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