YIJC 9758 2023 Prelim P2 Solution

1 ©YIJC 9758/02/Prelim Exam/2023 2023 JC2 H2MA Preliminary Examination P2 Solutions Solution 1a 41 0.754h −= = From the diagram, Total area of shaded rectangles < Area under the curve from 1x= to 4x= [ ] 4 1 f (1 ) Area of n nh h A = ∴+<∑ b [ ] 3 0 f (1 ) n nh h = +∑ or [ ] 4 1 f (0.25 ) n nh h = +∑ or [ ] 4 1 f (4 ) n nh h = −∑ or [ ] 4 1 f (1 ( 1) ) n n hh = +−∑ c By GC, [ ] [ ] 4 1 4 1 Lower Bound 0.75f (1 0.75 ) 0.75( 2ln(1 0.75 1) 7) 13.0 n n n n = = = + = − + ++ = ∑ ∑ y x 1 4 O h

2 ©YIJC 9758/02/Prelim Exam/2023 [ ] [ ] 3 0 3 0 Upper Bound 0.75f (1 0.75 ) 0.75( 2ln(1 0.75 1) 7) 14.4 n n n n = = = + = − + ++ = ∑ ∑ d (For students’ understanding only) Since translation by b units in the positive y-direction will result in an increase in the area by 4 rectangles (represented by c) with each length 0.75 units and breadth b units, 4 0.75 3 cb cb ∴=× = . Alternatively, [ ] [ ] [ ] [ ] 44 11 44 11 4 1 g(1 ) f (1 ) f (1 ) f (1 ) 4 nn nn n nh h nh b h nh h bh nh h bh = = = = = + = ++ = ++ = ++ ∑∑ ∑∑ ∑ 4 (0.75) 3 cb cb ∴= = 2a Method 1 3i 3 3 9 2 3− = += ( ) 1 32 πarg 3i 3 π tan 33 − −= − =  1i 11 2−= += ( ) 1 1 πarg 1 i tan 14 − −= − = −  y x 1 4 O b 0.75

3 ©YIJC 9758/02/Prelim Exam/2023 23 6 2 z = = 2π π 11πarg( ) 3 4 12z = −− = Method 2 (NOT recommended) ( )( ) ( )( ) ( ) ( ) 3i 3 1i 3i 3 1 i 1i1i 3 3 3 3i 2 z −= − −+ = −+ −−+− = ( ) ( ) 22 33 3 3 22 3963 3963 44 6 z   −− −  = +     ++ +−= + = ( ) ( ) 1 33 11πarg( ) π tan 1233 z −  −= −=  + b ( )( ) ( )( ) ( ) ( ) 2 3i 3 i 3i 3 i ii 3 3 3 3i 1 z k k kk kk k −= − −+ = −+ − −+ − = + ( )Im 0zz z ∗= ⇒= 2 3 1 3 0k k− + = 3 3k =

4 ©YIJC 9758/02/Prelim Exam/2023 3a ( )( ) ( ) ( ) ( ) ( ) ( ) 2 22 2 2 11ln 1 ln 11ln ln 1 ln 1 ln ln 1 2 ln l hnn o1 sw r rr rr r r rr r rr  −−=   −+=   = −+ +− = −− + + b ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 1ln 1 ln 1 2ln ln 1 ln1 2ln2 ln3 ln2 2ln3 ln4 ln3 2ln4 ln5 ln4 2ln5 ln6 ln 3 2ln 2 ln 1 ln 2 2ln 1 ln ln 1 2ln ln 1 ln2 ln ln 1 1ln 2 n r n n r S r r rr n nn n nn n nn nn n n = = = −  = −− + + = −+ +− + +− + +− + + + −− −+ − + −− −+ + −− + + = −−+ + +=  ∑ ∑  c 1 11ln ln2 22 n nS nn + = = +   As n→∞ , 1 2 0n → . Hence, 11 1lim ln ln22 2n S n ∞ →∞ = +=  . Considering 0.05nSS ∞−≤ , using the GC, n nSS ∞− 19 0.0513 (> 0.05) 20 0.0488 (≤0.05) 21 0.0465 (≤0.05) Smallest n = 20. 4ai 0vu×=  • Either 0v= • Or 0 (and given 0)vu≠≠  , v and u are parallel (accept , where 0v ku k= ≠ ) Combining both cases, , where v ku k= ∈ 

5 ©YIJC 9758/02/Prelim Exam/2023 aii 1 2, 2 11 11 22 3144 22 1 1Alternative answer : 2 3 2 v n n  = −−    = −=−  ++  −−   = −−−    bi ( ) 2243 3