VJC 9758 2024 Promo Solutions

1 2024/VJC/Math Dept 2024 H2 MATH PROMO SOLUTION No. Solution 1 Let x, y and z be the number of dresses, blouses and skirts produced per week respectively. 45 50 20 3150 70 60 25 4300 8634 8 0 xyz xyz xyz    By GC, x = 30, y = 20, z = 40. The shop produces 30 dresses, 20 blouses and 40 skirts each week. No. Solution 2(a)   2 2 22 2 23 1 0 2 23 1 0 2 0 51 0 01 xx xx xx x x xx x xx        2 0 or 1 xx    (b) Replace x in part (a) with ex  2 e 0 or e 1 (Reject since e 0 for all real ) 0 e 2 ln 2 : ln 2 xx x x x xx x             No. Solution 3(a)(i)    ' ,0 and ' 0,Aa B b 3(a)(ii)  '' 2, 0Bb  3(a)(iii) ''' 0, and ''' ,0 3 bAaB   3(b) 1 f( )y x 0 1 – + – + x x=b y y=0 O

2 2024/VJC/Math Dept No. Solution 4(a)(i)  1 1 1 23 1 32 2 21 1 1 1 3 3375 7555 33 55 33 55 3 75 455 330 5 5 nn nn nn nn n nn n n n n SS u                      This is a GP with common ratio r = 3 5 and first term a = 30 30 7531 5 S   OR 21 2 3375 75 2755 nn n nS      As n , 2 3 05 n   . Hence, 75S  4(b) 1 21 121 21 n i i nv nn   As 1, 021n n   . Hence, 1 1i i v    . 10 11 1 1 20 11 21 21 iii ii i vvv      

3 2024/VJC/Math Dept No. Solution 5(a) 223241xx y y Differentiating wrt to x: dd62 28 0 dd dd826 2dd d6 2 d82 yyxx yy xx yyyxx yxx yx y x yx        d3 d4 yx y x yx   (shown) 5(b) Gradient of tangent at (1, 1) = 31 4 41 3   Angle required = 1 4tan 53.13 − ⎛⎞⎜⎟ =°⎜⎟⎝⎠ (0.927 radian) 5(c) For tangent parallel to x-axis, d 0d y x  30 3x yy x  OR 1 3x y Sub into eqn of C: 2232 ( 3 ) 4 ( 3 ) 1xx x x   22 139 1 39xx    (no soln as 2 0x  for all x  ) OR 2 21132 ( ) 4 133yy y y    2213 3 131 3yy    (no soln as 2 0y  for all y  ) Since there is no solution for x (or y) such that d 0d y x  . There is no point on C at which the tangent to curve is parallel to the x-axis.

4 2024/VJC/Math Dept No. Solution 6(a)(i) 1 2 3 2.4 2412 2 2.4 2412 0 2 up u u      u4 will be undefined and we are not able to generate an infinite sequence. Hence, p cannot be 2.4. 6(a)(ii) If the sequence is cons tant, then all terms will be p.  2 2412 12 24 0 12 144 4 24 62 32 p p pp p       Since p > 5, 62 3p . 6(b)    10 19 11 11 1 35 39 51 8 32 759 0 26 3 vv vd v d vd vd vd         1212 642 642 n nSv n d n dn d nd n      Since d < 0 and Sn is a quadratic expression, Sn is largest when n = 64 322  . 32 32 32 64 5122 dSd    Alternative    2 2 642 642 32 10242 n nSd n d d n d n      Since d < 0, Sn is largest when n = 32 . 32 1024 5122 dSd  

5 2024/VJC/Math Dept No. Solution 7(a)(i) 7(a)(ii) 25 3 61 5vw vw   ---------- (1) 3* 1 5 ivw  ------ (2) (1) – (2): 6* 1 4 5 iww Let iwx y 6( i ) ( i ) 14 5ixy xy 75 i 1 4 5 ixy  Comparing real and imaginary parts: 71 4 2 55 1 xx yy      Hence, 2iw Sub 2iw into 25 :vw 52 ( 2i )12 iv  Alternatively 25 5 2vw v w    and sub into eqn (2): 3(5 2 ) * 1 5iww   15 6 * 1 5iww  Let iwxy : 15 6( i) ( i) 1 5ixy xy   75 i 1 4 5 ixy  and compare real and imaginary parts  2, 1W  

6 2024/VJC/Math Dept Transformation: W is rotated anti-clockwise through 2  about O to obtain V OR W is rotated clockwise through 3 2  about O to obtain V 7(b) Since 2222 1234zz zz < 0, the equation has at least one non-real root. In addition, since the coefficients of the polynomial equation are all real, by conjugate root theorem, there is at least 1 pair of complex conjugate roots. Hence, at most two of 123,,zzz and 4z are real. No. Solution 8(a) Since 22tan 1 sec ,  2 2 2 2 2 12 1 i.e. 1 and 22 y x yxa b      8(b) 22 22 22 (1 )(1 )4 4 1 21 xyxy     (1, 0) (-1, 0) (3, 0) y x O 2yx= 2yx=− (1,1) (1,-1)

7 2024/VJC/Math Dept 8(c) 22 22 2 2 (1 )(1 )4 4 1 12 xykx y k       To cut C2 at most twice, 2 2 k  . 1k  8(d) Translate the graph y = f(x) by 1 unit in the negative x-direction. Stretch the resultant graph by factor 0.5 parallel to x-axis, with y-axis invariant. OR Stretch the graph y = f(x) by factor 0.5 parallel to x-axis, with y- axis invariant. Then translate the resultant graph by 0.5 unit in the negative x-direction. No. Solution 9(a)  12 22 2 2 222 22 2 24 24 22 4 2 4 6 11 1 11 21 1 ... 2! 11 1 11 ... OR ... x xa a a xx aa a xx xxaa a a a a                  12 22 2 22 44 2 4 22 4 242 244 4 22 4 24 24 22 4 2 2 22 2 cos cos 1 1 1 ... 1 ...2! 4! 1 1 ... 2! 2! 4! 11 1 11 ... 22 2 4 11 1 1 2 ax x axxa a ax ax x x aa a xxa xxa x aa a aa xxaa a a xaa a                                    4 4 24 44 8 24 24 6 11 ...22 4 1 2 24 12 ...22 4 a xaa aa a xxaa a              4 12 24 12, 2 acc aa      and 48 3 6 24 12 24 aac a ++=

8 2024/VJC/Math Dept 9(b) Expansion is valid for 2 2 1x a    22 22 0 0 xa xa xaxa axa      9(c) 11 24 200 1 24 0 1 35 0 cos 2 1 1 4 1 1 1 1 16 d d44 4 2 4 4 1 6 2 2 4 19 5 9 d41 6 1 9 2 13 5 9 41 6 9 6 0 119 960 x x xx xx xx x xx x                No. Solution 10(a)(i)  tan tan 2d ee s e cd xx xx  10(a)(ii)  tan 3 tan 2e sec sin d e sec tan dxx x xx x xx tan tan 2e tan e sec dxx x xx  tan tanet a n exx x C Side working: 2 tan 2 tan dtan sec d d es e c ed x x uux x x v xvx     10(b) 14 ( 38 )x Ax B  By observation, 184 2AA    3131 1 22AB B    2 2 14 d 14 3 11(3 8 )22 d 14 3 x x xx x x xx        22 13 8 1 1 d d22 14 3 14 3 x x x xx xx   

9 2024/VJC/Math Dept  1 2 2 22 11 1(3 8 ) 1 4 3 d d22 532 88 x xx x x x               21 3 11 821 4 3 s i n 524 8 x x xC          21 18 314 3 s i n 45 xx xC      10(c) 1ux 2 1xu  d1 1 d2 21 u x ux   When x = 3, 13 4 2u   . When x = 1, 11 2u     222 3 3 32 12 1 d 2 d 1 ux x uuux    2 42 2 2 212 d uu uu   2 2 22 122 duu u  23 2 122 3 u u u   81 2 2 224 2 232 3 2         11 11 22 66   11 213 No. Solution 11(a) y x  0.2, 12.5  5, 0.5  0, 13  2.6,0 1x 1x 0y

10 2024/VJC/Math Dept 11(b) fg0R b u t 0D Since fgRD , gf does not exist. 11(c) Every horizontal line y = k cuts the graph of y = g(x) at most once. Hence, g is one-to-one. Inverse of g exists.  1 ln 1 e1 e1 g( ) e 1 y y x yx x x x         11(c) 1 ggRD  =[2, ) Let 1gR  be the new domain of f Using the graph in part (a), 1fg 2R0 , 3 3    Alternatively Sketch the graph   1 2 5e 1 1 3 fg ( )= e11 x x yx       for  1 ggDR , 0    and 1fg 2R0 , 3 3    y x y = g(x) (2, 0)