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 += +  ( )  22 1 tan ln 1 212 xx= +  +  + ( ) 22 112 yx=++ Rearranging, ( ) ( ) 2d1 2 2 1 d yxy x+ = + (shown) (b) Differentiating the given result, ( ) ( ) ( ) 2 2 d d d1 2 2 2 2d d d y y yxy x x x+ + = When 0x= , ( )tan ln 1 0 tan 0 0y=  + = = , ( ) ( )dd1 0 2 1 0 2dd yy xx+ = +  = , ( ) ( ) ( )( ) 22 22 dd1 0 2 2 2 0 2 4dd yy xx+ + =  =− , so ( ) ( ) ( ) ( ) 2 24tan ln 1 2 0 2 ... 2 2 ... 2!x x x x x − + = + + + = − + (c) ( ) 2 21 2 2 ... ...22e e e xxy xx −+ −+ == ( ) ( ) 22 2 ... 1 ... ... 2! xx xx −+ = + − + + + 2 .1 21 .. x x= + − + (d) 1 2 00 0 0 2 2 1 ... 3 1e1lim lim33 ... lim 1 2 1 2 11 ...3l 3 im 6 1 y xx x x x x x x xx x x →→ → →  + − + −−  =  −     −+=    =  = + 

Q9 (a) 2 2 d 2d d 22d d d d /d d d x tt y tat y y x t a x t t t = =− −== When ta= , 2 32 2 1 2 23 d 1d xa y a a y a a axa =− =− −= = − Equation of tangent: 3 2 22 2 ( 1)[ ( 1)]3y a a a x a− − = − − − 321( 1) 1 3y x a a aa = − + − − + − (b) From part (a), the tangent at point Q has gradient 1a− . 2 1ta at − =− 2t a at t− = − 2 0t t at a+ − − = ( )( 1) 0t a t− + = 1t=− (since ta= is point P) When 1t=− , ( ) 2 011x= − − = ( ) ( ) 322 2213 1 3y a a= − − − = − The coordinates of point Q are 302 2, a − . (c) When 0y = , 32 203 t at−= ( ) 22 303 t t a−= 0t= or 2 3ta= 1x=− or 31xa=− So we need 3 1 0a− , i.e. 1 3a 

Q10 (a) [0, 8] (b)(i) f (4) f (8) 8 == f is a many-to-one function and 1f− does not exist (b)(ii) 0 2c  (c)(i) [0, 9] (c)(ii) When 1c= : From graph, For 0 1x , f ( ) 2.750 x ; For 1 8x  , 4.5 f ( ) 8x . Hence, the range is   ( 80 4, 2.75 .5, . (c)(iii) For 2f to exist, we need the range of f to be a subset of [0, 8] the domain of f, i.e. ffR D [0,8]= From graph, we need 0 4c  . Q11 (ai) ( ) ( ) ( ) 21 32 43 11 3 2 3 2.6 or 55 1 1 883 2.6 3 3 5 .52 or 55 1 3 13 5 25 4 2 1 3.52 3 3.704 1or 3 55 6 uu uu uu = − = − =− = − − − − − − = − − =− = = − − = (aii) Since sequence converges to l, 1,nnu u l +  when n is large. Solving 35 ll =− , we get 153.75 4orl=− − . (bi) 4 1 n r r u + =  is the sum of n terms of an arithmetic progression with first term ( )5 2 3 4 10u = − =− and common difference 3d=− . ( ) ( )( ) ( ) 4 1 3 172 10 1 322 n r r nnnun+ = + =  − + − − =− . (bii) Since ( )( )1 3 32 5nu n n − − =+ −= , ( )34 53 3 4 7 9nu n n+ − +== −− ( ) ( )373 1 4 3 7 13 65 9nnu u n n +++ +== −−=− So ( ) ( )3 7 3 4 16 9 7 9 9nnu u n n++− − − − −= − =− This is a constant independent of n, so the sequence is an A.P. (biii) 7 100 34 1 10 304r r uu uu+ = + ++= is the sum of 100 terms of an A.P. with first term ( )7 2 3 6 16u = − =− and common difference 9− . ( ) ( ) 100 34 1 100 2 16 99 9 461502 r r u + =  =  − + − =−

Q12 (a) When x = 0, 62 2 3 12 ky k −=− + =− + =− . When y = 0, we have ( )( ) ( ) ( )( ) ( )( ) 2 2 2 2 2 3 2 0 2 2 3 0 20 2 x x k k x x x k k x x k x − − + − = − − + =  − + = = The coordinates are (0, 1) and (2, 0). (b) As x→ , 2yx→− Vertical asymptotes: 2 and 2x k x k= =− Oblique asymptote: 2yx=− (c) ( ) 2 322 2 xyx x −= − + − ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) 2 22 22 22 2 22 2 2 2d 13d 2 3 2 2 4 1 2 3 4 2 1 2 x x xy x x x x x x xx x  − − −=+  − − − + =+ − − + − =+ − At turning points, d 0d y x = . Hence we have ( ) ( ) ( ) 222 4 2 2 42 2 3 4 2 0 4 4 3 12 6 0 7 12 2 0 shown x x x x x x x x x x − + − + − = − + − + − =  − + − = Using GC, x = –3.29 or x = 0.187

(d) Q13 (a) 03 1400 5514        = =−      −    v.n . Thus 0 3 24 4 1 10 2 0 0 5 5 251 4 7              = − − =               −      w (b) 2211 24 7 576 49 125 25= + = + =w so w is a unit vector. (c) 10 5 10 MC OC OM  = − =  So 2 2 2 10 2 11 51 310 5 10 10 2        ==    ++        w (d) Let x y z  = n . Then 0 0 1 x yz z        = =−      −    v.n . , so substituting into (*) we get 1 2

( ) 2 3 1 3 2 3 0 02 1 x zy z          = − −        −    2 3 1 3 2 3 2 2 2 21 xz yz z    =  − Solving, 2 221 3z −= gives 5 6z= . Substituting each value into the other components to solve for x and y, 2 15 1 30 5 6 2 1 1 30 5   ==    n or 2 1 1 30 5  =−  n So 2 1 1 30 5    is a unit normal to plane P. (e) Since P passes through M, the equation of P is 2 10 2 1 5 1 5 0 5 −           =−                r. . 2 1 25 5 x y z         =−          . 2 5 25x y z+ + =− (f) Required angle is 11 20 1 . 0 51 5cos cos 24.1 2 5 30 0 10 1 −−                   = =                    (3 s.f.) (g) The incoming beam of sunlight cannot be para