2023 VJC Promo (Soln)
Uploaded by cy717 · 26 November 2024
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2023 VJC H2 Math Promo Solutions 1 (a) 1 12 2 2 2 2 2 2 2 d1 1tan 1 1 2d2 11 1 11 1 1 1 x xxx x x x x xx (b) cos cos ln cos ln 1d 1 cos ln sind dc o s sin lnd x x yx yx x y x xxyx x yx x xxxx
2 (1) replaced by ff yyyx y x (2) 1 replaced by 2 1ff 2 yy yx y x (b) f1yx . (1) replaced by 1ff 1 xxyx yx (2) replaced by f1 f 1 xxyx y x
3 2 2 2 2 11 21 01 2 01 2 01 17 24 01 ax a xa x ax a x a x xa x ax a x a x ax xa x xx xa x x xa x Since 2 17 7 024 4x for all x , 10xa x . 1 or xx a For 2 11 ax a xa x , replace x with x in the result above to get the following. 1x or xa 11 x or xa or xa : or 1 1 or xx a x x a
4 tan 2 tand es e c ed x xxx 4t a n 22 t a n 2t a n t a n 2t a n t a n 2 2 tan tan tan 2 tan 2 tan tan tan 2 sec e d sec sec e d sec e e 2sec sec tan d sec e 2 e sec tan d sec e 2 e tan e sec d es e c 2 et a n2 e e sec 2 tan 2 x x xx xx xx x xx x x xx xxx xx x x x xx x x x xx x xx C xx C
5 Replace with 22 2 2 2 2Replace with 222 2 44 4 42 4 yy xx xy x y xy x y (1) Scale the graph of C by a factor of 2, parallel to the y-axis, followed by (2) translating the resultant graph by 2 units in the positive x-direction.
6 22 22 21 dd 54 54 d 54 92 2 3 n 24 2 1 25 4 s i 42 1 3 xx xx xx x xx x xx xC xx x 43 4 22 3 3422 23 d 63 33 d 3 d 3322 99 9132 81 2 922 2 xxx x x x xx xx 3 22 22 2 1 d 1 x x x dsin cos d xx uu u π π π ππ π 23 2 4 π 23 4 π 23 4 ππ 33 44 π 3 π 4 1s i n 1s i n 1s i n d 1s i n d 1 d d cos d coscos ππcos cos 0cos 4 3 1c o s 2 3c o s 2 22 3s i n 2 24 3 ππ 12 ππsin sin23 4 4 3 2 π 31 884 u u uu u uu u u uu u uu uu uu uu
7 Since all the coefficients are real, 12 i is also a root. 2 2 Quadratic factor = 1 2 i 1 2 i 12 1 23 xx x xx Let 32 233 2 3x px qx x x ax b . By inspection or by comparing coefficient of 3x and constant term, 3a and 1b . 32 233 2 3 3 1xp xq x x x x By comparing coefficients, 16 5 29 7 p q 32 233 2 3 3 1 0xp xq x x x x Other than 12 i , the other roots are 12 i and 1 3 . Alternative 32 3 1 2i 1 2i 1 2i 3 0 3 132 i 6 22 i 1 22 i 2 1 2 i 3 0 12 3 2 2i 0 pq pq pq pq
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