2023 TJC NYJC VJC H2 FM 9649 P1 Ans

1 2023 TJC_NYJC_VJC H2 FM Prelim Paper 1 [Suggested Solution] Question 1 (a) Using Simpson’s Rule with 0.5h , 3 1 1 4 2 4 d (0.5)4 1 3 4(1) 1 4(1.5) 1 4(2) 1 4(2.5) 1 4(3) 1 k k k k k kxx           3386 1 4 2 4 31185 6 3 5 7 9 11 k k k k k     3386 1 6772 31185 6 3465 k  1 3k  (b) Volume required 33 211 11122 d 2 d4 4 1 5079 kkx x xx x x       Question 2 Let Pn be the proposition that 12 , for 2.n nn n 

2 2 1 1 1 11 12, i.e. 2 To show P is true. i.e. . 1i.e. 1 2 1 112 2 since 21 1When 2, 2 1.5 2 2 P is true. 1Assume P is true for 2 112 1 12 k k n k k k kk k nn kk k k kk k k k kkk k k                                         1 1 21 1 12 12 by inductive hypothesis 21 1 1 since 2 112 1 i.e. . P true P tru 11 e. Since P is true a 2 nd P true P true by mathematical inducti 1 k k k kk kk k k k k kk k k k k k k                                    12 , foon 2. r n nn n  Question 3 At the pole, 0.r cos3 1.  3 ,3 ,5 5,,33       

3 The cartesian forms are   5tan , tan and tan .33y x y x y x               π 23 0 π 23 0 1Area 2 3 d 2 3 1 cos3 d 4.712 r          Maximum value of r is 2 and it occurs when cos3 1.  3 0, 2 , 4 240, ,33       The points are   242,0 , 2, , 2, .33            Using cosine rule, Distance = 22 22 2 2(2)(2)cos 3  12 23   Or Distance between point 242, and 2, .33            = 242sin 2sin33           3322 22                23

4 Question 4 (a) 11 11 2 2.025 1.025 1.025( ) 2.025 1.025 0 1 or =1.02 0 5 n n n n n n n HH H m m HH HH m m           0 1 ( 50 54.9 1.025 19 9 1 6, 146 146 1 6(1.025 .02 ) 5)n n n n H A B H A B H A B BA H            (b) Let nI denote the new HPI n years after the start of 2020. 10 0 104.897IH 10.5nnI I  10nI  When n = 4, 6.56 10nI  Thus in the year 2024.

5 Question 5     1(1 i tan ) cos isincos = sec cos isin k kk k kk             11 00 cos sec Re (1 i tan ) nn kk kk k        1 0 1 i tan 1(1 i tan ) (sum of GP)1 i tan 1 nn k k         sin1 i 1cos i tan cos sin 1cos i tan sec cos sin 1 ( i) cot sec sin cot i 1 sec cos cot n n n nn i n i n nn                            11 00 cos sec Re (1 i tan ) nn kk kk k       1 0 cos sec sec sin cot n kn k kn        Let ,3   1 0 1 0 cos sec sec sin cot3 3 3 3 3 1cos 2 2 sin33 3 n kn k n kn k kn kn             