SJI 2024 Year 6 HL MAA Prelim Examination Paper 2 (MS)

SJI Year 6 HL MAA Prelim Exam 2024/P2 (Markscheme) Page 1 of 11 Year 6 HL MAA Preliminary Examination 2024 Paper 2 (Markscheme) Section A Qn Suggested solution Markscheme 1 Normal distribution with conditional probability [Marks: 6] (a)       2~ 3.24, 0.375 0.1 0.9 W N P W w P W w     By GDC (invNorm or nSolve), 3.72 (3 sf)w (A1) – cumulative prob (M1) – invNorm or nSolve A1 (b)      4 44 | 0.1 0.1 0.021348 0.1 0.213 (3 sf) P W W w P WP W W w          M1 A1 – conditional prob A1 – no follow through as   0.1P W w  is given 2 Vectors [Marks: 4]   2 2 2 2 0                                  a ba v b = a b b b a b= a b b b b a ba b b b a b a b = Since   0  a v b ,   a v b   sin 2       a v b a v b a v b M1 – distributive M1 – 2 b b = b A1 R1 – dot prod 0, perpendicular AG

SJI Year 6 HL MAA Prelim Exam 2024/P2 (Markscheme) Page 2 of 11 Qn Suggested solution Markscheme 3 Counting Techniques [Marks: 5] (a) (choose female P/VP and choose male P/VP and choose other 4 EXCO members)   4 4 28 1 1 42 655200 C C C   M1 M1 – combination & multiplication principle A1 (Accept 655000) (b) (arrange P/VP in middle two seats and arrange other 4 EXCO members in remaining seats)   2! 4! 48 M1 – arrangement A1 4 Sum and Product of Roots [Marks: 5] (a)   3 4 3( 3)( 6 10)P x x x x x      roots 0 6 6A    (M1)A1 (b)   7 roots 1 ( 30) 30G     M1A1 (c)  roots 1 31G   A1 5 Arithmetic sequence & series (with horizontal scaling [Marks: 8] (a) 2k  A1 (b) Method 1 1 2From graph, 3 and 7 7 3 4 x x a      Method 2 Period of  2 siny x  is 2 M1 A1 – consecutive min. pts. A1 M1 – use period of graphs

SJI Year 6 HL MAA Prelim Exam 2024/P2 (Markscheme) Page 3 of 11 Qn Suggested solution Markscheme Period of   22 4y f x      4a  Method 3 1 2 difference in values when ( )=0n nx x x f x    sin 0 , 2 ,...2 2x x           1 2 4 2 4n nx x      M1 – scaling of period A1 M1 – use of x-intercepts M1 A1 (c)     1 1 3 4 1 6 4 1 2 N N n n n Nx n N                Method 1 By GDC Graph   1 3 4 1 x n y n       and Table, 15 1 16 1 465 500 528 500 n n n n x x         Least 16N  Method 2 By GDC Graph  6 4 1 5002 xy x      15.563N  Least 16N  M1 A1 – AP with first term 3 and common diff. 4 Follow through from part (a) R1 A1 M1 A1

SJI Year 6 HL MAA Prelim Exam 2024/P2 (Markscheme) Page 4 of 11 Qn Suggested solution Markscheme 6 Domain, Area between Curves and Volume of Solid of revolution [Marks: 6] (a) For  f x to be defined,    2 7 0 7 7 0 7 or 7 x x x x x        Smallest 7a M1 Allow 7x A1 (b)   2 2 2 2 7 7 7 2 y x x y x x yx y        227 7 3 7Vol. of solid d 2 142 units (3 sf) y yy         M1 – Squaring M1 – making x the subject AG M1 – with square A1 – follow through only for incorrect positive value of a from part (a) 7 Area of sector + Roots of a complex number [Marks: 11] (a) 2 5   A1 (b) 2 2 1 2 52 5 25 Since 0, 5 r r r r          M1 (A1) A1 (c)i)   5 i 24 8 11 i 205 1 1 7 1 1 9 1 173 i i i ii5 5 20 5 20 5 20 5 204 1 i 2 e where 2, 1,0,1,2 e e , e , e , e , e n n kz k n z k k k k k k                        M1 A1 – for i 4ek  M1 A2, 1 (3 correct), 0 (c)ii) 1 5 5 5 5 3125 k r k     M1 A1

SJI Year 6 HL MAA Prelim Exam 2024/P2 (Markscheme) Page 5 of 11 Qn Suggested solution Markscheme 8 Sketch rational function (oblique asymptote) + Stationary points [Marks: 10] (a) 22 7 4 3 12 1 3 x xy x x x        (M1) – finding oblique asymptote A1 – shape (asymptotic behaviour), 2 branches A1 – 2 1y x  (O.A.) A1 – 3x (V.A.) A1 – Min.  2.29, 2.17 A1 – Max.  3.71,7.83 (b) Method 1:   2 2 2 7 4 1 2 13 3 d 1 2 o.e.d 3 x xy x x x y x x          At stationary points, d 0d y x  :   2 1 2 3 13 2 x x     x-coordinate of mid-point 1 1 13 32 2 2 3                  Therefore, mid-point of the turning points of the graph lie on its vertical asymptote 3x Method 2: A1 M1 A1 M1 AG A1

SJI Year 6 HL MAA Prelim Exam 2024/P2 (Markscheme) Page 6 of 11 Qn Suggested solution Markscheme   2 2 2 2 7 4 3 d 2 12 17 d 3 x xy x y x x x x       At stationary points, d 0d y x  :   2 2 2 2 12 17 0 3 2 12 17 0 x x x x x        Sum of roots 12 62   x-coordinate of mid-point 6 32  Therefore, mid-point of the turning points of the graph lie on its vertical asymptote 3x M1 A1 M1 AG

SJI Year 6 HL MAA Prelim Exam 2024/P2 (Markscheme) Page 7 of 11 Section B Qn Suggested solution Markscheme 9 Binomial Distribution + Discrete Random Variables [Marks: 19] (a) (i) ܺ~ܤ൬݊,1 4൰ ⟹ܧ(ܺ)= ݊ 4 (ii) ݊= 10 ⟹ܲ(ܺ≥ 3) < 0.5 ݊= 11 ⟹ܲ(ܺ≥ 3) > 0.5 Therefore, ݊= 10. M1 – evidence of ‘݌݊’ A1 M1 – valid attempt, e.g., use of “binomCdf” or “binomInvN” with correct parameters R1 – some comparison of probabilities must be seen A1 (b) 0 1 1 2 0 0 1 1 2 1 1 2 2 3 1 1 2 2 3 2 2 3 3 4 ݐ 0 1 2 3 4 ܲ(ܶ= ݐ) 1 16 4 16 6 16 4 16 1 16 A1 – ݌ଵ A1 – ݌ଶ A1 – ݌ଷ (c) ܵ= 0 1 1 2 0 0 1 1 2 1 1 2 2 3 1 1 2 2 3 2 2 3 3 4 ܲ(ܶ≥ 2ܵ)= 4 + 3 + 3 + 1 16 = 11 16 M1 – sample space A1

SJI Year 6 HL MAA Prelim Exam 2024/P2 (Markscheme) Page 8 of 11 Qn Suggested solution Markscheme (d) ⟹݉= 70 A1 – scatter plot must be seen OR R1 – logical argument must be seen A1 (e) ݉= 20 + 30 + ⋯ + 80 7 = 50 ݔ= 5 + 6 + 9 + 13 + 14 +ݔ+ 21 7 =ݔ+ 68 7 ⟹ 0.2834 × 50 − 1.741 =ݔ+ 68 7 ⟹ݔ= 19.003 ≈ 19 A1 A1 M1 A1 – 19 (integer) (f) ݎ= 0.989 Note: Allow FT2 for ݎ= 0.991 (ݔ= 18). A2 (g) The results are consistent (high r) and ௫ ௠ ≈ 0.25 A1

SJI Year 6 HL MAA Prelim Exam 2024/P2 (Markscheme) Page 9 of 11 Qn Suggested solution Markscheme 10 3D Geometry + Applications of Trigonometry [Marks: 19] (a) Let ℎ be the height of the pyramid. Consider a triangle with height ℎ, hypotenuse ݏ and base √ଶ ଶݏ. Thus, ݏଶ = ℎଶ + 1 2ݏଶ ⟹ ℎ = 1 √2 ݏ (M1) – correct triangle considered A1 – correct use of Pythagoras’ (b) Let ߠ be the angle between a triangular face and the base. Consider a triangle with height ℎ, base ଵ ଶݏ and hypotenuse connecting the midpoint of the base edge of the pyramid and its vertex. Thus, tanߠ= 1 √2ݏ 1 2ݏ =√2 ⟹ߠ≈ 0.955 rad or 54.7∘ (M1) – correct triangle considered M1A1 (c) Consider a triangle with height ℎ, base ݀+ ௦ ଶ. Here the hypotenuse is from M to the vertex. ℎ = 140 ⟹ݏ 2 = 1 √2 ℎ = 140 √2 ≈ 98.995 m Consider a triangle with height ℎ, base ݀+ ௦ ଶ and angle of elevation 10∘. ⟹ tan 10∘ = 140 ݀+ 140 √2 ⟹݀= 694.98 ≈ 695. (M1) – correct base (A1) M1 – use of correct trigo ratio, eg. tan A1 – valid intermediate step/s to find ݀ AG (d) The drone’s vertical distance from the ground is given by ݏ(ݐ)= නݒ(ݐ) dݐ= 2ݐଷ/ଶ +ܥ And since ݏ(0) = 140, we get ݏ(ݐ)= 2ݐଷ/ଶ. ݏ(ݐ)= 0 ⟹ݐ= 16.985 Therefore, ݒ(16.985) = 12.4 m/s. (M1)A1 – condone missing +ܥ A1) – do not award if ܥ= 0 was not considered (M1) A1 – 16.985 A1

SJI Year 6 HL MAA Prelim Exam 2024/P2 (Markscheme) Page 10 of 11 Qn Suggested solution Markschem