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 2nnS