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 larg