Find the domain and range of 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

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

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

8 f(x) 2 1 0 x 2r 4r 2r 8r - 1 3 3 3 - 2 - 3 - 4 - 5 - 6 8r The diagram shows the graph of f( x) = a cos bx + c for 0 G x G radians. 3 (a) Explain why f is a function. [1] (b) Write down the range of f. [1] (c) Find the value of each of the constants a, b and c. [4]

6 marks

Mark scheme: 8(a) Valid explanation e.g. B1 Each value of x is mapped to a unique value of y. 8(b) −5 f 1 B1 8(c) a = 3, b = 0.75 oe, c = − 2 B4 B1 for a = 3 B1 for c = −2 2π 8π M1 for = oe b 3 A1 for b = 0.75 oe

This question in 0606/22 Feb/March 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

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

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

8 (a) The functions f and g are defined by r 3 r f ( x) = sec x for 1 x 1 2 2 g ( x) = 3 ( x 2 - 1) for all real x. (i) Find the range of f. [1] - 1 2 r (ii) Solve the equation f ( )x = . [3] 3 (iii) Given that gf exists, state the domain of gf. [1] (iv) Solve the equation gf ( )x = 1. [5] (b) The function h is defined by h ( x) = ln ( 4 - x) for x 1 4 . Sketch the graph of y = h ( x) and hence sketch the graph of y = h -1 ( x) . Show the position of any asymptotes and any points of intersection with the coordinate axes. [4] y O x

14 marks

Mark scheme: 8(a)(i) f ⩽ 1 1 8(a)(ii) x = 2 nfww 3   2π    2π  M1 for x  f  sec or         3    3  1 2π sec x  3 1 A1 for  2π  cos    3  OR M1 for a complete attempt to find f x1 ( ) ; includes swapping the variables 1 1 1  A1 for f ( x )  cos    x  8(a)(iii) π 3π 1  x  2 2 8(a)(iv) gf ( x )  3(sec 2 x  1) B1 2 1 4 M1 3tan x  1 or  oe cos 2 x 3 3 M1 tan x 1 oe or cos x  oe and solves 3 4 for x, soi 5π 7π A2 A1 for one correct solution, condoning x  , and no other solutions extras 6 6 8(b) Correct diagram with intercepts indicated and 4 B1 for correct shape for h; may not be asymptotes shown. over correct domain but must have positive y-intercept and x-intercept and y appear to tend to an asymptote in the 4th quadrant h-1(x) 4 B1 for 3 and ln4 correctly marked; must 3 have attempted correct shape ln4 B1 for the position of the vertical O ln4 3 4 x asymptote indicated; must have attempted correct shape B1 for h1 the reflection of their h in the line h(x) y = x Maximum of 3 marks if not fully correct

This question in 0606/23 May/June 2023

2 (a) Write 3 + 4x - 2 x 2 in the form a + b ( x + c) 2 , where a, b and c are integers. [3] (b) Hence write down the range of the function f ( )x = 3 + 4x - 2x 2 , where x ! R . [1]

4 marks

Mark scheme: 2(a) 5 – 2(x  1)2 B3 Mark final expression B2 for – 2(x  1)2 or B1 for (x  1)2 or b = –2, c = –1 and B1 for 5 + b(x + c)2 oe with numerical values of b and c or a = 5 2(b) f  their 5 B1 STRICT FT of their 5 from (a)

This question in 0606/21 May/June 2024