x2 12x dx = 6x3 12x2 + c 3 2 f(x) = 2x3 6x2 + c f(1) = 4 4 = 2(13) 6(12) + c 4 = 26+c 4 = 4 + c ( add 4 both sides ) 8 = c c = 8 f(x) = 2x3 6x2 + c f(x) = 2x3 6x2 + 8 f(x) = 2(x3 3x2 + 4)