Cambridge IGCSE Mathematics - Additional 0606 — 2022 May/June Paper 2 · Variant 1

1 (a) Solve the equation 5 w - 1 = 12, giving your answer correct to 2 decimal places. [2] (b) Solve the equation x 2 1 3 - 5x 3 + 6 = 0. [3]

Mark scheme: Question Answer Marks Partial Marks 1(a) w 1 log 5 12 M1 log12 or w 1 log5 w  2.54 cao A1 1(b) Rewrites in quadratic form e.g.: M1 1 y  x 3 y 2  5 y  6  0 2 1 1 3   5 x 3  6  0 or  x and factorises or solves e.g. : M1 Factorising their 3 term (y – 2)(y – 3) = 0 quadratic x = 8 and A1 x = 27

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2 (a) Write 2 lg x - lg ( x + 6) + lg 3 as a single logarithm to base 10. [2] (b) Hence solve the equation 2 lg x - lg ( x + 6) + lg 3 = 0 . [4] ` j

Mark scheme: 2(a) x 2 B2 B1 for any two log laws lg oe, nfww applied correctly e.g. 3( x  6) x 2 lg  lg3 x  6 2(b) x 2 B1 x 2 lg  lg1 FT their lg 3( x  6) 3( x  6) 0 x 2 providing a single logarithm or 10  3( x  6) x 2  3 x  18  0 B1 dep on B2 in part (a) Factorises or solves their 3-term quadratic M1 x = 6 indicated as only solution A1 dep on B2 in part (a)

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3 Variables x and y are such that when 3 y is plotted against x2 , a straight line passing through the points (9, 8) and (16, 1) is obtained. Find y as a function of x. [4]

Mark scheme: 3 Valid method to find m M1 8  1 m =   1 9  16 Valid method to find c e.g. M1 FT their m 1  their ( 1)  16  c 3 y  their ( 1) x 2  their17 A1 Equation with correct variables and 3 y = 3 y  x 2  17  oe, isw A1

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4 The polynomial p ( )x = mx 3 - 17 x 2 + nx + 6 has a factor x - 3. It has a remainder of - 12 when divided by x + 1. Find the remainder when p ( )x is divided by x - 2. [6]

Mark scheme: 4 27m – 153 + 3n + 6 = 0 or better B1 m – 17 – n + 6 = 12 or better B1 Eliminates one unknown for a pair of linear equations in M1 m and n and solves for one unknown m = 6, n = 5 A2 A1 for either 24 cao A1

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5 (a) (i) Write down, in ascending powers of x, the first three terms in the expansion of ( 1 + 4x) n . Simplify each term. [2] (ii) In the expansion of ( 1 + 4x) n ( 1 - 4x) the coefficient of x2 is 6032. Given that n 2 0, find the value of n. [3] x 8 10 (b) Find the term independent of x in the expansion of - 4 . [2] e 2 x o

Mark scheme: 5(a)(i) 1 + 4nx + 8n(n – 1)x2 B2 B1 for any two correct terms or all 3 correct but listed not summed 5(a)(ii) 8n(n – 1) – 16n M1 FT from (i) identifying correct terms and combining their coefficient of x2 – 4  their coefficient of x Solves or factorises their 3-term quadratic in n only M1 Forms a 3-term quadratic = 0 and solves except allow ‘= a constant’ if they go on to complete to square n = 29 only A1 5(b) 8 2 M1 Must be clearly identified not 10  x   8  C 2     4  soi in expansion  2   x  45 A1 11.25 or 4

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6 (a) (i) A 5-digit number is to be formed from the seven digits 0, 1, 2, 3, 4, 5, 6. Each digit can be used at most once in any number and the number does not start with 0. Find the number of ways in which this can be done. [2] (ii) Find how many of these 5-digit numbers are even. [3] (b) A team of 7 people is to be selected from a group of 9 women and 6 men. Find the number of different teams that can be selected which include at least one man. [2] n n 1 3(c) (i) Show that C 3 + C 2 = ( n - n ) for n H 3. [5] 6 (ii) Hence solve the equation n C 3 n + C 2 = 4n where n H 3. [2]

Mark scheme: 6(a)(i) 6 × 6 × 5 × 4 × 3 oe M1 2160 A1 6(a)(ii) Full correct calculation (360 + 900) M2 M1 for any correct product ‘how many end with 0’+ soi ‘how many end with 2, 4, 6’ oe Eg (6 × 5 × 4 × 3 × 1) or 360 (5 × 5 × 4 × 3 × 1) or 300 (5 × 5 × 4 × 3 × 3) or 900 1260 cao A1 6(b) 15C7  9C7 M1 6399 A1 6(c)(i) n ! n ! B2 B1 for either expression  correct ( n  3)!3! ( n  2)!2! n ( n  1)( n  2) n ( n  1) M2 M1 for either expression  correct 6 2 n ( n  1)( n  2  3) A1 6 n 3  3n 2  2 n 3n 2  3n or  6 6 1 3 leading to ( n  n ) 6 6(c)(ii) n ( n 2  25)  0 oe M1 n = 5 as the only solution A1

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7 Variables x and y are such that y = . Use differentiation to find the approximate change x in y when x increases from 1.9 to 1.9+ ,h where h is small. [6]

Mark scheme: 7 d B1 (sin3 x )  3cos3 x soi d x u  (1  sin3 x ) 4 M1 FT their 3cos3x d u 3  4(1  sin3 x ) (3cos3 x ) dx soi 3 4 1  12 M1 FT their d u ortheir d v but x  4(1  sin3 x  (3cos3 x )   (1  sin3 x )  x d y 2 d x d x  2 not both dx x   Correct derivative A1 Evaluates their derivative at x = 1.9 and M1 multiplies by h 0.651h A1

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8 In this question, i is a unit vector due east and j is a unit vector due north. Distances are measured in kilometres and time is measured in hours. At 09 00, ship A leaves a point P with position vector 5i + 16 j relative to an origin O. It sails with a constant speed of 6 3 on a bearing of 120°. (a) Show that the velocity vector of A is 9i - 3 3 j . [2] (b) Find the position vector of A at 12 00. [1] (c) At 11 00 ship B leaves a point Q with position vector 29i + 16 j . It sails with constant velocity - 12 3 .j Write down the position vector of B, t hours after it starts sailing. [1] (d) Find the distance between the two ships at 12 00. [3]

Mark scheme: 8(a) B2 B1 for either x or y correct x  6 3sin60 y 6 3cos60 oe Allow SC1 for verification and completion to 9i  3 3j that 9i  3 3j has a bearing of 120 and that 9i  3 3j has a magnitude of 6 3 8(b) B1 (5i + 16 j)  3  9i  3 3 j  oe, isw 8(c) 29i  16 j  t ( 12 3 j) oe, isw B1   8(d) Forms AB or BA when t = 1 e.g. B1 FT their (b) and (c) with  t = 1 BA =(32i + (16  9 3) j)  (29i  (16  12 3) j) oe   M1 FT their AB or BA 32  (3 3) 2 6 (km) A1 cao

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9 In this question all lengths are in metres. x h 5 The diagram shows a water container in the shape of a triangular prism. The depth of water in the container is h. The container has length 5. The water in the container forms a prism with a uniform cross-section that is an equilateral triangle of side x. 5 3 h 2 (a) Show that the volume, V, of the water is given by V = . [4] 3 (b) Water is pumped into the container at a rate of 0.5 m3 per minute. Find the rate at which the depth of the water is increasing when the depth of the water is 0.1 m. [4]

Mark scheme: 9(a) 2 M1 Correct expression 2  x  h  x   connecting x and h  2 oe h or cos30  oe x 1 1 2 or xh = x sin60 2 2 2 h A1 Must be x = x  oe 3 2 M1 1  2 h  V =  their  sin60 5   2  3  1 2 h or h their  5 2 3 5 3 h 2 A1 Correct completion to given answer V  3 9(b) 10 3 B1 Correct derivative of V e.g. h 3 d h d V d h B1 V   soi Not d t d t d V h  10 3  M1 0.5    0.1   3 oe 0.866 (metres per minute) A1 3 or 0.8660[25…] rot to 4 or more sf Allow isw 2

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10 (a) Differentiate x ln x - 2x with respect to x. Simplify your answer. [2] d 2 y x + 1 2 dy e 2 e 3 2 (b) A curve is such that 2 = . It is given that = + 2e at the point e, + e . dx e x o dx 2 e 6 o Using your answer to part (a), find the exact equation of the curve. [8] Question 11 is printed on the next page.

Mark scheme: 10(a) ln x  1 B2 1 B1 for [1]ln x  x    2  oe x 10(b) 1 B1 x   2 x dy x 2 M2 d y   ln x  2 x   c  M1 for  ...  ln x  ... or dx 2 d x dy x 2 for   ...  2 x dx 2 Substitution to find c: M1 FT their attempt to integrate e 2 e 2 dependent on at least M1  2e   lne  2e  c 2 2 [c = 1]  x 2  A1 dx   2 x  ln x  1  y    2  Integrates and uses (a) M1 Dependent on M2M1 3 2 FT error in c only x 2 x y    x ln x  2 x  C 6 2 e 3 2 e 3 2 M1 Substitution to find C  e   e  elne  2e  C Dependent on previous M1 6 6 x 3 2 A1 y   x  x ln x  2 x  e 6

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11 y D x 2 y = e r x = 4 A y = cos 5x O B C x x r 2 The diagram shows part of the curves y = e and y = cos 5 x and part of the line x = . The 4 r curves intersect at A. The curve y = cos 5x cuts the x-axis at B. The line x = cuts the x-axis at C and 4 x 2 at D. Find the exact area of the shaded region, ABCD. [7] the curve y = e

Mark scheme: 11  π  B1 B  ,0  soi  10  π x π M2 x x 04 e 2 dx  010 cos5 x d x M1 Integrates e 2 to ke 2 , k  0 M1 Integrates cos5x to k sin5 x , 1 k  0 or k  5 π π A2 A1 for each part correct  x  4  sin5 x  10 2  2e  0     5  0 π M1 M1 for correct attempt at    1 5π 1  0 8  2e  2e   sin  sin0 subtraction and for one    5 10 5  correct use of limits dependent on at least M1for integration and B1 π A1 8 11 2e  isw 5

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