1 x 0 y - 3 - 5 1 5 The diagram shows the graph of the cubic function ( ) y x f = . The intercepts of the curve with the axes are all integers. (a) Find the set of values of x for which ( )x 0 f 1 . [1] (b) Find an expression for ( )x f . [3]

1(a) 3 1 − x B1 1(b) ( )( )( ) 1 3 1 5 3 − + − − x x x 3 B1 for a negative cubic function B1 for a cubic function multiplied by 1 3 B1 for ( )( )( ) 3 1 5 + − − x x x

3 The polynomial ( ) p x ax x bx 9 6 3 2 = - + - , where a and b are constants, has a factor of x 2 - . The polynomial has a remainder of 66 when divided by x 3 - . (a) Find the value of a and of b. [4] (b) Using your values of a and b, show that ( ) ( ) ( ) p q x x x 2 = - , where ( ) q x is a quadratic factor to be found. [2] (c) Hence show that the equation ( ) p x 0 = has only one real solution. [2]

3(a) ( ) p 2 : 8 36 2 6 0 − + − = a b B1 ( ) p 3 : 27 81 3 6 66 − + − = a b B1 M1 Dep on at least one of the previous B marks, for attempt to solve their equations and obtain a solution for both a and b 6, 3 = = − a b A1 For both 3(b) ( )( ) 2 2 6 3 3 − + + x x x 2 M1 for attempt at quadratic factor either by observation to obtain 2 6 3 + + x px or by algebraic long division to obtain at least 2 6 3 ... + x x A1 all correct 3(c) Discriminant of ( ) q x = 2 3 4 6 3 − × × 63 = − which is < 0 M1 For calculation of discriminant and confirmation that it is < 0 ( ) q 0 = x has no real solutions hence ( ) p 0 = x has only one real solution A1 For a correct conclusion from correct work.

5 The polynomial ( ) p ax bx x x 6 4 3 2 = + + + , where a and b are integers, is divisible by x 2 - . When ( ) p x l is divided by x 1 + the remainder is 7 - . (a) Find the value of a and of b. [5] (b) Using your answers to part (a), find the remainder when ( ) p x ll is divided by x. [2]

5(a) 0 8 4 12 4 = + + + a b B1 For ( ) p 2 ( ) 2 p 3 2 6 ′ = + + x ax bx M1 For an attempt to obtain ( ) p′ x 3 2 6 7 − + = − a b M1 Dep for ( ) p 1 ′ − 0 2 4 = + + a b 13 3 2 − = − a b M1 Dep on both previous M marks for solution of equations to obtain both a and b 3 2 = − = a b A1 5(b) ( ) p 18 4 ′′ = − + x x M1 For differentiation of their ( ) p′ x to obtain ( ) p′′ x 4 A1 FT on twice their b.

6 DO NOT USE A CALCULATOR IN THIS QUESTION. The polynomial ( )x x ax x b 10 10 p 3 2 = + - + , where a and b are integers, is divisible by x 2 1 + . When ( )x p is divided by x 1 + , the remainder is 24 - . (a) Find the value of a and of b. [4] (b) Find an expression for ( )x p as the product of three linear factors. [4] (c) Write down the remainder when ( )x p is divided by x. [1]

6(a) 5 0 5 4 4 = − + + + a b M1 For use of the factor theorem 24 10 10 − = − + + + a b M1 For use of the remainder theorem 4 15 + = − a b 24 + = − a b leading to M1 Dep on both previous M marks for solution of their equations without using a calculator 27 = − a , 3 = b A1 6(b) ( )( ) 2 2 1 5 ........... + + x x their b M1 Allow for observation or algebraic long division. Their a and b must be integers. ( )( ) 2 2 1 5 16 3 + − + x x x A1 ( )( )( ) 2 1 5 1 3 + − − x x x 2 M1 for attempt to factorise their 3-term quadratic A1 all correct from fully correct working 6(c) 3 B1 FT on their (integer) b

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.