Additional Mathematics 0606

1 Solve the equations y x 4 - = , x y x y 8 4 16 0 2 2 + - - - = . [5]

1 It is given that tan y x 1 3 = + . (i) State the period of y. [1] (ii) On the axes below, sketch the graph of tan y x 1 3 = + for ° x 0 180 ° ° G G . [3] 0 30 60 90 120 150 180 y x

1 Solve the equations y x 4 - = , x y x y 8 4 16 0 2 2 + - - - = . [5]

1 The remainder obtained when the polynomial ( )x x ax x b 3 p 3 2 = + - + is divided by x 3 + is twice the remainder obtained when ( )x p is divided by x 2 - . Given also that ( )x p is divisible by x 1 + , find the value of a and of b. [5]

1 (a) On the axes below, sketch the graph of cos y x 3 2 1 = - , for ° ° ° x 0 360 G G . 5 4 3 2 1 0 –1 –2 –3 –4 –5 –6 x y 90 180 270 360 [3] (b) Given that sin y x 4 6 = , write down (i) the amplitude of y, [1] (ii) the period of y. [1]

1 Solve ( °) sin x 1 2 50 0 + + = for ° ° x 180 180 G G - . [4]

1 (a) In the expansion of ( ) px 2 5 + the coefficient of x3 is equal to 25 8 - . Find the value of the constant p. [3] (b) Find the term independent of x in the expansion of x x 2 4 1 2 2 8 + e o . [3]

2 Find the equation of the perpendicular bisector of the line joining the points (1, 3) and (4, -5). Give your answer in the form ax + by + c = 0, where a, b and c are integers. [5]

2 Find the values of k for which the line y kx 1 2 = - does not meet the curve ( ) y x k x 9 3 1 5 2 = - + + . [5]

2 Find the equation of the perpendicular bisector of the line joining the points (1, 3) and (4, -5). Give your answer in the form ax + by + c = 0, where a, b and c are integers. [5]

2 A curve has equation sin y x 4 5 3 = + . (i) Find x y d d . [2] (ii) Hence find the equation of the tangent to the curve sin y x 4 5 3 = + at the point where r x 3 = . [3]

2 ( )x x x x a 2 5 4 p 3 2 = + + + ( )x x ax b 4 3 q 2 = + + Given that p(x) has a remainder of 2 when divided by x 2 1 + and that q(x) is divisible by x 2 + , (i) find the value of each of the constants a and b. [3] Given that ( ) ( ) ( ) x x x r p q = - and using your values of a and b, (ii) find the exact remainder when r(x) is divided by x 3 2 - . [3]

2 Find the equation of the curve which has a gradient of 4 at the point ( , ) 0 3 - and is such that x y 5 d d e x 2 2 2 = + . [5]

2 r cm O i rad P Q The diagram shows a sector POQ of a circle, centre O, radius r cm, where angle POQ i = radians. The perimeter of the sector is 20 cm. (i) Show that the area, A cm2, of the sector is given by A r r 10 2 = - . [3] It is given that r can vary and that A has a maximum value. (ii) Find the value of i for which A has a maximum value. [3]

3 Diagrams A to D show four different graphs. In each case the whole graph is shown and the scales on the two axes are the same. y x A O y x B O y x C O y x D O Place ticks in the boxes in the table to indicate which descriptions, if any, apply to each graph. There may be more than one tick in any row or column of the table. [4] A B C D Not a function One-one function A function that is its own inverse A function with no inverse

3 The variables x and y are such that when e y is plotted against x 2, a straight line graph passing through the points (5, 3) and (3, 1) is obtained. Find y in terms of x. [5]

3 Diagrams A to D show four different graphs. In each case the whole graph is shown and the scales on the two axes are the same. y x A O y x B O y x C O y x D O Place ticks in the boxes in the table to indicate which descriptions, if any, apply to each graph. There may be more than one tick in any row or column of the table. [4] A B C D Not a function One-one function A function that is its own inverse A function with no inverse

3 Do not use a calculator in this question. (a) Simplify 3 2 5 4 5 6 2 5 + - - ^ ^ ^h h h, giving your answer in the form a b 5 + , where a and b are integers. [3] (b) In this part, all lengths are in centimetres. B A C AB 6 2 3 = - BC 6 2 3 = + The diagram shows the triangle ABC with AB 6 2 3 = - and BC 6 2 3 = + . Given that cosABC 2 1 =- , find the length of AC in the form c d , where c and d are integers. [3]

3 The coefficient of x2 in the expansion of ( )( ) x kx 2 3 6 - + is equal to 972. Find the possible values of the constant k. [6]

3 (i) On the axes below, sketch the graph of y x 6 3 = - , showing the coordinates of the points where the graph meets the coordinate axes. [2] 10 12 8 4 6 2 0 –4 –2 –6 –8 –10 –12 x y 5 6 4 3 2 1 –1 –2 (ii) Solve x 6 3 2 - = . [3] (iii) Hence find the values of x for which x 6 3 2 2 - . [1]