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 (a) To earn this mark, you must correctly identify the range of x values where the function is less than or equal to zero, including the endpoints. (b) To earn these marks, you need to show that the function is a negative cubic, that it has a factor of 1/3, and that the factors (x+3), (x-1), and (x-5) are present.

1 (a) Write the expression x x 6 1 2 - + in the form ( ) x a b 2 + + , where a and b are constants. [2] (b) Hence write down the coordinates of the minimum point on the curve y x x 6 1 2 = - + . [1]

1(a) (x – 3)2 – 8 B2 B1 for (x – 3)2 + k where k ≠ −8 or a = −3 or (x + m)2 − 8 where m ≠ −3 or b = −8 1(b) (3, −8) B1 strict FT their a and b

3 (a) Solve the inequality x 4 1 9 2 - . [3] (b) Solve the equation x x 2 11 12 0 - + = . [3]

3(a) 4x – 1 *9 oe and 4x − 1*−9 oe OR 16x2 − 8x − 80*0 oe soi M1 where * could be = or any inequality sign x > 5 2 , x < −2 only; mark final answer A2 not from wrong working A1 for CV 5 2 , −2 oe If M0 then SC1 for any correct inequality with at most one extra inequality 3(b) ( )( ) 2 3 4 − − x x or x = u2 and (2u – 3)(u – 4) oe soi M1 3, 4 2 = = x x oe A1 x = 9 4 , x = 16 B1 FT their x Alternative ( ) 2 1 2 2 2 12 11   + =     x x simplified to 2 4 73 144 0 − + = x x (M1) solves 3 term quadratic in x (M1) x = 9 4 , x = 16 (A1)

3 Find the values of the constant k for which ( ) k x x k 2 1 6 1 0 2 - + + + = has real roots. [5]

3 Uses 2 4 − b ac : 62 − 4(2k – 1)(k + 1) M1 2 8 4 40*0 − − + k k oe M1 dep on first M1 where * is = or any inequality sign condone one sign or arithmetic slip in simplification Factorises or solves their 3-term quadratic expression or equation for CVs e.g. (5 + 2k)(8 – 4k) oe M1 Finds correct CVs: −2.5 oe, 2 A1 −2.5 ⩽ k ⩽ 2 A1 mark final answer

7 Find the exact values of the constant k for which the line y x 2 1 = + is a tangent to the curve y x kx k 4 2 2 = + + - . [6]

7 2 4 2 2 1 + + − = + x kx k x M1 2 4 ( 2) 3[*0] + − + − x k x k soi A1 * can be , =, ≤, ≥ ( ) 2 2 4(4)( 3) − − − k k M1 2 20 52 0 − + ∗ k k A1 2 ( 20) ( 20) 4(1)(52) 2 −− ± − − = k M1 k = 10 48 ± oe isw A1 Alternative (using calculus) 2 8 = + x k oe (M1) ( ) 2 4 2 8 2 8 2 = + − + − − y x x x x or 2 4 6 = − − y x x (M1) 0 = 4x2 + 8x +1 (A1) 2 8 8 4(4)(1) 8 −± − = x (M1) 48 1 8 = −± x oe (A1) for k = 10 48 ± oe (A1)

Cambridge O Level Additional Mathematics · 4037

O-Level Additional Mathematics (4037) Quadratic functions past paper questions

Revise O-Level Additional Mathematics (4037) — Quadratic functions with real Cambridge past-paper questions grouped by topic, each shown with its marking scheme. Working one topic at a time targets exactly the skill each question tests.

61 free questions from 2016–2021, each with its marking scheme. Open the full topical bank to filter by subtopic, paper, and year.