HCI 9758 2023 Prelim P2 Solution

2023 HCI C2 H2 Mathematics Prelim P2 Solutions 1. Solution ( ) ( ) 23 0 3 2 2 0 3 2 0 3 0 3 1π cos dcos 1π cos 2 dcos 1π sec cos 2 1 2 d2 15π tan sin 242 π1 2 5π tan sin3 4 3 23 5π3π3 68 93 5 ππ unit86 AV xxx xxx xxx x xx π π π π ππ = +  = ++  = + ++ = ++    = ++       = ++ = + ⌠⌡ ⌠⌡ ⌠⌡ 2 a. Solution 32 d13 d uux x x= −⇒ = − 53 2 33 33 31 22 31 22 1d ( 3 )d311 11 d3 11 d3 22 39 22(1 ) (1 )39 xx x xx xx u u u uu u u uC x xC = −− −− −=− = −− = −++ = − −+ −+ ∫∫ ∫ ∫ O 2

2023 HCI C2 H2 Mathematics Prelim P2 Solutions with Comments © HCI 2023 2b. Solution Let 3ux= and 5 3 ' 1 xv x = − 2'3ux∴= and v= 33 31 2222(1 ) (1 )39 xx−− +− 8 3 5 3 3 d 1 d 1 x x x xxx x − =  − ∫ ∫ ( ) ( ) 33 3 23 3 31 22 31 22 22(1 ) (1 )39 223 (1 ) (1 ) d39 xx x x x xx =−− +− − −− +−∫ 33 33 23 23 33 33 33 31 22 31 22 31 22 35 22 22 (1 ) (1 )39 22 ( 3 )(1 ) d ( 3 )(1 ) d39 22 (1 ) (1 )39 44(1 ) (1 )9 45 xx xx x xx x xx xx xx x xC = − −+ − − −− + −− = − −+ − − −+−+ ∫∫ 3ai. Solution 11arg( 2 ) arg( 2) arg( ) π zz θ − = −+ = + 3aii. Solution 1 2 ( 3) i 2 ( 3) i za aa a aa − = +− − = −+ − 1arg( 2 ) arg[ ( 3) i] = π π( ) π za aa θ θ θ − = −+ − −+ = −+− = −− 3b. Solution Re Im O

3 © HCI 2023 DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN 12 ( ( 3)i)(1 3i) ( 3 9) (3 3)i ( 2 9) (4 3)i zz a a a a aa aa =+− + = − + + +− = −++ − 12Im ( ) 4 3zz a∴= − 2 22 2 22 2 12 1 2 2 ( ( 3) )(1 3 ) 10(2 6 9) zz z z a a aa = = +− + = −+ 2 12 12 2 2 2 2 2 10Im ( ) 10(2 6 9) 1043 2 69 143 2 69 1043 2 10 12 043 2( 5 6) 043 2( 2)( 3) 043 zz zz aa a aa a aa a aa a aa a aa a ∴≤ −+ ≤− −+ ≤− −+ −≤− −+ ≤− −+ ≤− −− ≤− 3 4a∴< or 23 a≤≤ Since 1 π arg 02 z−< < , 1Im ( ) 3 0za∴ =−< Also, it is given that 0a> Hence 03 a<< ∴ required range of values of a is 30 4a<< or 23 a≤< 4a. Solution When 0x= , 0t = and 1y= When 0y= , 1t =− and 1x= (0,1) (1,0) x y O C

2023 HCI C2 H2 Mathematics Prelim P2 Solutions with Comments © HCI 2023 4b. Solution d 2d x t t= , 2d 3d y tt = 2d d d d d 33 d 22 y t x t y t txt∴=== When 1t = , 1x= , 2y= and d 3 d2 y x = ∴ equation of tangent at P is 32 ( 1)2 31 ...(*)22 yx yx −= − = + 4c. Solution Substitute 2xt= and 3 1yt= + into (*):: 32 32 32 2 311 22 2 23 1 2 3 10 ( 1)(2 1) 0 ( 1)(2 1)( 1) 0 tt tt tt t tt t tt += + += + − += − −− = − + −= 1 or 1 (reject since this is ) 2tt P∴= − = When 1 2t =− , 17 and 48xy= = Hence coordinates of Q is 17,48  . Alternatively: Solving 31 22yx= + and 3 2 1yx= ±+ since tx= ± Notice that the tangent meets at Q at 3 2 1yx= −+ . 3 231 122xx+= −+ 3 2312 2xx+= − + 324 9 6 10xxx− + −= ( )( ) 214 51 0xx− −+ = ( ) ( ) 2 1410xx− −= 11 (rej) or 4xx= =

5 © HCI 2023 DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN 3 217 148y = − += or 31 1 7 24 2 8y = += Hence coordinates of Q is 17,48  . 4d. Solution 2 1 22 1 4 15 2 7 8 1 2 1 2 1 2 11 9Area 1 d24 8 45 3d64 45 3d64 45 364 5 45 3 1 164 5 32 45 99 64 160 27 unit320 xy tt t tt t − − − = +− = − = − = −   = − −−   = − = ∫ ∫ ∫ Some alternative methods: ( ) 22 37 8 11 7 12 1 d24 8 yy  + −− −     ∫ OR 13311 422 100 4 311d 1d d22xx xx x x+− − +− +∫∫ ∫ OR ( )( ) 11 3 11 24 3112 d d 22t tt x x − + −+∫∫ (Full credit is awarded for this method only if the graph in part (a) is accurate) (0,1) (1,0) x y O C

2023 HCI C2 H2 Mathematics Prelim P2 Solutions with Comments © HCI 2023 5a. Solution 2 1 32 25 13 4 s s s +     ×− = −   −−   5b. Solution xz-plane 0y⇒= 2 3 ...(1) 3 1 ...(2) xz xz += − += 2 (2) (1) :×− 55 1zz=⇒= 2x∴= − 2 0 1 OA → − ∴=  Hence coordinates of A is ( 2,0,1 )− . 5c. Solution 1 2 :. 3 1 prs   =− In standard form, 1 2 22 2 22 2 21 13: . ( 3) 21 21 5 1 pr s s ss  − = −=++ ++ + ∴ shortest distance between O and 1p is 2 2 2 2 2 2 33 65 33 65 6 15 6 15 65 1 1 (reject 1 since 0) s s s s s s s ss − = + = + =+ =+ = + = = ±= < 1s∴= −

7 © HCI 2023 DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN 5d. Solution Given 0 0 1 OB k →  = =  Method 1: Using angles 1 2 :. 2 3 1 pr  −= − ( )0, 0, 3Q − is a point that lies in 1p . Since 1 30OQ⋅ = −<n   , then the angle between OQ and 1n is obtuse. 1 02 0. 2 40 41 BQ    ⋅ = − = −< −  n   , then the angle between BQ and 1n is obtuse. ∴ O and B are on same side of 1p . ( )1−−− 2 1 :. 2 1 3 pr  −= ( )1 ,0 ,0T is a point that lies in 2p . Since 2 10OT⋅= >n   , then the angle between OT and 2n is acute. 2 11 0 2 20 13 BT      ⋅ = ⋅− = −<  −  n   , then the angle between BT and 2n is obtuse. ∴ O and B are on opposite side of 2p . ( )2−−− Combining (1) and (2) with reference to given diagram, B∴ lies in 2R . Method 2: (using shortest distance of planes to origin O) Let 3p be plane containing B and parallel to 1p . Equation of 3p is 2 02 .2 0 .2 1 1 11 r        −= −=       ∴ equation of 3p in standard form is 3 2 22 2 22 21 11:. 2 032 ( 2) 1 2 ( 2) 11 pr  −= =>+− + +− + O B Q O B T

2023 HCI C2 H2 Mathematics Prelim P2 Solutions with Comments © HCI 2023 Since equation of 1p in standard form is 1 213: . 2 1033 1 pr  −− = = −< , 1p∴ and 3p are on opposite sides of origin O. …(1) Let 4p be plane containing B and parallel to 2p . Equation of 4p is 1 01 .2 0 .2 3 3 13 r        −= −=       ∴ equation of 4p in standard form is 4 2 22 2 22 11 33:. 2 0 141 ( 2) 3 1 ( 2) 33 pr  −= = >+− + +− + Since equation of 2p in standard form is 2 111:. 2 0 14 143 pr  −= > , 2p∴ and 4p are on same side of origin O. …(2) Combining (1) and (2) with reference to given diagram, B∴ lies in 2R . 6. Solution Group the two first prize awardees as one object, and the two second prize awardees as another object. 1st Prize 2nd Prize GOH S S S S There are 7 units to be arranged in a row. Within the unit of 1 st Prize and the unit of 2 nd Prize, the respective prize winners may arrange themselves. Number of ways required = 7!2!2! 20160= Solution For this arrangement, it means the guest of honour’s seat opposite is an empty seat. Thus the 8 students sit at the 8 remaining chairs. Number of seating arrangement = 8! = 40320 7. Solution Let X be the number of students who study H2 Biology out of 40 students.

9 © HCI 2023 DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN ~ 40, 100 pXB   ( ) ( ) ( ) P 9 20 0.25 P 20 P 8 0.25 X XX ≤≤ = ≤ − ≤=