VJC 2023 JC1 MYE 9649 (Solutions)
Uploaded by fwyr · 14 September 2024
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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
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