DHS 9758 2023 Prelim P1 Solution

1 DHS 2023 Year 6 H2 Mathematics Preliminary Examination DHS 2023 Year 6 H2 Mathematics Prelim Paper 1 Solutions and Comments Qn Suggested Solution 1 2 0 2 3 2 d a xx x−−∫ 2 22 02 (2 3 2) d (2 3 2) d a xx x xx x=− −− + −−∫∫ 0 32 32 22 23 23 2232 32 a x xx x xx = −−+ −−   3214 2 3 14()( 2 )332 3 a aa= + − −+ 322 3 28 232 3a aa= − −+ 321(4 9 12 56)6 aa a= −−+ Qn Suggested Solutions 2(a) 2 33 1 211 xxyx xx −+= =−+−− Asymptotes: y = 2 – x , x = 1 x =1 y =2-x y x (0,3) (2,-1) O

s (b) ( ) ( ) ( ) 2 2 2 33 1 33 1 1 3 30 xx kxx x x kx x kx kx −+ =− − += − + − + += For two points of intersection, discriminant > 0. ( ) ( )( ) ( ) 2 2 2 2 3 41 3 0 9 6 12 12 0 6 30 3 12 0 323o r 323 kk kk k kk k kk +−+ > ++−− > − −> − −> <− >+ 2 33 1 211 xxyx xx −+= =−+−− Consider the oblique asymptote of the curve C is 2yx= − , for two points of intersection between the curve and the line, the set of values of k is { }: 3 23 o r 3 23 , 1 .kk k k∈ <− >+ ≠ − Qn Suggested Solution 3(a) By Conjugate Root Theorem, another root is 1iza= − . (b) Let 3 2 ( [1 i])( [1 i])( )z z k z az az c− += −+ −− − where c is a real constant. 3 2 ([ 1] i)([ 1] i)( ) z zk z a z a zc −+ = −− −+ − 22 22 ([ 1] [ i] )( ) ( 2 [1 ])( ) z a zc z z a zc =−− − = − ++ − Comparing the coefficients of z2: 20c−−= ⇒ 2c=− Comparing the coefficients of z: 2 12 2ac++ = − ⇒ 1a= since 0a> So, 2(1 ) 4kca= −+ = (c) Area 1 (2)(3) 32= = square units

Qn Suggested Solution 4(a) ( )( ) 2 2 2 isin 2 1 2isin 2 1 2i sin 2 1 2i sin 2 2 isin 2 1 2isin 2 1 4sin 2 2 2sin 2 5isin 2 1 4sin 2 αα αα αα α αα α −− ×+− −−= + −−= + Since the expression is real, 2 5i sin 2 01 4sin 2 5i sin 2 0 sin 2 0 2 π, π 2 π| 2 kk k k α α α α α α αα − =+ −= = = ∈ = ∴∈ =   (b) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) * * 2* ** ** * * 3 1 3 1 3 1 3 3= =3 3 wz zw wz zw wz ww www zw wz ww z wz w wz wz wz wzwz − − −= − −= = =− −= − − − − −= −− =   Alternative 1 Let iezr θ= , iewr φ= ,

( ) ii * ii )i i( i( i( i i( ) ) )* 3 311 (1 )3 1 13 (1 ) 3 ee ee ee e ee e rwz z r r r r r w φθ θφ φ θφ θφ θφ φ θφ − − − − − − − −=−− −= − −= − = Alternative 2 ( ) ( ) ( ) * * ** * 2* * * 1 3 1 1 3 1,1 1 3 1 3 so wz zw w zw ww wzw w zw w zw w − − − = = =−  − − = = = 

Qn Suggested Solution 5(a) Since A, B and C are collinear BC  → // AB  → = −ba ( )33BC µ  → −= = −b a ba 3µ∴= ( )3 43 OC OB BC= + = +− = − b 3b a ba       (b) Given OQ tOC=     Using RT: ( ) ( ) 1 1 1 1 31 1 4 131 OQ OAOP t OC t tt λ λ λλ λλ λλλ += + = ++ = −+ + = +−  + a 4b a a ba        Since OP   // b 13 0 1 3 t t λ λ −= ∴= (c) When 5λ = [ ]1 14 241 153 9OP t λλ = = = ++  b bb   27: : 2:7 99OP PB∴= = A C B O Q P t 3

Qn Suggested Solution 6(a) 3(3 )sin 3 (3 ) ... 3! xxx= −+