Cambridge IGCSE Mathematics - Additional 0606 — 2021 Oct/Nov Paper 2 · Variant 1

1 Solve the inequality ( x + 5)( x - 2) 2 3x + 6 . [3]

Mark scheme: Question Answer Marks Partial Marks 1 x2 + 3x – 10 − 3x − 6 * 0 oe M1 Condone one sign or arithmetic error * can be = or any inequality sign Critical Values: 4 and −4 A1 x > 4 or x < −4 A1 Mark final answer

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2 Solve the following simultaneous equations. xy + x 2 = 15 y + 3 x = 11 [5]

Mark scheme: 2 Eliminate one unknown M1 x(11− 3x) + x2 =15 2 A1 2x −11x + 15[ = 0] Factorises or solves their 3-term quadratic M1 5 7 A2 5 x = , y = A1 for x = , x = 3 nfww 2 2 2 x = 3, y = 2 7 or y = , y = 2 nfww 2

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3 A curve has equation y = . x + 1 dy r k (a) Show that the exact value of at the point where x = can be written as 2 , where k dx 6 r is an integer. + 1 [5] b 6 l (b) Find the equation of the normal to the curve at the point where x = 0 . [4]

Mark scheme: 3(a) d B1 ( sin3 x ) = 3cos3 x soi d x Applies the correct form of the quotient rule M1 dy ( x + 1)(3cos3 x) − (2 + sin3 x) [1] A1 d FT their ( sin3 x ) = 2 dx dx ( x + 1)  π  3π   3π  M1  + 1  3cos  −  2 + sin  [1] dy  6  6   6  = dx  π  2  + 1   6   dy  −3 A1 not from wrong working  =  2  d x   π   + 1   6  3(b) [When x = 0 ] y = 2 B1 dy B1 dy [When x = 0 ] = 1 FT their dx dx [ m⊥=] = −1 M1 FT −1 their1 y – 2 = −x oe A1 FT their m⊥

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5 It is given that y = 3 tan 2 x for 0° 1 x 1 360° . dy 2 (a) Show that = m tan x sec x where m is an integer to be found. [2] dx dy (b) Find all values of x such that = 3 sec x cosec x . [5] dx

Mark scheme: 5(a) dy 2 B2 B1 for = 6tan x sec x dx d 2 1 2 (tan x ) = 2(tan x ) sec x dx 5(b) 6 tan x sec 2 x − 3sec x cosec x = 0 B1 NB division by secx is B0 3sec x (2 tan x sec x − cosec x ) = 0 oe 2 tan 2 x = 1 oe B1 1 M1 FT tan 2 x = k where k > 0 tan x = [ ± ] or [±] 0.707[1…] 2 35.3 or 35.2643… rot to 2 or more dp A2 no extras in range 215.3 or 215.2643… rot to 2 or more dp 144.7 or 144.7356… rot to 2 or more dp A1 for any two correct answers 324.7 or 324.7356… rot to 2 or more dp

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6 Find the values of m for which the line y = m x - 2 does not touch or cut the curve y = ( m + 1) x 2 + 8x + 1. [6]

Mark scheme: 6 2 B1 (m +1) x + (8 − m) x + 3 = 0 oe, soi 2 M1 (8 − m) − 4(m +1)(3) m 2 − 28 m + 52 [*0] oe M1 dep on previous M1; condone one sign error where * is = or any inequality sign Factorises or solves their 3-term quadratic M1 dep on use of b2 – 4ac expression or equation for CVs Finds correct CVs: 2, 26 A1 2 < m < 26 A1 Mark final answer

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7 (a) Use logarithms to solve the following equation, giving your answer correct to 1 decimal place. 5 x - 2 = 3 # 2 2 x + 3 [4] (b) Solve the equation log 3 (y 2 + 11 ) - 2 = log 3 ( y - 1 ) . [5]

Mark scheme: 7(a) x−2 2 x+3 M1 log5 = log3 + log2 soi ( x − 2)log5 = log3 + (2 x + 3)log 2 oe M1 dep on previous M1; Condone one sign or bracketing error log 3 + 3log 2 + 2 log 5 A1 x = soi log 5 − 2 log 2 x = 28.7 A1 7(b)  y 2 + 11  B1 log 3   = log 3 ( y − 1)  9   y 2 + 11  or log 3   = 2 oe  y − 1  y 2 + 11 y 2 + 11 M1 = y − 1 or = 9 oe 9 y − 1 2 A1 y − 9y + 20 = 0 Solves their 3-term quadratic M1 dep on previous M1 y = 4, y = 5 A1

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8 Marc chooses 5 people from 4 men, 4 women and 2 children. Find the number of ways that Marc can do this (a) if there are no restrictions, [1] (b) if at least 2 men are chosen, [3] (c) if at least 1 man, at least 1 woman and at least 1 child are chosen. [3]

Mark scheme: 8(a) 252 B1 8(b) [2 men and 3 others = ] 120 M2 M1 for any two correct [3 men and 2 others = ] 60 [4 men and 1 other = ] 6 186 A1 Alternative method [0 men =] 6 (M1) [1 man and 4 others = ] 60 (their 252) – (6 + 60) (M1) 186 (A1) 8(c) M W C M2 for at least four out of five 1 3 1 32 correct values soi 1 2 2 24 or M1 for any two or three 2 2 1 72 correct values soi 2 1 2 24 3 1 1 32 184 A1 Alternative method [0 men =] 6 (M1) [0 women ] 6 [0 children] 56 (their 252) – (6 + 6 + 56) (M1) 184 (A1)

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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]

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

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10 y 5 2 y = + x - x x O x 5 2 The diagram shows part of the curve y = + x - x . x (a) Find, in the form y = mx + c , the equation of the tangent to the curve at the point where x = 1. [5]

Mark scheme: 10(a) dy −2 M2 M1 for any two correct terms = −5 x + 2 x − 1 oe dx d y A1 [When x = 1 ] = − 4 and y = 5 d x y – 5 = −4(x – 1) oe M1 d y FT their and y; dep on at d x x =1 least M1 y = −4x + 9 A1 FT 10(b) x 3 x 2 B2 B1 for 5lnx and one other term F( x ) = 5ln x + − (+c) correct 3 2 F(3) − F(1) M1 dep on at least B1 for integration 14 A1 5ln 3 + 3

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11 The volume, V, of a cone with base radius r and vertical height h is given by rr 2 h . 3 The curved surface area of a cone with base radius r and slant height l is given by rrl . A cone has base radius r cm, vertical height h cm and volume V cm3. The curved surface area of the cone is 4r cm2. 2 16 2 (a) Show that h = 2 - r . [4] r (b) Show that V = r 16r 2 - r 6 . [2] 3 (c) Given that r can vary and that V has a maximum value, find the value of r that gives the maximum volume. [5]

Mark scheme: 11(a) 4 B1 l = r h 2 = l 2 − r 2 or l 2 = r 2 + h 2 M1 2 2 M1 FT their l ; dep on previous M1 2 4 2 4 2 2 = h =  − r or  r + h r r 2 16 2 2 2 or l = and h = l − r r 2 2 16 2 A1 Correct, convincing completion to h = − r r 2 Alternative method 2 2 (B1) l = r + h 2 2 (M1) πr r + h = 4π 2 (M1) 2 4 2 2 ( r + h ) = r 2 16 2 (A1) Correct, convincing completion to h = − r r 2 11(b) π 2 16 2 M1 r 2 − r 3 r A1 π 4  16 2  r  2 − r  3  r  π 2 6 and correct completion to 16r − r 3 111(c) − d V π  1 2 6  5 B3 B2 for 2 1 =  (16 r 32 r − 6 r ) oe − r )  ( 2 6 − d r 3  2 16 r − r 32 r − 6 r   k ( 2 5 ) ( ) where k is a constant and k ≠ 0 or B1 for 1 2 6 − 2 k 16 r − r × ( f ( r ) ) where ( ) f(r) ≠ 32 r − 6 r 5 d V M1 FT their f(r) = ar +br5 Equates their to 0 and solves as far as for a, b ≠ 0 d r r4 = … r = 1.52 or 1.519[67…] rot to 4 or more sf or A1 2 oe 4 3

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