Find the inverse of a one–one function — 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

9 A function f is defined, for x G , by f ( x) = 2 x 2 - 6x + 5 . 2 (i) Express f ( x) in the form a ( x - b) 2 + c , where a, b and c are constants. [3] (ii) On the same axes, sketch the graphs of y = f ( x) and y = f -1 ( x) , showing the geometrical relationship between them. [3] y O x (iii) Using your answer from part (i), find an expression for f -1 ( x) , stating its domain. [3]

9 marks

Mark scheme: 9(i) 2 B3 or B3 for a = 2 and b = 1.5 and c = 0.5 2 ( x − 1.5 ) + 0.5 isw provided not from wrong format isw or B2 for 2 ( x − 1.5 ) 2 + c where c ≠ 0.5 or a = 2 and b = 1.5 or SC2 for 2 ( x − 1.5 ) + 0.5 or  2 1  seen 2  ( x − 1.5 ) +   4  or B1 for ( x − 1.5 ) 2 seen or for b = 1.5 or for c = 0.5 or SC1 for 3 correct values seen in incorrect format e.g. 2 ( x − 1.5 x ) + 0.5 or 2 ( x 2 − 1.5 ) + 0.5 9(ii) y B3 B1 for correct graph for f over correct domain or correct graph for f – 1 over 5 correct domain B1 for vertex marked for f or f – 1 and intercept marked for f or f – 1 B1 for idea of symmetry – either symmetrical by eye, ignoring any scale or 1.5 line y = x drawn and labelled 0.5 x Maximum of 2 marks if not fully correct 0 0.5 1.5 5 9(iii) x – 0.5 2 M1 FT their a,b,c, provided their a ≠1 and = ( y − 1.5 ) a,b,c are all non-zero constants 2 y – 0.5 2 or = ( x − 1.5 ) and reverses 2 variables at some point −1 x – 0. 5 A1 must have selected negative square root f ( x ) = 1.5 − oe only; condone y = ... etc.; must be in terms 2 of x −1 6 − 8 x – 4 If M0 then SC2 for f ( )x = 4 oe or SC1 for − 1 −−( 6) ± 36 – 4( 2)(5 − x ) f ( x ) = oe 2(2) 1 B1 x ≥ oe 2

This question in 0606/22 May/June 2017

2 y y A B O x O x C y D y O x O x The four graphs above are labelled A, B, C and D. (i) Write down the letter of each graph that represents a function, giving a reason for your choice. [2] (ii) Write down the letter of each graph that represents a function which has an inverse, giving a reason for your choice. [2]

4 marks

Mark scheme: 2(i) B and C with valid reason B2 B1 for one graph and valid reason or both graphs and no reason 2(ii) B only with valid reason B2 B1 for graph B or valid reason

This question in 0606/23 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

5 Given that A = and B = , find 1 4 - 2 5 (i) A-1 , [2] (ii) the matrix C such that CA = B , [2] (iii) the matrix D such that A-1 D + B = I . [3]

7 marks

Mark scheme: 5(i) 1  4 −3  2  4 − 3    B1   5  −1 2   − 1 2  1 B1 5 5(ii) post multiply by A–1 M1 C = BA–1 1  0 5  A1   5  −13 16  5(iii)  0 − 4   − 4 23  B1 I − B =   or AB =    2 − 4   − 7 24  D = A ( I − B ) or D = A – AB M1  6 − 20  A1 D =    8 − 20 

This question in 0606/23 Oct/Nov 2018

9 (a) It is given that g ()x = 6x 4 + 5 for all real x. (i) Explain why g is a function but does not have an inverse. [2] (ii) Find g2 ()x and state its domain. [2] It is given that h ()x = 6x 4 + 5 for x G k . (iii) State the greatest value of k such that h-1 exists. [1] (iv) For this value of k, find h-1(x). [3] (b) The function p is defined by p ()x = 3ex + 2 for all real x. (i) State the range of p. [1] (ii) On the axes below, sketch and label the graphs of y = p (x) and y = p -1 (x) . State the coordinates of any points of intersection with the coordinate axes. [3] y y = x O x (iii) Hence explain why the equation p (x) = p -1 (x) has no solutions. [1]

13 marks

Mark scheme: 9(a)(i) Valid explanation e.g. B2 B1 for either each x is mapped to a Each x is mapped to a unique value of y [and so g unique value of y oe or for inverse does is a function] but the inverse does not exist not exist because it is many to one oe because it is many to one oe 9(a)(ii)  g 2 ( x ) =  6(6 x 4 + 5) 4 + 5 isw B2 B1 for  g 2 ( x ) =  6(6 x 4 + 5) 4 + 5 isw     for all real x B1 for correct domain 9(a)(iii) [k = ] 0 B1 9(a)(iv) 4 y − 5 M1 4 x − 5 x = soi or y = 6 6 y − 5 A1 x − 5 x = ± 4 or y = ± 4 6 6 −1 x − 5 A1 If M1 A0 A0, allow SC1 for an answer h ( x ) = − 4 x − 5 x − 5 6 of h −1 ( x ) = 4 or y = 4 6 6 9(b)(i) p > 2 B1 9(b)(ii) For p: B2 B1 for each Correct exponential shape tending to y = 2 passing through (0, 5) For the inverse function: B1 Approximate reflection of p in the dotted line passing through (their 5, 0) 9(b)(iii) Valid explanation e.g. B1 The graphs do not intersect and so there are no solutions oe

This question in 0606/22 Feb/March 2019

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

10 (a) g( )x = 3 + for x H 1. x (i) Find an expression for g -1 ( x). [2] (ii) Write down the range of g -1 . [1] (iii) Find the domain of g -1 . [2] 2(b) h( )x = 21n( 3x - 1) for x H . 3 The graph of y = h( )x intersects the line y = x at two distinct points. On the axes below, sketch the graph of y = h( )x and hence sketch the graph of y = h -1 ( x). [4] y O x

9 marks

Mark scheme: 10(a)(i) Correct method to find inverse M1 −1 1 A1 g ( x ) = oe x − 3 10(a)(ii) g−1 ⩾ 1 or [1, ∞) B1 10(a)(iii) 3 < x  4 or (3, 4] B2 B1 for 3 and 4 in an incorrect inequality or for x > 3 or x ⩽ 4 10(b) Correct graph for h B1 h−1 the reflection of h in y = x B1 FT their h Both graphs drawn over the correct domain B1 FT their h and h−1 h−1 B1 Correct graphs intersecting twice h 2 3 2 3

This question in 0606/22 Feb/March 2020

11 The function f is defined by f( x) = nl ( 2 x + 1) for x H 0. (a) Sketch the graph of y = f( x) and hence sketch the graph of y = f -1 ( x) on the axes below. [3] y O x The function g is defined by g( x) = ( x - 4) 2 + 1 for x G 4. (b) (i) Find an expression for g -1 ( )x and state its domain and range. [4] (ii) Find and simplify an expression for fg( x). [2] (iii) Explain why the function gf does not exist. [1]

10 marks

Mark scheme: 11(a) y B3 B1 for correct shape of f or f1 B1 for symmetry B1 for drawn over correct domain Maximum of 2 marks if not fully correct x 11(b)(i) [ ±] x −=1 y − 4 soi M1 −1 A1 g ( x ) = 4 − x −1 [Range] g–1 ⩽ 4 B1 [Domain] x ⩾ 1 B1 11(b)(ii) ln(2[(x – 4)2 + 1] + 1) M1 ln(2x2 – 16x + 35) A1 11(b)(iii) Valid explanation, e.g. some of the B1 values in the range of f are outside the domain of g

This question in 0606/21 May/June 2020

10 The function f is defined by f ( )x = for 0.5 G x G 1 .5 . 2x The diagram shows a sketch of y = f ( x) . y 4x 2 - 1 y = 2x 0 x 0.5 1.5 (a) (i) It is given that f -1 exists. Find the domain and range of f -1 . [3] (ii) Find an expression for f -1 ( )x . [3] a 1 - 2 (b) The function g is defined by g ( )x = ex2 for all real x. Show that gf ( )x = e e bx o, where a and b are integers. [2]

8 marks

Mark scheme: 10(a)(i) Range f−1: 0.5 ⩽ f−1 ⩽ 1.5 B1 2 2 2 2 Domain f−1: 0 ⩽ x ⩽ oe B2 B1 for 0 and in an incorrect inequality 3 3 2 2 or for x ⩾ 0 or x ⩽ 3 10(a)(ii) Correctly collects terms ready to M1 factorise e.g. 4 x 2 − 4 x 2 y 2 = 1 or 4 y 2 x 2 − 4 y 2 = −1 or simplifies to subject in one term 1 2 only e.g. = 1 −x or 4 y 2 1 2 − = y − 1 oe 4 x 2 Correctly factorises and/or M1 FT only if of equivalent difficulty rearranges at least as far as: 2 1 2 −1 x = or y = oe 4 − 4 y 2 4 x 2 − 4 −1 1 A1  f ( x ) =  or   2 4 − 4 x −1 [ y = ] 2 oe, isw 4 x − 4 10(b) Correct order of composition: M1 2 −1   4 x 2   gf(x) = e  2 x  1   A1  1−  2 gf ( x ) = e  4 x  isw

This question in 0606/22 Feb/March 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 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