Form and use composite functions — Paper 2 · IGCSE Mathematics - Additional 0606

11 The functions f and g are defined by x 2 - 2 f ( x) = for x H 2 , x x 2 - 1 g ( x) = for x H 0 . 2 (i) State the range of g. [1] (ii) Explain why fg(1) does not exist. [2] 2 c (iii) Show that gf ( x) = ax + b + 2 , where a, b and c are constants to be found. [3] x (iv) State the domain of gf. [1] 2 -1 x + x + 8 (v) Show that f ( x) = . [4] 2

11 marks

This question in 0606/22 Feb/March 2017

12 The function g is defined, for x 2- , by g ( x) = . 2 2x + 1 (i) Show that g l ( x) is always negative. [2] (ii) Write down the range of g. [1] The function h is defined, for all real x, by h ( x) = kx + 3 , where k is a constant. (iii) Find an expression for hg ( x) . [1] (iv) Given that hg ( 0) = 5 , find the value of k. [2] (v) State the domain of hg. [1]

7 marks

Mark scheme: 12(i) B1 −2 −×3 2 − 2 − 6 Allow − 3(2 x + 1) × 2 or 2 oe −6(2 x + 1) or 2 oe isw ( 2 x + 1) ( 2 x + 1) Denominator or (2 x + 1) 2 is positive [and B1 − k FT their g ′( x ) of the form 2 oe numerator negative therefore g ′( x ) is always ( 2 x + 1) negative] oe where k > 0; Allow (2 x + 1) −2 is always positive 12(ii) g > 0 B1 12(iii) 3k B1 + 3 oe isw 2 x + 1 12(iv) 3k B1 + 3 = 5 2(0) + 1 2 B1 implies the first B1 k = isw 3 12(v) 1 B1 x > − 2

This question in 0606/22 May/June 2017

9 The functions f and g are defined, for x 2 1 , by f ( x) = 9 x - 1 , g ( )x = x 2 + 2 . (i) Find an expression for f -1 ( )x , stating its domain. [3] (ii) Find the exact value of fg(7). [2] (iii) Solve gf ( )x = 5 x 2 + 83x - 95 . [4]

9 marks

Mark scheme: 9(i) y M1 attempt to swop; may be in later work that = x − 1 with attempt to swop x and y at contains an error 9 some point x or = y − 1 9 2 A1 condone y = ... etc; must be a function of x −1  x   f ( x ) =    + 1 oe   9  x > 0 B1 9(ii) f(51) M1 2 or fg( x ) = 9 x + 1 9 50 oe A1 9 x − 1 + 29(iii) [ gf ( x ) = ]( 2 M1 ) [ gf( x ) = ] 81( x − 1) + 2 or better A1 their ( 81x − 79 ) = 5 x 2 + 83 x − 95 → M1 provided their (81x − 79) of the form 2 ax + b for non-zero a and b their 5 x + 2 x − 16[ = 0] ( ) 1.6 oe only A1 must disregard other solution

This question in 0606/23 May/June 2017

6 The functions f and g are defined for real values of x by f x = x + 2 2 + 1 , ^ h ^ h x - 2 1 g x = , x ! . ^ h 2x - 1 2 (i) Find f 2 - 3 [2] ^ h. (ii) Show that g -1 x = g x [3] ^ h ^ h. 8 (iii) Solve gf x = . [4] ^ h 19

9 marks

Mark scheme: 6(i) f 2 = f(f) used M1 numerical or algebraic algebraic ([(x + 2)2 + 1] + 2)2 + 1 17 A1 6(ii) y − 2 M1 change x and y x = 2 y − 1 2 xy − x = y − 2 → y ( 2 x − 1) = x − 2 M1 M1dep multiply, collect y terms, factorise x − 2 A1 correct completion y = =  g ( x )  2 x − 1 6(iii)  ( x + 2 ) 2 + 1 − 2 B1   oe gf ( x ) = 2 2  ( x + 2 ) + 1 − 1   ( x + 2 ) 2 − 1 8 M1 their gf = 8 and simplify to = 2 19 2 ( x + 2 ) + 1 19 2 quadratic equation 3 ( x + 2 ) = 27 oe 3x2 + 12x – 15 = 0 solve quadratic M1 M1dep Must be of equivalent form x = 1 x = −5 A1

This question in 0606/23 Oct/Nov 2017

5 The function f is defined by f x = for x 2 2.5 . 2x - 5 (i) Find an expression for f -1 x [2] ^ h. (ii) State the domain of f -1 x [1] ^ h. ax + b (iii) Find an expression for f x giving your answer in the form , where a, b, c and d are 2^ h, cx + d integers to be found. [3]

6 marks

Mark scheme: 5(i) Putting y = f(x), changing subject to x M1 and swopping x and y or vice versa − 1 1  1  5 x + 1 A1 f ( x ) =  + 5  or oe isw 2  x  2 x 5(ii) x > 0 oe B1 5(iii) 1 B1  1 2  − 5  2 x − 5  1 M1 FT if expression of equivalent difficulty oe 2 − 5(2 x − 5) 1 e.g. 2 x − 5  1  − 5  2 x − 5  2 x − 5 A1 Completes to oe −10 x + 27 final answer

This question in 0606/21 May/June 2018

5 The function f is defined by f x = for x 2 2.5 . 2x - 5 (i) Find an expression for f -1 x [2] ^ h. (ii) State the domain of f -1 x [1] ^ h. ax + b (iii) Find an expression for f x giving your answer in the form , where a, b, c and d are 2^ h, cx + d integers to be found. [3]

6 marks

Mark scheme: 5(i) Putting y = f(x), changing subject to x M1 and swopping x and y or vice versa − 1 1  1  5 x + 1 A1 f ( x ) =  + 5  or oe isw 2  x  2 x 5(ii) x > 0 oe B1 5(iii) 1 B1  1 2  − 5  2 x − 5  1 M1 FT if expression of equivalent difficulty oe 2 − 5(2 x − 5) 1 e.g. 2 x − 5  1  − 5  2 x − 5  2 x − 5 A1 Completes to oe −10 x + 27 final answer

This question in 0606/23 May/June 2018

11 The functions f and g are defined for real values of x H 1 by f ()x = 4x - 3 , 2x + 1 g ()x = . 3x - 1 (i) Find gf ()x . [2] (ii) Find g -1 ()x . [3] (iii) Solve fg ()x = x - 1. [4]

9 marks

Mark scheme: 11(i) 2 ( 4 x − 3 ) + 1 M1 gf ( x ) = 3 ( 4 x − 3 ) − 1 8 x − 5 A1 = 12 x − 10 11(ii) y ( 3 x − 1) = 2 x + 1 B1 or x ( 3 y − 1) = 2 y + 1 ( 3 y − 2 ) x = y + 1 M1 or ( 3 x − 2 ) y = x + 1 -1 x + 1 A1 g ( x ) = 3 x − 2 11(iii)  2 x + 1 B1 − = x − 1] 4   3 [  3 x − 1  3 x 2 − 3 x − 6 oe B1 3 ( x + 1)( x − 2 ) M1 x = 2 only A1

This question in 0606/22 Oct/Nov 2018

9 Given that A = and B = , find - 9 - 3 6 5 (i) A -1 , [2] (ii) B2, [2] (iii) the matrix C, where B -1 C + A = B , [3] (iv) the matrix D, where B -2 DA = I . [3]

10 marks

Mark scheme: 9(i) 1 B1 3  − 3 − 2  B1 ×   9 5  9 (ii) 2  10 7  B2 Minus one each error B =    42 31  9(iii) C = B 2 − BA M1  1 1  A1 BA =    − 15 − 3   9 6  A1 C =    57 34  9(iv) D = B2A–1 M1 1  33 15  A2 Minus one each error D =   3  153 71 

This question in 0606/21 Oct/Nov 2019

10 The functions f and g are defined by 2 f (x) = ln (3x + 2) for x 2 - , 3 g (x) = e 2x - 4 for x ! R . (i) Solve gf ()x = 5 . [5] (ii) Find f -1 ()x . [2] (iii) Solve f -1 ( x) = g ( x) . [4]

11 marks

Mark scheme: 10(i) 2 ( ln ( 3 x + 2 ) ) B1 gf ( x ) = e − 4 their gf = 5 M1 use lnap = plna or elna = a or lnea = a B1 correct use of log/exponential relationship seen anywhere 3x + 2 = 3 or (3x + 2)2= 9 A1 3 may take the form of e0.5ln9 9 may take the form of eln9 1 A1 x = only 3 10(ii) e y − 2 M1 find x in terms of y x = 3 e x − 2 ( −1 A1 interchange x and y = f ( x ) or = y ) correct completion 3 10(iii) e x − 2 2 x M1 their f -1 ( x ) = g ( x ) = e − 4 3 3e 2 x − e x − 10 ( = 0 ) A1 obtain quadratic in ex must be arranged as a three term quadratic in order shown 3e x + 5 e x − 2 = 0 ) M1 solve for ex ( )( ) ( x = ln2 or 0.693 only A1

This question in 0606/23 Oct/Nov 2019

13 The functions f and g are defined, for x 2 0 , by 2x 2 - 1 f ( x) = , 3x 1 g ( x) = . x (a) Find and simplify an expression for fg ( x) . [2] (b) (i) Given that f -1 exists, write down the range of f -1 . [1] 2 -1 px + qx + r (ii) Show that f ( x) = , where p, q and r are integers. [4] 4

7 marks

Mark scheme: 13(a) 2 M1  2  1 − 1  x  [ fg( x ) = ] oe  1  3    x  2 − x 2 2 −x A1 mark final answer or [ fg( x ) = ] 3 x 3 x 3 13(b)(i) f−1 > 0 B1 13(b)(ii) 2 x 2 − 3 xy −=1 0 B1 or 2 y 2 − 3 xy −=1 0 Correctly applies quadratic formula: M1 FT their 2 x 2 − 3 xy −=1 0 or 2 y 2 − 3 xy −=1 0 with at most one sign error −−( 3 y ) ± ( − 3 y ) 2 − 4(2)( − 1) [ x = ] oe in the equation 2(2) or −−( 3 x ) ± ( − 3 x ) 2 − 4(2)( − 1) [ y = ] oe 2(2) Justifies the positive square root at B1 some point 2 A1 must be a function of x −1 3 x + 9 x + 8 f ( x ) = cao 4

This question in 0606/22 May/June 2021

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

9 marks

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

This question in 0606/21 Oct/Nov 2021

9 The functions f ( x) and g ( x) are defined as follows for x 2- by 3 f ( x) = x 2 + 1, g ( x) = ln ( 3 x + 2) . (a) Find fg ( x) . [1] (b) Solve the equation fg ( x) = 5 giving your answer in exact form. [3]

10 marks

Mark scheme: 9(a) (ln(3x + 2))2 + 1 oe isw B1 9(b) ln(3x + 2) = [] 2 B1 e2 = 3x + 2 M1 FT ln(3x + 2) = k, where k > 0 2e − 2 A1 x = as only solution 3 9(c) ln(3ln(3x + 2) + 2) B1 their(3ln(3x + 2) + 2) = e M1 FT their gg(x) with at most one error 3ln(3x + 2) + 2 = e A1 e − 2 M1 FT their aln(3x + 2) + b = e, where a and ln(3x + 2) = or 0.239[42…] b are non-zero constants 3 e − 2 A1 3x + 2 = e 3 or 3x + 2 = 1.270[52…] awrt −0.243 A1

This question in 0606/21 Oct/Nov 2022

8 The function f is defined for x H 0 by f ( x) = 5 - 2e -x . (a) (i) Find the domain of f -1 . [2] (ii) Solve ff - 1 ( x) = 5 x - 4 . [3] (iii) On the axes, sketch the graph of y = f ( x) and hence sketch the graph of y = f -1 ( x) . Show clearly the positions of any points where your graphs meet the coordinate axes and the positions of any asymptotes. [4] y x O 3(b) The function g is defined for 0 G x G 0.2 by g ( x) = . 1 - x Find and simplify an expression for f - 1 g ( x) . [4]

13 marks

Mark scheme: 8(a)(i) 3 ⩽ x < 5 B2 B1 for x ⩾ 3 or for x < 5 or for 3 and 5 in an incorrect inequality 8(a)(ii) x = 5 x − 4 and rearrangement to B1 x 2 − 5 x + 4  = 0 Factorises x 2 − 5 x + 4 or solves their x 2 − 5 x + 4 = 0 M1 x = 4 only, nfww A1 8(a)(iii) Correct pair of graphs. 4 B1 for correct shape for f; may not be over correct y y = f − 1 ( x ) domain but must have positive y-intercept and appear to tend to an asymptote in the 1st quadrant 5 y = f ( x ) B1 for (0, 3) and f in 1st 3 quadrant only; must have attempted correct shape x O 3 5 B1 for asymptote at y = 5; must have attempted correct y = x shape B1 for a correct reflection of their f in the line y = x Maximum of 3 marks if not fully correct 8(b) −1 5 − x −1 2 2 M1 for a complete attempt to f ( x ) = − ln or f ( x ) = ln oe find the inverse function with 2 5 − x at most one sign or arithmetic error: Putting y = f(x) and changing subject to x and swopping x and y or swopping x and y and changing subject to y −1 2 − 5 x 2 M1 FT for a correct Correct simplified form e.g.  f g( x ) =  − ln   unsimplified form of the 2(1 − x ) function; FT providing of −1 2 − 2 x or  f g( x ) =  ln equivalent difficulty   2 − 5 x

This question in 0606/22 Feb/March 2023

9 The functions f and g are defined as follows, for all real values of x. f (x) = 2 x 2 - 1 g (x) = e x + 1 (a) Solve the equation fg ( x) = 8 . [3] (b) For each of the functions f and g, either explain why the inverse function does not exist or find the inverse function, stating its domain. [4]

7 marks

Mark scheme: 9(a) 2(ex + 1)2 − 1 [= 8] M1 9 A1 ex = –1+ oe 2  3  A1 x = ln − 1   isw or 0.115  2  or 0.1145[06…] rot to 4 or more dp 9(b) f is not one-one, hence f–1 does not exist oe B1 g−1(x) = ln(x − 1) 2 M1 for x = ln(y − 1) and a swop of variables at some point or y = ln(x + 1) or ex = y – 1 and y = lnx – 1 x > 1 B1

This question in 0606/21 Oct/Nov 2023

1 The functions f and g are defined as follows, for all real values of x. f ( x) = 2 sin x + 3 cos x g ( x) = e 3 x - 1 (a) Find fg(0). [2] (b) Find gg(x). [1] -1 1 (c) Solve the equation g ( x) = ln 5 . [3] 3

6 marks

Mark scheme: Question Answer Marks Guidance 1(a) 3 B2 B1 for g(0) = 0 or [fg(x) =] 2sin(e3x – 1) + 3cos(e3x – 1) soi 1(b) 3( e 3 x −1) B1 gg ( x ) = e − 1 oe, isw 1(c) 3 y = ln( x + 1) M1 condone one error or 3x = ln( y + 1) and swops the variables at some point g−1(x) = 13 ln( x + 1) soi A1 [x =] 4 A1 Alternative method x = g( 13 ln5 ) soi (B1) e3(13ln 5) −1 oe (M1) [x =] 4 (A1)

This question in 0606/23 Oct/Nov 2023

9 The functions f and g are defined by 3x 2 f ( )x = for x 1 0 4x - 1 1 g ( )x = 2 for x 1 0 . x (a) Explain why the function fg does not exist. [1] (b) Given that the function gf does exist, find and simplify an expression for gf ( )x . [2] -1 px - x ( qx + r) (c) Show that f ( )x can be written as where p, q and r are integers. [4] 3

7 marks

Mark scheme: 9(a) Valid explanation: Range of g is g > 0 oe B1 9(b) 1 M1 2   3 x 2    4 x  1  (4 x  1) 2 A1 or simplified equivalent, isw 9 x 4 9(c) 3 x 2  4 xy  y   0  or 3 y 2  4 xy  x   0  B1 2 M1 FT their expression providing it has at ( 4 y )  ( 4 y )  4(3)( y )  x   oe most one sign error 2(3) or ( 4 x )  ( 4 x ) 2  4(3)( x )  y   oe 2(3) Justifies the negative square root B1 A1 1 2 x  x (4 x  3) f ( x )  3

This question in 0606/21 May/June 2024