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 r