YIJC 9758 2023 Prelim P1 Solution

1 ©YIJC 9758/01/JC2PE/23 2023 JC2 H2MA Prelim Examination Paper 1 (Solutions) 1 ( ) f ( ) ln 3 = ln ln 3 ax xa a xa =  + −+ ( ) ( ) ( ) ln ln 3 ln 3 ln ln 3 A BC y x y xa y xa y a xa =  → = +  → =− +  → = − + Sequence of transformations: A: A translation of 3a units in the negative x-direction B: A reflection in the x-axis C: A translation of ln a units in the positive y-direction OR ( ) ( ) (1) (2) (3) ln ln ln 3 ln ln 3 yx y x y xa y a xa =  → = −  → =− +  → = − + Sequence of transformations: (1): A reflection in the x-axis (2): A translation of 3a units in the negative x-direction (3): A translation of ln a units in the positive y-direction 2a Since 2 3i+ is a root and all the coefficients are real, 2 3i− is also a root. A quadratic factor is: [ ][ ](2 3i) (2 3i)zz−+ −− ( ) ( ) 22 2 3iz= −− ( ) 2 2 4 49 4 13 zz zz = −++ =−+ ( )( ) 32 23 13 4 13 1z z kz z z z− ++= −+ + Comparing coefficient of z: 13 4 9k = −= The other roots are 2 3i and 1.zz= −= − 2b Let i,zw= then we get ( ) ( ) ( ) 32 i 3 i i 13 0w w kw− + += 32 i 3 i 13 0w w kw⇒− + + + = Replace z with iw , i 2 3i, i 2 3i and i 1 2 3i 2 3i 13 2i, 3 2i and iii i ww w ww w = += − = − +−= =− = = −− = −=

2 ©YIJC 9758/01/JC2PE/23 3a ( ) 2 22 22 2 3 8 3 3 ------ (*) 3 83 3 o r 3 833 3 9 6 0 or 3 7 0 3 2 0 or (3 7) 0 3 9 4(1)( 2) 70 or 23 3 17 2 xx x xx x xx x xx xx x x xx xx + −=− +− = − − +−= − + −= + = + −= + = −± − −= = − −±= (Alternative method: Squaring both sides and so on) 3b As seen from the graphs, for 23 8 33xx x+ −≥− 3 17 7 3 17 or 0 or 23 2x xx−− −+≤ −≤≤ ≥ 4a 2 d3 sin 2 2 3 cos 2d d4cos 8cos ( sin ) 4sin 2d d 4sin 2 d 2 3 cos 2 2 = tan 2 3 23 tan 2 3 tan 23 2where . 3 xxt t t yy t tt t t yt x t t tk t k = ⇒= = ⇒= − = − −∴= − = −≡ =− y x

3 ©YIJC 9758/01/JC2PE/23 4b When , 4t π= 23sin2 3, 4cos 2,44xy ππ = = = =   d 23 tan , which is undefined.d 32 y x π=− ⇒The tangent is parallel to the y-axis. Hence the equation of the tangent at the point where 4t π= is 3x= . 2 333sin2 3 , 3 22 When = , 3 4cos 1 3 t x y π π π = = =    = =  ( ) d 23 2 23 tan 3 2d 33 3 y x π= − = − −= OR From GC, 3d, 1, 2.2when = d,3 yxyt x π = = = Equation of the tangent is: 312 2 22 yx yx −= −  = − 4c Let θ be the angle in which the tangent 22yx= − makes with the positive x-axis. Then tan 2θ = . Hence, acute angle between the 2 tangents 1 90 90 tan 2 26.6 θ − = °− = °− ≈° 5a Sum of the all the terms after the nth term ( )1 11 1 n n n araSS rr ar r ∞ − =−= − −− = − Given 2nnSS u∞ −= , therefore 121 2(1 ) 2 3 n nar arr rr r −=− = − =

4 ©YIJC 9758/01/JC2PE/23 Hence 3 (Shown)21 1 3 aaSa r ∞ = = =− − 5bi Total number of integers in the first (r−1)th brackets is ( ) ( )111 2 3 ... ( 1) 1 ( 1) 22 rrrrr −−+ + ++−= +− = Hence, first integer in the rth bracket ( ) 21 2122 rr rr− −+= += Last integer in the rth bracket 2 2 2 2 ( 1)2 22 2 2 2 rr r rr r rr −+= +− −++ −= += Alternative method: Last integer in the rth bracket = First integer in the (r+1)th bracket minus 1 = ( )( ) 21 1122 rr rr + ++ −=  5bii There are r integers in the rth bracket. First integer in the rth bracket = 2 2 2 rr−+ Last integer in the rth bracket = 2 2 rr+ Sum of all the integers in the rth bracket = 22 2 2 22 2 2 2 22 rr r r r r r  −+ + ++=      = ( ) 21 12 rr + (Shown)

5 ©YIJC 9758/01/JC2PE/23 6a Required area ( ) ( ) 2 0 0 0 0 2 1 cos2 1 cos22 1 1 cos d cos2 d d 11 sin224 cos2 2 nits2 2 u x x xx x x x x xx π π π π π = + = = = − −  = + −  −+∫ ∫ ∫ 6b ( ) ( ) ( ) 242 2 2 cos cos 1 cos 2 2 1 1 2cos 2 cos 24 1 1 cos 41 2cos 242 1 3 4cos 2 cos 48 (shown) xx x xx xx xx = += = ++ += ++  = ++ 6c Required volume ( ) ( ) ( ) ( ) ( ) 2 22 00 24 2 0 0 0 2 3 0 2cos 2cos 2 2 111 cos 2 3 4cos 2 cos 4 2cos 2 1 cos 482 317 cos 2828 71 3 sin 1 cos d 1 cos 2 d 1 cos 2 sin 1 cos d 48 4 32 d cos 4 d 7 units8 xx xx x x x xx x x xx xx x xx x x x ππ π π π π π π π π π π =  = +− + ++ + −+ = −−+ = −− + −+ + −− = − = ∫∫ ∫ ∫ ∫

6 ©YIJC 9758/01/JC2PE/23 7ai dd 4d4 dvx vy xxy⇒== −− ( ) ( ) 2 2 d d d d 42 42 v x v x v v −= + = −+ 7aii ( ) 2 42d d v x v= −+ ( ) 2 d1 1d 42 vx v− = + ∫∫ 44 44 1 2 ( 2)ln2(2) 2 ( 2) 4ln 4 4 41 41e e wh, ere e xC xC A v xCv v xCv v Av + ++ = +−+ + = +− = = ± =± + + When 0,x= 2y=− , ( )0 22v∴= −− = . Hence 041e 32 AA+= ⇒= 4 4 4 4 41 3e 4 3e 1 44 3e 1 44 3e 1 x x x x v v xy yx += = − −= − = − − 7bi 2 2 5 22 2 3 2212 d ed d e23 14 4 15 d e x x xy y xC y x Cx D xx x − − −= + = − ++ = + ++ 7bii When x = 0, y = 0. 110 44 DD=+⇒= − When x = 0, d 2d y x = . 2 5 2 1 2 C C+ ⇒==− Particular solution is 5 221 4 51 4 15e 24 xy xx −= + +− .

7 ©YIJC 9758/01/JC2PE/23 8a 22 1f( ) 45 x x ax a = −+ ( ) 222 24f '( ) 45 xax x ax a −= −+ For maximum point, ( ) 222 24 0 45 240 2 xa x ax a xa xa − = −+ −= = Hence largest 2ka= Alternative method: ( ) ( ) ( ) 22 22 2 2 2 45 2 25 2 x ax a xa a a xa a −+ = −− + = −+ Since 2xa= gives the minimum value of 22 45x ax a−+ , it gives the maximum value for 22 1f( ) 45 x x ax a = −+ . Hence largest 2ka= 8b When xa= , 2 22 11f( ) 4() 5 2 a a aa a a = = −+ From the graph, 2 10 f( ) 2 x a << ∴ 2f 10, 2a R = To show 2f exists, ffR D⊆ . Since, 22 1 11 11 222 1 aa aa a ⇒ < ⇒< ⇒<> , ∴ ( )2f f 10, , 2 Da a R = ⊆ = −∞  . Thus 2f exists. (Shown)

8 ©YIJC 9758/01/JC2PE/23 8c Let 22 1 45x ax y a−+ = ( ) 2 22 1 xa y a−+ = ( ) 2 2 12 yxa a−+ = 212 ya ax ±−−= 212a ax y±−= Since xa< , hence 212a ax y−−= Hence 1 2f () 2 1xa ax − −= − 8d 9ai ( ) ( ) ( ) 1 22 3 2 3 23 3 2 3 3 1dd 3 1 3 1 31 2 1 2 1 3 9 xx x x x xx C C x ++ + =  = +   += +  ∫∫ x y y = x O

9 ©YIJC 9758/01/JC2PE/23 9aii ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) 53 32 3 33 2 3 33 2 0 1 0 1 033 0 22 1 1 3 302 2 1 05 1 3 2 3 1 1 213 1 9 211 d d 2 d9 201 d9 20 9 4 1045 4 45 319 1 5 2 xx xx x x xx xx x x xx x x − − − − − − = ⋅ + +  + − ⋅+ −+ − ⋅+  +    = =  =−−  = = − −− − ∫ ∫ ∫ ∫ 9b d d1 de e11 d1 xx ux uxu uu⇒= = − ⇒=+ −= ( ) ( ) ( ) ( ) 2 1 2 1 2 53 22 5 2 3 3 2 1e1 1 1 1 22 53 22 1e 1e53 ed d d d x xx x uu u uu uu x u u u u uC C +− − − − −+ = +− + = = = = + ∫∫ ∫ ∫ 9c ( ) ( ) ( ) 2 tan5 cos5 sin3 d d sin52cos5 sin3 cos5 sin3 d cos5 2 c o s 5s i n 3d s i n 5s i n 3d 1sin8 sin2 d 2sin5 sin3 d2 cos8 cos2 1 cos8 cos2 d8 22 cos8 cos2 1 sin8 sin2 8 2 28 x x xxx xxx xx x x x xx x xx x x x x xx xx x xx x x xx + = + = + = − −− = −+ − − = −+ − − ∫ ∫ ∫∫ ∫∫ ∫ ( ) 2 1 8cos 2 4sin 2 2cos8 sin 816 C x x x xC  + = + − −+

10 ©YIJC 9758/01/JC2PE/23 10a Let A, B and C be the points (6, 9, 3) , ( 2 , 1 3 , 1 )− and (4, 10, 0) . 26 8 4 13 9 4 2 2 132 1 AB −−         = − = = −−         −       46 2 10 9 1 03 3 AC −     = −=     −     4 2 61 5 1 2 1 ( 12 ( 2)) 10 5 2 1 3 44 0 0 n −−         = − × = −− −− = =        −−      1 61 : 2 9 2 24 0 30 rπ    = =     2 24xy+= (shown) 10b Let l represent the path of the laser beam. 20 2 1 32 21 1 d      =− −− =        02 : 3 2 , 11 lr λλ      = −+ ∈        Let θ be the angle between the laser beam and the reflective shield. 1 22 2 22 21 2