5 The functions f and g are defined as follows. ) (f x x x 4 2 = + for x R ! ( ) g e x 1 x 2 = + for x R ! (a) Find the range of f. [2] (b) Write down the range of g. [1] (c) Find the exact solution of the equation ) ( fg x 21 = , giving your answer as a single logarithm. [4]

5 2 5 10 1 1 ... 3 3 9   − = − +     x x x 2 M1 allow one sign error or one arithmetic slip 3 27, 3 = = a a B1 Term in x: 2 3 5 3 3 − = a a b M1 For multiplying their terms, must have sum of 2 relevant products = b 18 = − b A1 Term in 2 : x ( ) 2 3 5 10 3 3 3 9 − + = a a a c M1 For multiplying their terms, must have sum of 3 relevant products = c This question is incomplete and appears to be a mark scheme for a different question, likely involving polynomial expansion or algebraic manipulation, not the stated functions f and g. Based on the provided mark scheme, marks are earned by correctly identifying coefficients after expanding and simplifying a polynomial expression. You need to show a correct method for multiplying terms to find the coefficient of x, and a correct method for multiplying terms to find the coefficient of x^2, allowing for one sign or arithmetic error for the initial expansion.

4 Find rational values a and b such that a b 5 2 5 2 1 + + - = . [5]

4 ( ) ( ) 5 2 5 2 1 a b − + + = oe, soi M1 2 2 1 b a − = A1 0 a b + = or 5 5 0 a b + = A1 Solves their linear simultaneous equations in a and b as far as a = … or b = … M1 dep on previous M1 1 1 , 4 4 a b = − = A1

7 A curve has equation ( ) y x p = , where ( )x x x x 4 6 1 p 3 2 = - + - . (a) Find the equation of the tangent to the curve at the point (3, 8). Give your answer in the form y mx c = + . [5] (b) (i) Given that p 1 - exists, write down the gradient of the tangent to the curve ( ) y x p 1 = - at the point (8, 3). [1] (ii) Find the coordinates of the point of intersection of these two tangents. [2]

7(a) 2 d 3 8 6 d = − + y x x x B1 Finds their 3 d d = x y x M1 condone one slip 1 9 = T m A1 y – 8 = their 9(x – 3) or y = 9x + c and 8 = 9(3) + c M1 y = 9x – 19 cao A1 7(b)(i) 2 1 9 = T m their B1 FT their 9 7(b)(ii) [Uses y = x in their ( y = 9x – 19) to form] their ( x = 9x – 19) or their ( y = 9y – 19) oe and solves for x or y or solves e.g. their (9x – 19) = their 19 9 + x M1 19 19 , 8 8       oe A1 FT equal x and y coordinates providing at least 3 marks earned in (a)

9 (a) The function f is defined, for all real x, by ( ) f x x x 13 4 2 2 = - - . (i) Write ( ) f x in the form ( ) a b x c 2 + + , where a, b and c are constants. [3] (ii) Hence write down the range of f. [1] (b) The function g is defined, for x 1 H , by ( ) g x x x 2 1 2 = + - . (i) Given that ( )x g 1 - exists, write down the domain and range of g 1 - . [2] (ii) Show that ( )x px q 1 g 1 2 =- + + - , where p and q are integers. [4]

9(a)(i) 15 – 2(x + 1)2 isw B3 B1 for ( ) 2 1 + x B1 for a = 15 9(a)(ii) f ⩽ 15 B1 STRICT FT their a 9(b)(i) Domain: x ⩾ 2 B1 Range: g−1 ⩾ 1 B1 9(b)(ii) 2 2 2 ( 1 ) 0 + + −− = x x y or 2 2 2 ( 1 ) 0 + + −− = y y x B1 Correctly applies quadratic formula: [ ] 2 2 2 2 4(1)( 1 ) 2 −± − −− = y x or [ ] 2 2 2 2 4(1)( 1 ) 2 −± − −− = x y M1 FT their 2 2 2 ( 1 ) 0 + + −− = x x y or 2 2 2 ( 1 ) 0 + + −− = y y x with at most one sign error in the equation Justifies the positive square root at some point B1 Correct completion to 1 2 g ( ) 1 2 − = −+ + x x A1

9 The following functions are defined for x 1 2 . f( )x x x 1 3 = + - g( )x x 1 2 = + (a) Find f ( )x g . [2] (b) Find ( )x g 1 - . [2] (c) DO NOT USE A CALCULATOR IN THIS PART OF THE QUESTION. Solve the equation f( ) g( ) x x = . [5]

9(a) [ ] 2 2 4 fg( ) x x x + = oe, final answer 2 B1 for an attempt at the correct order of composition with at most one error 9(b) Complete, correct method to find the inverse M1 1 1 g ( ) x x − −   =   final answer A1 9(c) 3 2 4 0 x x − − = M1 condone one sign or arithmetic error Shows x – 2 is a factor or shows that x = 2 is a solution M1 Uses x – 2 is a factor to find 2 2 x x + + B2 B1 for a quadratic factor with 2 terms correct Indicates that 2 2 x x + + has no real roots and states x = 2 as the only solution A1 dep on all previous marks awarded

Cambridge IGCSE Additional Mathematics · 0606

IGCSE Additional Mathematics (0606) Functions past paper questions

These questions cover functions — domain and range, composite and inverse functions, and the modulus — the language the rest of the 0606 course is written in.

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