Cambridge IGCSE Mathematics - Additional 0606 — 2022 Feb/March Paper 2 · Variant 2

1 A line, L, has equation 4x + 5y = 9 . Points A and B have coordinates ( - 6 , 7) and (1, 9) respectively. Find the equation of the line parallel to L which passes through the mid‑point of AB. [3]

Mark scheme: Question Answer Marks Partial Marks 1  5  B1 Mid-point:  − , 8  soi  2  4 B1 Gradient: − soi or substitution of mid- 5 point into e.g. 4 x + 5 y = k 4  5  B1 y − 8 = −  x +  5  2  or 4 x + 5 y = 30 oe, isw

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3 A group of students, 4 girls and 3 boys, stand in line. (a) Find the number of different ways the students can stand in line if there are no restrictions. [1] (b) Find the number of different ways the students can stand in line if the 3 boys are next to each other. [2] (c) Cam and Dea are 2 of the girls. Find the number of ways the students can stand in line if Cam and Dea are not next to each other. [2]

Mark scheme: 3(a) [7! = ] 5040 B1 3(b) 3! × 5! oe M1 720 A1 3(c) 5040 – (2! × 6!) oe M1 3600 A1

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4 Find the x‑coordinates of the points of intersection of the curves + = 1 and y = . 4 9 2x Give your answers correct to 3 decimal places. [5]

Mark scheme: 4 Eliminates one unknown M1 x 2 1  3  2 e.g. +   = 1 4 9  2 x  Rearranges to solvable form A1 e.g. x 4 − 4 x 2 + 1 = 0 Solves : M1 dep on attempt to eliminate one unknown and 2 having a 3-term quadratic in x2 −−( 4) ± ( −4) − 4(1)(1) 2 x 2 = 2 ± 3 oe isw A1 or 3.7320[5...] and 0.2679[4...] x = ± 1.932 or x = ± 0.518 A1

5 (a) y 4 0 1 4 x The diagram shows the graph of y = f ( x) , where f ( x) is a quadratic function. Write down the two possible expressions for f ( x) . [2] 3 2 1 (b) The three roots of p ( x) = 0 , where p ( x) = 5x + ax + bx - 2 are x = , x = n and x = n + 1, 5 where a and b are positive integers and n is a negative integer. Find p ( x) , simplifying your coefficients. [5]

Mark scheme: 5(a) f ( x ) = ( x − 1)( x − 4) oe and B2 B1 for either correct f ( x ) = − ( x − 1)( x − 4) oe 5(b) Factorised form: B1 (5 x − 1)( x − n )( x − ( n + 1)) oe their −×1 ( − n ) × ( − ( n + 1)) = −2 M1 n = −2 as the only valid solution A1 Multiplies out (5 x − 1)( x + 2)( x + 1) M1 5 x 3 + 14 x 2 + 7 x − 2 A1 or a = 14 and b = 7 following n = −2 Alternative method 1: Factorised form: (B1) (5 x − 1)( x − n )( x − ( n + 1)) oe Multiplies out (M1) 5 x 3 + ( −10 n − 6) x 2 + (5 n 2 + 7 n + 1) x − ( n 2 + n ) their (n2 + n) = 2 oe (M1) n = −2 as the only valid solution (A1) 5 x 3 + 14 x 2 + 7 x − 2 (A1) or a = 14 and b = 7 following n = −2 5(b) Alternative method 2: Using product of roots: (M1) 1  − 2  n ( n + 1) = −  oe 5  5  n = −2 as the only valid solution (A1) Using sum of roots and/or sum of products (M1) of pairs of roots, solves to find a or b: 1 a + n + n + 1 = − oe and/or 5 5 1 1 b ( n ) + n ( n + 1) + ( n + 1) = 5 5 5 a = 14 or b = 7 (A1) 5 x 3 + 14 x 2 + 7 x − 2 (A1) or a = 14 and b = 7 following n = −2 If 0 scored for any method, SC3 for  1  3  1  2  1  5   + a   + b   − 2 = 0 oe  5   5   5  and 5 n 3 + an 2 + bn − 2 = 0 oe or 5 ( n + 1) 3 + a ( n + 1) 2 + b ( n + 1) − 2 = 0 oe leading to a = 14, b = 7 with n = −1 or n not stated  1  3  1  2  1  or SC1 for 5   + a   + b   − 2 = 0 oe or  5   5   5  5 n 3 + an 2 + bn − 2 = 0 oe or or 5 ( n + 1) 3 + a ( n + 1) 2 + b ( n + 1) − 2 = 0 oe

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6 (a) (i) Use the binomial theorem to expand ( 1 + 3)x 7 in ascending powers of x, as far as the term in x3. Simplify each term. [2] (ii) Show that your expansion from part (i) gives the value of .1037 as 1.23 to 2 decimal places. [2] 15 x 4 2 (b) Find the term independent of x in the expansion of + . [2] e 2 x o

Mark scheme: 6(a)(i) 1 + 21x + 189 x 2 + 945 x 3 B2 B1 for three out of the four terms correct or for a correct answer seen then spoilt If 0 scored then SC1 for 1, 21x ,189 x 2 , 945 x 3 seen but not summed 6(a)(ii) 1 + 21(0.01) + 189(0.01) 2 + 945(0.01) 3 B2 M1 for use of x = 0.01 oe in their expansion seen or implied by or 1 + 0.21 + 0.0189 + 0.000945 oe e.g. 1 + 21(0.01) + 189(0.01) 2 + 945(0.01) 3 leading to 1.229[845 = 1.23] cao or 1 + 0.21 + 0.0189 + 0.000945 OR 1.229845 without working or from working that is not fully correct 6(b) 4 3 12 M1 15  x   2  C12 ×   ×   oe  2   x  232960 A1

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7 In this question, all angles are in radians. (a) Solve the equation sec 2i = tan i + 3 for - r 1 i 1 r . [5] r tan z (b) Show that, for 0 1 z 1 , = sec z . [3] 2 1 - cos 2z 17 3 r (c) Given that cosecx =- and that 1 x 1 2 r , find the exact value of cotx. [2] 8 2

Mark scheme: 7(a) Uses a valid Pythagorean identity to write B1 in terms of a single trig ratio e.g. 1 + tan 2 θ = tanθ + 3 Rearranges and factorises/solves M1 e.g. tan 2 θ − tan θ − 2 = 0 (tanθ − 2)(tanθ + 1) = 0 tanθ = 2 tanθ = −1 soi A1 π 3π A2 A1 for any 3 correct, ignoring extras 1.11, −2.03, − , and no extras in 4 4 range cao 7(b) sin φ M1 Use of tan φ = in a correct cosφ expression or correct working Use of 1 − cos 2 φ = sin 2 φ in a correct M1 expression or correct working 1 A1 nfww Completion with = secφ cosφ 7(c) 2 M1 cot x = [ − ] cosec x − 1 soi 8 8 or sin x = − and tan x = − soi 17 15 8 15 or sin x = − and cos x = soi 17 17 1 or  −1  8   tan  sin  −     17   15 A1 − or −1.875 cao, isw 8

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8 A B 15 cm a cm C r rad 6 O r The diagram shows the sector AOB of a circle, centre O and radius 15 cm. Angle AOB is radians. 6 Point C lies on OB such that CB is a cm. AC is a straight line. (a) Find the exact value of a such that the area of triangle AOC is equal to the area of the shaded region ACB. [4] (b) For the value of a found in part (a), find the perimeter of the shaded region. Give your answer correct to 1 decimal place. [3]

Mark scheme: 8(a) 1 2  π  B1 [Area of sector = ] (15)   soi 2  6  1  π  B1 [Area of triangle = ] (15)(15 − a )sin   2  6  soi Forms correct equation and attempts to M1 solve for a or 15 – a or OC 1  e.g. (15) 2  π − 15(15 − a ) = 15(15 − a ) 2  6  4 4 75π 15 or = OC 8 4 and solves as far as a = ... or 15 – a = ... or OC = ... 5 A1 15 − π (cm) or exact equivalent 2 8(b) [CA + arc AB + BC = ] M2 FT their(15 −5 π ) and 5 π 2 2 2  5  2 5 π 15 +  π  − 2 × 15 × π × cos π  2  2 6 M1 for 15 × oe seen or 6  π   5  +  15 ×  +  15 − π  oe, soi 2  6   2  2  5  5 π  5  15 +  π  − 2 × 15 × π × cos +  15 − π   2  2 6  2  oe seen 24.1 (cm) A1

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9 (a) A vehicle travels along a straight, horizontal road. At time t = 0 seconds, the vehicle, travelling at a velocity of w ms -1 , passes point O. The vehicle travels at this constant velocity for 12 seconds. It then slows down, with constant deceleration, for 10 seconds until it reaches a velocity of ( w - 14) ms -1 . It continues to travel at this velocity for 28 seconds until it reaches point A, 458 m from O. Find the value of w. [4] (b) A particle moves in a straight line. The velocity, v ms -1 , of the particle at time t seconds, where t H 0 , is given by v = ( t - 4 )( t - 5 ) . (i) Find the value of t for which the acceleration of the particle is 0ms - 2 . [2] (ii) Find the set of values of t for which the velocity of the particle is negative. [2] (iii) Find the distance travelled by the particle in the first 5 seconds of its motion. [4]

Mark scheme: 9(a) Correct v-t graph soi e.g. B1 v w w – 14 t O 12 22 50 [A] Equates ‘area’ and distance M2 implies B1 1 e.g. 12 w + × 10 × ( w + w − 14) + M1 for 2 12w + ... + 28(w – 14) = 458 soi 28( w − 14) = 458 or 12w + 70 + 10(w – 14) + ... = 458 soi or ... + 70 + 10(w – 14) + 28(w – 14) = 458 18.4 (m s–1) A1 9(b)(i) v = t 2 − 9 t + 20 , a = 2t – 9 M1 or a = t – 5 + t – 4 from product rule t = 4.5 A1 9(b)(ii) Critical values 4, 5 M1 4 < t < 5 A1 9(b)(iii) 4 2 5 2 M1 FT their t 2 − 9t + 20 providing 3 terms (t − 9t + 20)d t + ( t − 9t + 20)d t 0 4 oe, soi 4 5 B1 3 2 3 2  t 9t   t 9t   − + 20t  +  − + 20t   3 2  0  3 2  4 64 M1 dep on an attempt to integrate − 72 + 80 + 3 125 225  64  − + 100 −  − 72 + 80  3 2  3  59 A1 oe or 29.5 2

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10 Relative to an origin O, the position vector of point P is 3i - 2j and the position vector of point Q is 8i + 13j . (a) The point R is such that PQ = 5PR . Find the unit vector in the direction OR. [5] (b) The position vector of S relative to O is mj. Given that RS is parallel to PQ, find the value of m. [3]

Mark scheme:  10(a) PQ = 5i + 15j B1     or OQ − OP = 5 ( OR − OP )  1 M1  1 OR = 3i − 2 j + (their (5i + 15 j)) FT PR = (their (5i + 15 j)) 5 5  or OR = xi + yj   PR = i + 3 j and PR = ( x − 3) i + ( y + 2) j x – 3 = 1 and y + 2 = 3 oe    or 5OR = OQ + 4OP = 8i + 13j + 4(3i −2j) oe  OR = 4i + j A1  2 2 M1 FT their ai + bj theirOR = their (4 ) + their (1 ) 4 i + j A1 oe 17  10(b) RS = λj − their (4 i + j) = −4 i + (λ− 1) j soi M1 or finds [equation RS is] y = 3x + c Correct method to find λ: M1 dep on prev M1  −4 λ− 1 FT their PQ = oe 5 15 or [for some scalar t, t (λ− 1) = 15 and 5 − 4t = 5 , therefore] − (λ− 1) = 15 oe 4 or finds e.g. −2 = 3(3) + c oe λ = − 11 cao A1

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11 y A B y = 6 + e 4x − 5 0 2 x The diagram shows part of the graphs of y = 6 + e 4 x - 5 and x = 2 . The line x = 2 meets the curve at the point B(2, b) and the line AB is parallel to the x‑axis. Find the area of the shaded region. [7]

Mark scheme: 11 Length of rectangle: 6 + e3 M1 2 or  (6 + e 3 )x  0 Area of rectangle: A1 2(6 + e3) soi Area between curve and x-axis: M2 1 4 x− 5 2 M1 for e 2 4 x − 5  1 4 x − 5  4 (6 + e )d x = 6 x + e  0  4  0 1 4 x − 5 M1 dep on an attempt to integrate that results in Correct use of limits in their 6 x + e : – 5 ax + be4x 4 F(2) – F(0) their area of rectangle – their area between M1 dep on an attempt to integrate using correct curve and x-axis limits 7 3 1 A1 nfww e + or 35.2 4 4e 5 or 35.15.... isw 11 Alternative method Length of rectangle: 6 + e3 soi (M1) 2 3 4 x − 5 (M1) 0 ( their (6 + e ) − (6 + e ) ) dx 2 3 4 x − 5 (A1) 0 ( (6 + e ) − (6 + e ) ) dx oe 2 3 4 x − 5  3 1 4 x − 5  2 (M2) 1 4 x− 5  0 (e − e )d x =  e x − 4 e oe M1 for e  0 4 3 1 4 x − 5 (M1) dep on an attempt to integrate that results in Correct use of limits in their e x − e – ax + be4x 5; may be unsimplified 4 F(2) – F(0) 7 3 1 (A1) nfww e + or 35.2 4 4e 5 or 35.151374.... rot to 4 or more sf

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12 In this question all lengths are in centimetres. O R 12 P Q C h A B 8 The diagram shows a right triangular prism of height h inside a right pyramid. The pyramid has a height of 12 and a base that is an equilateral triangle, ABC, of side 8. The base of the prism sits on the base of the pyramid. Points P, Q and R lie on the edges OA, OB and OC, respectively, of the pyramid OABC. Pyramids OABC and OPQR are similar. 3 3 2 (a) Show that the volume, V, of the triangular prism is given by V = ( ah + bh + ch ) where a, 9 b and c are integers to be found. [4] (b) It is given that, as h varies, V has a maximum value. Find the value of h that gives this maximum value of V. [3]

Mark scheme: 12(a) PQ 12 − h M1 = oe 8 12 2  or  12 − h = Area Δ PQR oe  12  16 3 8(12 − h ) A1 PQ = oe 12  12 − h  2 or Area ΔPQR = 16 3    12 oe 1  8(12 − h )  2 π M1 FT their PQ or Area ΔPQR providing of correct V = ×   × sin × h oe structure 2  12  3 2  or 16 3  12 − h × h  12  3 3 2 A1 V = ( h − 24 h + 144 h ) 9 12(b) d V 3 2 B1 3 3 2 = ( 3h − 48 h + 144 ) FT their V = ( h − 24 h + 144 h ) if of the d h 9 9 same structure 3 2 M1 their ( 3h − 48 h + 144 ) = 0 and must be a 3-term quadratic; must be an attempt 9 at a derivative factorises/solves 4 oe identified as the only solution; cao A1

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