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                                                    Let 2 cos 1sec cot 1 x x x x        1 0 cos sec sec sin cot n kn k kn       

6       1 11 20 1 12 11cos cos sin cos 1 sin cos = 1 n kn k n xk x n x xx x nx xx          Question 6 Minimum value will be the perpendicular distance from origin to the line segment CB. Let the point of intersection of this perpendicular and CB be D. By observing the right angle triangle formed by O, C, and D, OD = | a | sin 4  = || 2 a . It is clear that the locus of z satisfying both relations is the line segment CB, not including C. The argument of the point represented by B is arg( ) 2a  , while the argument of the point represented by C is arg( )a  . Hence the range of values is arg( ) arg( ) arg( ) 2a z a    . i 2 2 i2 2Let e , e (cos 2 isin 2 )z r z r r     2 3Re( ) 0 cos 2 0 2 22 3 44 z               Since arg( )42 a , arg( ) arg( ) arg( ) 2a z a    can satisfy the criteria of 3 44   

7 Question 7 (a) 2( 1) (3 )e 0 xk x x     The roots occur at the point of intersections of (3 )e xyx  and 2( 1)y k x . Since the y-intercept of 2( 1)y k x is k and that of (3 )e xyx  is 3, for the equation to have (i) two negative roots is 3k  , (ii) one negative and one positive root is 03 k . (b) If 1k  , then 2f ( ) ( 1) (3 )e xx x x     f (0) 2 0  4f (1) 4 0 e   Thus the positive root lies in (0,1) and so 0n . Using linear interpolation, 1 f (0) 2e 0.44164 0.44f (0) f (1) 6e 4a     f ( ) 2( 1) (3 )e e 2( 1) (2 )e 0x x xx x x x x              within (0,1) f ( ) 2 (2 )e e 2 e (1 ) 0x x xx x x            within (0,1) The graph of f is increasing and concave upwards in (0,1) so 1a is an underestimation. (c) 1 21 1 f ( ) 0.47186 0.47f ( ) aaa a    (d) For 2( 1) (3 )e 0 xxx     , 2 2 2 ( 1) (3 )e 3 ( 1) e ( 1) e 3 x x x xx xx xx          2 1 (1 ) e 3 nx nnxx    (shown) Let 2F( ) ( 1) e 3 xxx   and so 22F ( ) 2( 1)e e ( 1) e ( 4 3)x x xx x x x x        [B1] Since 2e ( 4 3) 0x xx   in the interval  0,1 , and  min F ( ) 3x  , the sequence is not convergent.

8 Question 8 (a) y A path T (a2 – x2) a O x Since AB is tangential to T, gradient of T at B(x,y) 22d gradient of d y a x ABxx   . (b) 2 2 2 2d dd y a x a x yxx x x     . Using the substitution sinxa  gives   2 2 2 sin cos dsin aaya a           2 2 cos dsin 1 sin dsin cos ec d sin d ln cos ec cot cos ln cos ec cot cos a a a aC a a C                               2 2 2 2 sin sin cos ,cot x a x a xxa a a x          Substitute above into the above gives  ln cosec cot cosy a a C       2 2 2 2 22 22 ln ln a a x a xa a Cx x a a a xa a x C x                       Since  ,0a lies on T, 22 220 ln 0 a a aa a a C C a        . x B(x,y)

9 Hence 22 22ln a a xy a a x x     . (c) Height of A above the point where the x-coordinate is b is 22ab . Therefore the height of A above O is 22 2 2 2 2 22 ln ln a a ba a b a b b a a ba b          (d) Length of arc   2 2 22 22 2 d1 d ( )d 1 d 1 d d ln ln a b a b a b a b a b y x b ax ax xx ax xx a xx ax aa b                   Question 9 (a) 2 2 2 2 2 21 tan sec 1 ( 3) ( 3) 1 822 yy xx           For hyperbola, 2 2 2 1 8 9c a b     26c   (b) 3ce a (c) Since the coastline is 18 km in total distance, 18OP PF Using cosine rule on OPF , we have 2 2 2 2 2 2 2( )( )cos (18 ) 6 2( )(6)cos 324 36 36 12 cos 24 3 cos PF OP OF OP OF r r r rr r                

10 (d) When 0x , 2 2(0 3) 1 8 8 y y     *check that 24 83 cosr   when 2   Cartesian coordinates of A is (0, 8) which means coordinates of B is (6, 8) 1 4tan 3BOF  