EJC_9758_2024_Promo_Solutions

2024 EJC JC1 Promo Solutions Q1 Let S, C and T be the number of stools, chairs and tables produced respectively. 68 kg of metal is used in all, 4 10 68S C T+ + = . 32 kg of plastic is used in all, 2 2 4 32S C T+ + = . Solving this system of linear equations produces 42 33 52 8 33 ST CT  =− +  =− Since 1S , 42 1 3.533 TT+   − Since 1C , 55 1 6.123 28 3TT − So 4,5 or 6T= . Testing all cases, • When 4T = , S is not integer. • When 5T = , 2S = , 4C= . • When 6T = , S is not integer. So the only solution is 2, 4, 5S C T= = = .  The number of stools, chairs and tables that could be produced is 2, 4, and 5 respectively. Q2 1.5 0.5x−  − or 1x= .

Q3 (a) (b) Q4 Let the length of BC be x cm and angle BAC be  . By cosine rule 2 2 2 5 4 2(5)(4)cosx = + − 2 41 40cosx =− --- (1) When π 3 = , ( )12 241 40 21x = − = , 21x= (since x is positive) Differentiating (1) with respect to time t, ( )0dd2 sd i d4nxx tt −=− Substituting in π 3 = , 21x= , d 0.2dt  =− , ( )d2 40 0.21 d2 32 x t =− − − From GC, 6d 0d .75x t =− (3 s.f.) At that instant, BC is decreasing at a rate of 0.756 cm/s. x y x y

Q5 (a) Required length of projection is 2cos 25π 3 6 62 −==a (b) ( ) ( ) 2 3 2 3 2 3 2+ = + +a b a b . a b ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )3 3 2 3 3 2 2 2= + + +a . a b . a a . b b . b 22 9 12 4= + +a a.b b ( ) ( )( ) ( )9 2 12 2 4 6 6 2 3−= + +  6= So 3 2 6+=ab Q6 (a) ( ) 1d sin cos sind nn n   −= (b) 22cosx = d 4cos sind x  =− ( ) ( ) 22 32 32 22 22 42 1d 2 2cos 1 cos 4cos sin d 8cos sin sin d 8cos sin d 8cos cos sin d 8cos 1 sin sin d 8 cos sin cos sin d (Shown) xxx                          − = − − =− =− =− = − − =−         (c) Using part (a), 53 42 sin sin8 cos sin cos sin d 8 53 C     − = − +   Since 22cosx = ,

22sin 1 cos 1 2 sin 1 2 x x    = − = −  = − 53 22881 d 1 12 5 2 3 2 x x xx x C     − = − − − +         Q7 (ai) If 1r = , all the terms of the sequence will be equal to the first term a. Then the sum of the first 10 terms will be 10a , which is only 2 times the sum of the first 5 terms 5a . (aii) Using sum of G.P. formula, 10 511 3311 rra rr a−− − = − 5101 33 33rr− = − ( )( ) ( ) 5 5 51 1 33 1r r r− + = − ( )( ) ( )( ) 55 55 1 1 33 0 1 32 0 rr rr − + − = − − = 5 1r = (reject since 1r  ) or 5 32r = 5 32 2r== (shown) (aiii) 6 11u  and 7 11u  ( ) 5 12 1a  and ( ) 6 2 11a  32 11a and 64 11a 11 11 64 32a (b) ( )( )4 1 2 3 2 5 n r rr= ++ ( )( ) 2 24 1 2 1 2 1 rn r rr −= −= = −+ (replace r with 2r− ) ( )( ) 2 6 1 2 1 2 1 n r rr + = = −+ ( )( ) ( )( ) 25 11 11 2 1 2 1 2 1 2 1 n rr r r r r + == =− − + − +

1 1 1 1 2 4 10 2 22n    = − − −   +    11 22 4 10n=− + Q8 (a) ( ) 2 2d 1sec lnd 21 12y xx x += +  (