7 (a) O A B C a b c The diagram shows triangle OAC, where OA a = , OB b = and OC c = . The point B lies on the line AC such that : : AB BC m n = , where m and n are constants. (i) Write down AB in terms of a and b. [1] (ii) Write down BC in terms of b and c. [1] (iii) Hence show that ( ) n m m n a c b + = + . [2] (b) Given that ( ) ( ) 2 1 1 4 7 1 4 2 m n m + - - = + - e e e o o o, find the value of each of the constants m and n. [4]

7(a)(i) − b a B1 7(a)(ii) − c b B1 7(a)(iii) =   nAB mBC M1 For substitution of their (i) and (ii) into =   nAB mBC ( ) + = + n m m n a c b A1 For correct manipulation to obtain the given answer 7(b) 2 4 4 4 4 λ μ λ − + = + or 7 7 2 2 λ μ λ + − = − − M1 For equating like components at least once, allow unsimplified M1 Dep for solving their equations to obtain both λ and μ 5 μ = A1 10 λ = − A1 (a)(i) You must express vector AB as the difference between position vector B and position vector A. (a)(ii) You must express vector BC as the difference between position vector C and position vector B. (a)(iii) You earn the mark by correctly substituting your expressions from (i) and (ii) into the relationship nAB = mBC and then manipulating it to match the given equation. (b) You need to equate the corresponding components of the given vector equation to form two simultaneous equations. You then solve these equations to find the values of m and n.

8 The curves y x x 1 2 = + - and y x x 2 6 2 2 = + - intersect at the points A and B. (a) Show that the mid-point of the line AB is (2, 9). [5] The line l is the perpendicular bisector of AB. (b) Show that the point C (12, 7) lies on the line l. [3] (c) The point D also lies on l, such that the distance of D from AB is two times the distance of C from AB. Find the coordinates of the two possible positions of D. [4]

8(a) 2 2 2 2 2 6 2 + − = + − x x x x 2 4 0 − = x x ( )4 0 − = x x 0, 4 = = x x M1 For obtaining an equation in one variable M1 Dep for a correct attempt to obtain at least one solution ( ) 0, 1 − A1 nfww ( ) 4, 19 A1 nfww Mid-point ( ) 2, 9 with sufficient detail B1 AG 8(b) Either Gradient of perpendicular 1 5 = − M1 ( ) 1 9 2 5 − = − − y x M1 Dep on previous M mark for perpendicular bisector using their mid- point and their perpendicular gradient ( ) 1 7 9 12 2 5 − = − − oe A1 For checking by substitution, must see evidence. Or Alternative 1 Gradient of perpendicular 1 5 = − (M1) ( ) 1 7 12 5 − = − − y x (M1) Dep on previous M mark for perpendicular bisector using ( ) 12, 7 and their perpendicular gradient ( ) 1 9 7 2 12 5 − = − − oe (A1) For checking by substitution, must see evidence Or Alternative 2 Gradient of perpendicular 1 5 = − (M1) Gradient of line joining their ( ) 2, 9 to ( ) 12, 7 = 1 5 = − (M1) ( ) 2, 9 is a common point and gradients of perpendicular bisector and l are the same so C lies on l. (A1) 8(c) ( ) 22,5 2 B1 for 22 B1 for 5 ( ) 18,13 − 2 B1 for 18 − B1 for 13

9 The polynomial ( )x x x x 2 3 1 p 3 2 = - - + has a factor x 2 1 - . (a) Find ( )x p in the form ( ) ( ) x x 2 1 q - , where ( )x q is a quadratic factor. [2] y x O B A y x 1 = y x x 2 3 1 2 =- + + The diagram shows the graph of y x 1 = for x 0 2 , and the graph of y x x 2 3 1 2 =- + + . The curves intersect at the points A and B. (b) Using your answer to part (a), find the exact x-coordinate of A and of B. [4] (c) Find the exact area of the shaded region. [6]

9(a) ( )( ) 2 2 1 1 − − − x x x M1 For attempt at factorisation by observation or by algebraic long division ( )( ) 2 2 1 1 − − − x x x A1 cao 9(b) At A 1 2 = x B1 2 1 0 − −= x x M1 For a valid attempt to solve their quadratic equation, allow for decimal solutions 1 5 2 ± = x soi A1 At B 1 5 2 + = x A1 9(c) 1 d ln =  x x x B1 [ ] ( ) 1 5 2 1 2 ln ln 1 5 + = + x B1 Allow 1 5 1 ln ln 2 2   + −       2 2 3 2 3 2 3 1 d 3 2 x x x x x x   − + + = − + +        M1 M1 for attempt at 2 3 2 3 3 2 − + + x x x , must have 2 correct terms. 1 2 2 3 0 2 3 3 2 x x x   − + +     2 1 3 1 1 3 8 2 4 2     = − × + × +         oe M1 Dep for correct application of the limits 0 and 1 2 and attempt to evaluate – may be implied by 0.792 or 19 . 24 19 24 A1 ( ) 19 ln 1 5 24 + + A1 isw To earn marks, you must either perform algebraic long division of p(x) by (x-1/2) or use observation to find the quadratic factor q(x) such that p(x) = (x-1/2)q(x). To earn marks, you must recognize that one intersection point is where x = 1/2. Then, you need to equate the two given equations (y = 1/x and y = -x^2 + 3x + 1) and use your quadratic factor from part (a) to solve for the other x-coordinates, stating the exact values for A and B. To earn marks, you must integrate 1/x and the quadratic function separately. You need to correctly apply the limits of integration (from 1/2 to (1+sqrt(5))/2 for both functions) and then subtract the area under the quadratic from the area under 1/x to find the shaded area, giving an exact answer.

11 The line x y 2 10 + = intersects the two lines satisfying the equation x y 2 + = at the points A and B. (a) Show that the point ( , ) C 5 20 - lies on the perpendicular bisector of the line AB. [8] (b) The point D also lies on this perpendicular bisector. M is the mid-point of AB. The distance CD is three times the distance of CM. Find the possible coordinates of D. [4]

11(a) 2 10 + = x y 2 + = x y M1 For attempt to solve simultaneously ( ) 6, 8 − A1 2 10 + = x y 2 + = − x y M1 For attempt to solve simultaneously ( ) 14, 12 − A1 Alternative ( ) ( ) 2 2 10 10 4 4 − + − + = x x x x or ( ) ( ) 2 2 10 2 2 10 2 4 − + − + = y y y y (M1) For attempt to eliminate one of the variables using ( ) 2 4 + = x y 2 20 84 0 + + = x x or 2 20 96 0 − + = y y (M1) Dep for attempt to obtain a 3 term quadratic equation = 0 and solve to obtain at least one solution, allow 1 arithmetic error ( ) 14, 12 − (A1) ( ) 6, 8 − (A1) Mid-point of AB: ( ) 10, 10 − M1 For attempt to obtain the mid-point using their coordinates for A and B. Gradient of line perpendicular to AB = 2 M1 For attempt to obtain the perpendicular gradient using their coordinates for A and B. ( ) ( ) 10 2 10 − = − − y their their x their M1 ( ) 20 10 2 5 10 − = −+ oe A1 For verification 11(b) ( ) 10, 50 2 FT on their midpoint B1 for each coordinate ( ) 20, 10 − − 2 FT on their midpoint B1 for each coordinate

2 Solve the following simultaneous equations. xy x 15 2 + = y x 3 11 + = [5]

2 Eliminate one unknown 2 (11 3 ) 15 x x x − + = M1 [ ] 2 2 11 15 0 x x − + = A1 Factorises or solves their 3-term quadratic M1 5 7 , 2 2 3, 2 x y x y = = = = A2 A1 for 5 , 3 2 x x = = nfww or 7 , 2 2 y y = = nfww To earn marks, you must first eliminate one variable to form a quadratic equation, then correctly solve that quadratic equation, and finally provide both pairs of x and y values.