HCI 9758 2023 Prelim P1 Solution

2023 HCI C2 H2 Mathematics Prelim P1 Solutions 1 Solution 2 2 2 d 2ed xy x −= 2d ed xy Cx −= −+ 21e2 xy Cx D−= ++ Hence a possible solution curve with an oblique asymptote is 21e12 xyx −= ++ , where 1C = , 1D= Oblique asymptote is 1yx= + 2 Solution By sine rule, ππ 33sin ( ) sin PQ PR θ =− 2 1 2 2 22 22 2 2 2 π 3 π 3 π 3 ππ 33 3 2 3 1 2 22 3 2 3 1 22 3 11 23 11 23 11 11 2233 11 1 233 5511 6633 sin sin ( ) sin sin cos cos sin (1 ) [(1 ) ] 1 1 1( ) 1 () () 1 1 where , (shown) PR PQ θ θ θ θθ θ θ θθ θθ θθ θθ θθ θ θθ α β − ∴= − = − ≈ −− = −− = −− = −+ ≈ ++++ ≈+ + + = ++ = =

2 © HCI 2023 3 Solution (a) 24 5xy y+= 23dd2 40 dd yyxy x y xx++ = 23 d2 d 4 y xy x xy −∴= + 3 Solution (b) 23 d 0d 2 04 y x xy xy = −∴= + 0 or 0xy∴= = [reject since points of the form ( ,0)x does not lie on C ] When 0x= , 1 45y=± ∴ required distance 1 42(5 )= 4 Solution 2 22 2 d24 x xxx− − −+∫ 2 22 2 d24 x xxx− −= −+∫ 22 2222 1 22 1 dd2 24 24 x xxxx xx−− −=−+ −+ −+∫∫ 22 2 2 2 1 2 2 2 11ln( 2 4) d2 ( 1) 3 1 11[ln4 ln12] tan2 33 xx xx x − − − − = − −+ + −+   −= − −+   ∫ 11111 1 1ln tan tan ( 3)23 3 33 −−= −+ − −   11 π1πln 32 63 33   = + −−      11 ππln 32 36 3 1 π 11ln 3 where , (shown)22 23 23 pq = ++  = += = x y 2 –2 –1 1 O 1 –1

3 © HCI 2023 5 Solution (a) Method 1: Replace with Replace with Replace with 2 3 f (3 2) f (3 2) f ( 2) f( ) y ya x x x x yx a yx yx yx + + = − +    → = −    → = −    → = 1. Translate in negative y -direction by a units. 2. Scale parallel to x -axis with scale factor 3. 3. Translate in negative x -direction by 2 units. Method 2: Replace with Replace with Replace with 2 3 3 f (3 2) f (3 2) f (3 ) f( ) y ya x x xx yx a yx yx yx + + = − +    → = −  → =    → = 1. Translate in negative y -direction by a units. 2. Translate in negative x -direction by 2 3 units. 3. Scale parallel to x -axis with scale factor 3. 5 Solution (b) Method 1: 1 f( ) 1 Translate in negative -direction by uni ts Scale parallel to -axis with scale facto r 3 Translate in negative -direction by 2 un its Transform f ( ) to ( ,0) (, ) (3 , ) (3 2, ) (3 2, x a ya x x y xy b ba ba ba b = = ↓ − ↓ − ↓ −− ↓ −− ) 32cb∴= − , 1d a=− Method 2: 2 3 2 3 1 f( ) Translate in negative -direction by uni ts Translate in negative -direction by uni ts Scale parallel to -axis with scale facto r 3 Transform f( ) to ( ,0) (, ) ( ,) (3 2, ) (3 2, x ya x x y xy b ba ba ba b = = ↓ − ↓ −− ↓ −− ↓ − 1)a−

4 © HCI 2023 5 Solution 32cb∴= − , 1d a=− 6 Solution (a) 1 11 (2 1)(2 3) 2(2 1) 2(2 3)rr r r = −++ + + 1 1 1 (2 1)(2 3) 111 2 2 12 3 11 1 23 5 11 57 11 79 11 2 12 3 11 1 23 2 3 11 64 6 n r n r rr rr nn n n = = ++ = − ++ = − +− +− +− ++  = − + = − + ∑ ∑  6 Solution (b) 1 11() 3 (2 1)(2 3) r r n rr= −+ ++∑ 11 11() 3 (2 1)(2 3) n r rr n rr= = =−+ ++∑∑ 1 1() 3 r n r= −∑ is a GP with 1st term 1 3− and common ratio 1 3− . 1 11 33 1 3 () [ 1 () ]1 11() [ 1 () ]3 431( ) n rn r n = − −−− = = − −−−−∑ As n→∞ , 1() 03 n−→ and 1 11() 34 r r n = ∴ − →−∑ Hence it is a convergent series. From (a), as n→∞ , 1 046n →+ and 1 1 11 1 (2 1)(2 3) 6 4 6 6 n r rr n= ∴ = −→++ +∑ Hence it is also a convergent series.

5 © HCI 2023 6 Solution 1 11() 3 (2 1)(2 3) r n r rr= ∴ −+ ++∑ is a sum of 2 convergent series, and hence is also convergent. 1 1 3 1 3 11lim ( ) 34 1( ) r n n r→∞ = −−= = − −−∑ 1 11lim (2 1)(2 3) 6 n n r rr→∞ = =++∑ , ∴ required sum to infinity 11 1 4 6 12= −+= − 7 Solution (a) 7 Solution (b) From sketch in (a), f( ) f( )xx−= − 2a O 1 –1 y x 4a 6a 7a -2a -a 3a a 5a -3a

6 © HCI 2023 7 Solution (c) 4 2 34 23 30 2 30 2 03 2 ππ 42 ππ 24 3π 2 f ( ) d f ( ) d f ( ) d f ( ) d f ( ) d f ( ) d 0 0 f ( ) d ππsin d tan d24 ln sec ( )cos ( ) 2 cos ( )π a a aa a a a aa a a aa a aa a a a aa aa xx xx xx xx xx xx xx xx xxaa xx a − − −− − − − = +++ =+++  = +   − = +   = −+ ∫ ∫∫∫∫ ∫∫ ∫∫ [ ] [ ] π 4 cos (π) 4 ln sec (0) ln sec( )π 24 1 ln 2ππ 2 (1 2 ln 2 )π 2 (1 ln 2) where 2, 2 (shown)π a aa a a km  + −− = −+ − −= + −= + = −= 8 Solution (a) Method 1: 20 . 1. 0 0 0 OA OB α → →      = −=     ∴ OA and OB are perpendicular. 90AOB∴=  8 Solution (b) 1sin 2θ = 30θ∴=  or 150θ =  Method 1: (using right-angle OAB ) Since OA OB⊥ , OAB∴  is a right-angle  . Hence OBA and OAB must be acute. 30θ∴=  Method 2: (using dot product .AB OB → → ) 02 2 0 11 0 AB OB OA αα → → → −     = − = −− =     Since 2 20 . 1.0 0AB OB α αα → → −     = = >     , and . cosAB OB AB OB θ → → → → =

7 © HCI 2023 8 Solution cos 0 is acuteθθ∴ >⇒ Hence 30θ =  8 Solution (c) 22 22 22 2 2 2 22 2 . cos30 3(5 ) ( ) 2 2 53 2 15 3 2 15 3 2 15 3 4 15 3 15 15 (reject 15 since 0) 15 AB OB AB OB α αα αα α αα α αα α αα αα α αα α → → → → = = + = + = + = + = + ⇒= + = =±+< ∴= −  8 Solution (d) cba OC OB OA AB → → → → = − =−=  OC AB → → ∴  () CB OB OC OB AB OB OB OA OA → → → → → → → → → = − = − = −− = CB OA → → ∴  Since OC AB → →  and CB OA → →  , ∴ OABC is a parallelogram. B A O B A O C

8 © HCI 2023 8 Solution 2 2 22 11 0 15 15 2 15 0 15 2 (15) 0 75 5 3 unit OA OC OA AB → → → → ×=×  − =−×  −   =  = ++ = = 9 Solution (a) Let f( )yx= . From graph, since any horizontal line yk= , k∈  cuts f( )yx= at most once, ∴ f is 11− and hence 1f− exists. (shown) 9 Solution (b) 2 22 2 2 2 2 ln ( 1) 2 ( 1) e e1 e 1 (reject e 1 since 0) y y y y yx x x xx − − − − = ++ += ∴= ± − = − = −− ≥ 1 2 2f () e 1 x x− − ∴= − 1 ff [2, )DR− = = ∞ 9 Solution (c) 1 ff [0, )RD− = = ∞ 1 gg \{ 1}DR− = = − Since 11fgRD−− ⊂ , 11gf−−∴ exists. y x O 1 1

9 © HCI 2023 9 Solution Let 1 1 xy x −= + 1 ( 1) 1 1 1 xy y x xy y yx y += − += − −∴= + 1 1g() 1 xx x − −∴= + Using 1f [0, )R− = ∞ as restricted domain of 1g− , 1g ( 1,1]R − = − Hence 11gf ( 1,1]R −− = − 9 Solution (d) 11 1gf () 2x−− = 11 1gg f ( ) g( ) 2x−− = 1 1 2 1 2 1 1f () 31x− −∴= = + 1 2 1ff ( ) f( ) 3 1 16ln( 1) 2 ln 239 x x − = ⇒ = + += + 10 Solution (a) 1 (1 3 i)4z= + 221 3i 1 ( 3) 4 2+= + = = 1 3 πarg (1 3 i) tan 13 −+= = ππ 33ii11(2e ) e42z∴= = 10 Solution (b) ( ) 2 2πi2 3 π 3i11ee24z = = 1 1

10 © HCI 2023 10 Solution ( ) 3 3i ππ 3i11ee28z = = 10 Solution (c) Area of 2 01 11 π3(1) sin unit2 2 38OP P ∆= =  Area of 2 12 11 1 π3sin unit2 24 33 2OPP  ∆= =    Area of 2 23 11 1 π3sin unit2 4 8 3 128OP P   ∆= =      ∴ Area of (3 1)-polygon+ 2 333 8 32 128 21 3 unit128 =++ = 10 Solution (d) Area of ( 1)-polygonn+ 1 1 1 3 5 21 terms 2 2 11 π 11 1 π 11 1 π( 1 ) () s i n () () s i n () () s i n2 2 22 4 24 8 11 1 π... ( )( )sin2 22 1 11 11 1 1 1 π1 ... sin2 22 44 8 22 11 1 1 1 π... sin2 222 2 1112 21 12 1 2 nn nn n n n nnn n n n − − − = ++ ++ = ⋅+⋅+⋅++ ⋅ = + + ++   −  = −    πsin 11 π 111 sin where ,3 4 34 n n abn       = −= = 