VJC 2023 JC1 MYE 9649 (Solutions)

S/N Solution 1 2(i) 21 (1 ) 0nn nur u r u    Auxiliary equation: 2 (1 ) 0rr  (1 ) ( ) 0 1 or r r      General solution is , , are arbitrary constantsn nuA B r A B (ii) 0 00 As , provided 1. Hence, = . n n n uA B nu A L r BL uLL r         5(i) (ii) Curve intersects itself when 3 0yt t  . 20 or . 1 tt x       (iii) The curve is symmetrical about the x-axis. required area bounded by curve is 1 0 2 d yx    . 2 3 2 2 22 ,,1 11 1 2dd (1 ) txy t t t t txt t       y x

1 3 2200 22 220 22 d 2 () d (1 ) ()=4 d (1 ) tyx t t t t tt tt              22 2 22 ()f( ) , g ( ) . (1 ) ttt t      4        1 Let f 1 cos 2 . 0, f 0 2 1, f 1 2 02 1 2 0.522 x xx x x x         By Newton-Raphson method,       1 f f' 1c o s 2 1sin 1c o s 2 1sin n nn n nn n n n nn n n n xxx x xxx x x xxx x x           Using 1 0.5,x  2 3 0.4091 0.4096 x x   Since f(0.405) = 00212 > 0 and f(0.415) = 0.0245 < 0, 0.405 0.415  0.41 (to 2 d.p) ** 22 * 2 0, is undefined. f ( ) is undefined. Hence Newton -Raphson method fails. xx x   * 2x •  is obtained from 1 For 0, we consider 0. Hence, we take . xt yt t       

5(a) 00 1 221 1 11 ds i n ( ) 1 xa xaax       1 1 10s i n ( ) 1 sin ( ) 1 aa aa a        (b) 2 2 2 22 2 12tan : d sec d d d22 2 1 22 1sin , cos , tan : sin 211 1 xxtt x x t t tx t txt t x tt t         22 22 2 2 2 2 2 22 12 2 2 cos 1 1 2 2d1 d1c o s s i n 11 1 1 (1 ) d (1 )(1 ) 1 d ( s h o w n ) 1 11 2 d d 211 11tan ln(1 )22 11 ln 1 tan22 2 11 ln sec22 2 1or = xt t t xtxx tt t t t t tt t t t ttt tt tt C xxC xxC                             ln sec22 xxC 6(i) 22 2( ) ( 1 ) ( 1 )xk x k x kk k k kyx k xk xk xk        The asymptotes are and .xk yxk  (ii) 2 d( 1 )2 stationary points 1 0 has 2 distinct real roots.d( ) yk k xx k     2()( 1 ) 0 01 xk k k k    The set of k is  :0 1 .kk  (iii)

(iv) l bisects the angle between the asymptotes. 32 48   Acute angle between l and x-axis = 3 288   gradient of l = tan .8m  7or tan 8   7(a)        iiee c o s i s i n c o s i s i n cos isin cos isin cos isin cos isin 2isin (shown) nn nn n n nn nn nn nn n                 (b)       5i i i5 i4 i i3 i2 i2 i3 i i4 5 i5 i5 i3 i3 i i 5 5 e e e 5e e 10e e 10e e 5e e 2isin e e 5 e e 10 e e 32isin 2isin 5 10isin 3 20isin 15 5sin sin5 sin 3 sin16 16 8                                    55cos sin 2 15 5s i n5 s i n3 s i n16 2 16 2 8 2 15 53 5sin 5 sin 3 sin16 2 16 2 8 2 15 5cos5 cos3 cos16 16 8                                    (c)   ii 2i 3 i ii i sin sin 2 sin 3 sin Im e e e e ee 1 Im e1 xxx N x xN x x xxx N x             Line of symmetry, l

 ii ii 22 2 ii i ii i22 2 (1 )i 2 (1 )i 2 ee e eee 1 e1 ee e e2 i s i n 2 2isin 2 es i n 2 sin 2 (1 ) (1 )cosec cos isin sin22 2 2 Nx Nx Nx x xN x x xx x Nx Nx Nx x Nx x x Nx Nx N x               sin sin 2 sin3 sin 11cosec sin sin (shown)22 2 xxx N x NNxx x             8(i)   2 1cross-sectional area 2 1 2 0 2(0.88 0.24 0.04)2 1o r 202 ( 20 . 8 80 . 2 40 . 0 4 )2 4.16 m          (ii) The builder will have enough cement as trapezium rule over-estimates the cross-sectional area since the curve is concave upwards. (iii) Note: Simpson’s rule applies for an even number of strips.  2 1cross-sectional area 2 1 2 0 2(0.24) 4(0.88 0.04)3 4.05 m (3sf)      (iv) 21p( ) ( 5)8xx (v) Method I: Shell method Method II: Disc method 5 2 1 5 2 1 Volume = (1) (2) 2 d 122 ( 5 ) d 8 39.794 xyx x xx         The builder needs to make at least 40 m3 of cement. 9(i) 2 2 0 0 0 cos 2 d 1 (cos 4 1) d2 1s i n 4 24 2 I                 Volume = 2 2 0 dx y 2 2 0 58d 39.794 yy    

(ii) 0 cos (2 ) dn nI     1 0 12 00 22 0 22 0 2 cos(2 )cos (2 ) d sin(2 ) sin(2 )cos (2 ) ( 1)cos (2 ) 2sin(2 ) d22 0( 1 ) s i n ( 2 ) c o s ( 2 ) d (1 )[ 1 c o s ( 2 ) ] c o s ( 2 ) d (1 ) (1 ) n nn n n nn n n n nI nI                                   2 2 (1 1) ( 1) 1 (shown) nn nn nI nI nII n       (iii) For odd integers n, nI will be reduced to 1.I 1 0 0 1cos 2 d sin 2 0 2I     0nI - independent of n (iv) 2 1 nn nI In   1 2 2 4 6 2 (2 ) (4 ) 2 (2 ) (4 ) 2 2 13 2 135 24 135 3 24 4 (1 ) (3 ) (5 ) 3 . . 1 (2 ) (4 ) 4 2 ! 2 21 2 2 . 2 21 2 12222 n n n n nn Inn nn n Inn n nn n Inn n nn n nn n n nn n nn n nn n nn n                                     2 ! (shown) 2! 2 n n n   10(i) 1 (1 )nnM pM k   (ii)    2 2 2 12 1 1 1 1 1 (1 ) (1 ) (1 ) (1 ) 1 (1 ) (1 ) (1 ) 1 1( 1 )(1 ) 500 500 (1 ) nn n nn n n n n Mp p M k k pM k p pM k p p ppk p kk ppp                          

(iii) 5 6 80 80500 (1 ) 750 From GC, 0.0495803 1 0.95042 Mp pp p p       Hence the club needs to retain about 95% of its members each month. (6v) 5 6 500 (1 0.10) 7500.10 0.10 kkM    From GC, least value of k = 112. 11 (a) 2 (angle at centre = 2 angle at circum)63QCR      33 6sin 3 232 2QR QC      13 2cos 3 232 2CR QC     Thus 32 363, 3 22Q  So, the complex number is 32 363i 3 22     Alternatively, 23 2c o s 3 66PQ  39 2cos 3 662 2PR PQ     13 6sin 3 662 2QR PQ    k M6 Im C (–3, 3) O Re 32 3 , 3P  Q R 5 6 6  (i) (ii)

Thus 92 36 32 3632 3 , 3 , i . e . 3 , 322 2 2QQ             So, the complex number is 32 363i 3 22     (b)      33 arg 042 i 3 arg 3 arg 2i 04 3 arg 3 042 arg 342 z z z z                  From the diagram, 2233 3 2AC   Greatest 3i 3 2 5zA B   Least 3i 3zA N  C (3,0) 8 –2 Re Im 4  4 –4 A (0,3) N B