RI 9758 2023 Prelim P2 Solution

Raffles Institution H2 Mathematics 2023 Year 6 _____________________________________________________________________________________________ 2023 Year 6 H2 Mathematics Preliminary Examination Paper 2: Solutions 1 Solutions (a) [2] 2 22 2 2 11 22 11 24 xt t tt t = −  = −+ −   = −− Since 22 1 111 0, .2 244tt − ≥ − − ≥−   Hence 1 4x≥− for all values of t. (b) [3] d 3d y t = , d 21d x tt = − d dd 3 d d d 21 y yt x txt∴ =×= − When 1t =− , 2x= , 1y=− , d 1d y x =− & gradient of normal = 1. Equation of normal at the point where 1t =− is 12yx+=− 3yx⇒= − (c) [4] The curve and the line 1 4x=− intersect when 1 2t = . Required Area 11 44 11 24 22 12 d ( 3) d (3 2)(2 1) d ( 3) d 5.90625 5.906 (3.d.p) yx x x t tt xx −− − − = −− = + −− − = = ∫∫ ∫∫ Alternative Solution 1

Raffles Institution H2 Mathematics 2023 Year 6 _____________________________________________________________________________________________ Required Area 13 4 13 4 7 12 1 1 122 1 11( )d ( 3 )d44 113( ) d ( 3 ) d24 5.90625 5.906 (3.d.p) xy y y t ty y − − − − − − = + + ++ = − + ++ = = ∫∫ ∫∫ Note that 13 4 1 1 199( 3 ) d can be seen as .4 244yy − − ++ ××∫ Alternative Solution 2 Required Area ( ) ( ) 1 4 1 2 10 2 9 10 9 2 1 3 2 3 22 1 13 9d 1d 24 4 1 13 96 2 dt 1 6 2 dt24 4 yx yx tt tt − − − −  = + × +×−   = +− + × + × − +−  ∫∫ ∫∫

Raffles Institution H2 Mathematics 2023 Year 6 _____________________________________________________________________________________________ 2 Solutions (a) [2] 1 , 3 , 5 , 7 , . . . Number of terms in each bracket follows an AP with first term 1 and common difference 2. ( ) 2 Number of integers in the first sets 2(1) 1 22 n n nn ∴ = +− = (b) [4] Last integer in the nth set is the ( ) 2n th term of the AP 1, 4, 7, 10, 13, 16, ... which has first term 1 and common difference 3. ( ) 22Last integer of the th set 1 1 3 3 2n nn = +−=− From GC, n 232n − 25 1873 26 2026 ∴ 26k = OR: Given that 2023 occurs in the kth set, ( ) ( ) 2 2 2 2 first term in the th set 2023 last term in the th set 3 1 2 3 2023 3 2 2022 2025 1 and 33 24.961 26.961 and 25.980 or 25.980 25.980 26.961 kk kk kk k kk k ≤≤  − − +≤ ≤ − −≤ ≥ − ≤ ≤ ≤− ≥ ∴ ≤≤ Since k∈ + , 26k = OR: ( ) 2 23 1 2 3 2023 3 2kk − − +≤ ≤ − From GC, 24.961 26.961 and 25.980 or 25 .980k kk− ≤ ≤ ≤− ≥ 25.980 26.961k∴ ≤≤ Since k∈ + , 26k = (c) [3] Required sum

Raffles Institution H2 Mathematics 2023 Year 6 _____________________________________________________________________________________________ ( ) ( ) ( ) ( ) 22 22 22 Sum of first 10 th terms Sum of first 4 th terms 10 42(1) 10 1 3 2(1) 4 1 322 14 574 = −   = + − − +−   = OR: Last term in the 10th set 23(10) 2 298= −= Last term in the 4th set 23(4) 2 46= −= First term in the 5th set 46 3 49= += To find the sum of the AP : 49,52,55,...,298 with first term 49 and common difference 3: ( )298 49 1 3 84 m m =+− = ( )84Required sum 49 298 14 5742∴ = += OR using GC: Since the 1st integer in the 5th set is the 2(4 1)+ th term in the AP and the last integer in the 10th set is the 210 th term, required sum = ( ) 100 17 13 1 r r = + −=∑ 14574

Raffles Institution H2 Mathematics 2023 Year 6 _____________________________________________________________________________________________ 3 Solutions (a) [6] e cos3xyx= d e cos3 e ( 3sin 3 )d 3e sin 3 xx x y xxx yx = +− = − ( ) 2 2 dd 3e sin 3 3e 3cos 3dd dd 9dd d2 10 (shown)d xxyy xxxx yy yyxx y yx = −− =+ −− = − 32 32 dd d 2 10 ddd y yy xxx = − 23 23 When 0, dd d1 , 1 , 8 , 2 6d dd x yy yy x xx = = == −= − By Maclaurin expansion, ( ) ( )23 23 8 261 ...2! 3! 131 4 ... 3 yx x x xx x −−=++ + + =+− − + (b) [2] Using standard series expansion, 2311e 1 ...2! 3! x xx x=++ + + 2 2(3 ) 9cos3 1 ... 1 ...2! 2 xxx= − += − +

Raffles Institution H2 Mathematics 2023 Year 6 _____________________________________________________________________________________________ 23 2 23 23 11 9e cos3 1 ... 1 ...26 2 19 191 ...22 62 131 4 ... (verified)3 x x xx x x xx x xx x   = ++ + + − +       =++ − + − +     =+− − + (c) [3] ( ) ( ) 2 2 2 2 2 2 2 2 2 2 ln(1+e cos3 ) ln 1 1 4 ln 2 4 ln 2 1 2 2 ln 2 ln 1 2 2 22ln 2 2 ...22 1ln 2 2 ...2 24 1 17ln 2 ... (shown)28 x x xx xx x x x x x xx x xx x xx ≈ ++− = +− = +− = + +−   − = +− − +  = +− − +  =+− + 4 Solutions (a) [4] If n is perpendicular to a, 0⋅=na . [ ]( )() () () 0 since (shown) ⋅ −⋅ ⋅ = ⋅⋅ − ⋅⋅ = ⋅=⋅ ab a aa b a ab aa aa ba ab ba If n is parallel to plane OAB, it is perpendicular to ×ab since ×ab is perpendicular to plane OAB. i.e ( ) 0⋅×=nab .

Raffles Institution H2 Mathematics 2023 Year 6 _____________________________________________________________________________________________ [ ] ( ) ( ) ( ) ( )() () () 0 since and (shown) ⋅ −⋅ ⋅× =⋅ ⋅×−⋅ ⋅× = ⊥× ⊥× ab a aa b a b ab a a b aa b a b a ab b ab OR: Since n is a linear combination of non-zero and non-parallel vectors a and b, n lies on the same plane as a and b ie plane OAB. Hence n is parallel to the plane OAB. (b) [2] ( )() 12 1 11 2 111 1 11 13 1 11 3 120 0 61 31 3 3 1 13 3 1 =⋅ −⋅        ⋅ −⋅                     =−==          −−      n ab a aa b = 0 1 1 2 1  = − m or 0 1 1 2 1  = −  m OR: Let vector parallel to m be x y z    . Then, 1 1 0 0 (1) 1 x y xyz z    ⋅ = ⇒ + + = −−−    Also, 12 11 0 13 2 1 0 2 0 (2) 1 x y z x y xyz z      ⋅× =            ⋅ − = ⇒ − − = −−−    −   (1) (2) : 0x+= Then 0yz y z+=⇒= − If 1y= , then 1z=− Hence 0 1 1 2 1  = − m .

Raffles Institution H2 Mathematics 2023 Year 6 _____________________________________________________________________________________________ (c) [2] Line OA : 1 1, 1 λλ  = ∈ r  Let l be the line which passes through point B and is parallel to m. l : 20 1 1, 31 µµ      = +∈   −   r  When line OA intersects l, 12 0 11 1 13 1 2 11 31 λµ λ λ µµ λ µµ        = +    −    = = +⇒= =−⇒= ∴Coordinates of point of intersection are ( )2, 2, 2 . OR : Find foot of perpendicular of B to line OA (d) [3] Note that m // n, and n is perpendicular to a. Thus, line l is perpendicular to line OA. By Ratio Theorem, ' 2 '2 22 2 22 1 3 231 OB OBOF OB OF OB  →  →  →  →  →  → += = −  = −= Equation of line of reflection is 2 3, 1 ββ  = ∈ r  ∴A cartesian equation of the line of reflection is 23 xy z= = . O B l F

Raffles Institution H2 Mathematics 2023 Year 6 _____________________________________________________________________________________________ 5 Solutions (a) [2] Let X denote the mass of a small massage ball in grams. ( ) 2~N 200,X σ P (195 < X < 205) = 0.98273 195 200 205 200P 0.98273Zσσ −− << = 55P 0.98273 5 2.3809 2.1000 2.1 (1dp) Zσσ σ σ −<< = = = = (b) [3] Let Y denote the mass of a medium massage ball in grams. ( ) 2~N 200, 2.1X and ( ) 2~N 500,1.4Y ( ) ( ) ( )16E 2 6E 2E 200X XY X Y+ −= − = ( ) ( ) ( )16Var 2 6Var 4Var 34.3X XY X Y+ −= + = 16 2 ~ (200,34.3)X X YN+− ( )16P 2 210X XY+ −> = 0.0439 (3sf) (c) [1] Assume that the mass of a massage ball is independent of the mass of another massage ball. 6 Solutions (a) [3] Arrange the 5 boys in ( 51− )! =24 ways. Then slot in each of the 3 girls into the 5 spaces between the boy