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= −+ 393 ... 2 xx= −+ 23 sin 3 (sin 3 ) (sin 3 )f ( ) e 1 (sin 3 ) ...2! 3! x xxxx= = ++ + + 33 3 239 19 191 (3 ...) (3 ...) (3 ...) ...2 22 62 xx xxx x= + −+ + −++ −++ 3 2391 11 3 [(3 ) ] [(3 ) ] ...22 6 xx xx= +− + + + 23913 0 2xx x≈+ + + (independent of x3) Alternative (by differentiation) sin 3 sin 3 3 3 2 2 2 2 e d 3cos3 e 3cos3d dd 3cos3 9s let d in 3dd d dd 3cos3 9sin 3 9sin 3 27 cos3dddd x xy x xyx yy xx xx y yy x y y y xyxxx xxx = ⋅⋅ ⋅ ⋅− = = = − = − ⋅− When 0,x= 32 2 3 dd d1, 3, 9, 0dd d yyyy xx x= = = = 23913 0 2xxy x++ +∴= +… (b) sin 3 21 2 e9 d ( 3 ) d 2 x xx x x x −−≈ ++∫∫ 1 93ln 2x x xC−= −+ ++ where C is an arbitrary constant sin 30.2 0.22 sin 3 20.1 0.1 1 0.2 0.1 2 4e( )e d d 94[ 3 ln ] 2 30.1178 (4 d.p.) x x xxx x x xx− = = −+ + = ∫∫ (c) Using GC, 0.2 2 sin 3 0.1 2( ) e d 29.9995 (4 d.p.)x xx =∫ (d) The approximation is accurate as the values of x (between 0.1 and 0.2) are close to 0 for the magnitude of x4 and higher powers of x to be neglected. Alternative:

% error | 30.1178 29.9995 | 2 1009 0.3943%.9995 ×=−= Since percentage error is small, approximation is accurate.

Qn Suggested Solutions 7(a) (b) At ( )6, 8 , 2.t = 2 2 133 22 When 2, dd 2 and 3dd d d d d 3 tt t t xy tttt y x y x =×= = = = = Equation of tangent is ( )83 6yx−= − i.e., 3 10yx= − (c) Since tangent to C at the point ( )6, 8 meets the curve C again at point P, ( ) 23 3 2 10tt = +− ( )( ) ( )( ) ( )( )( ) 32 2 3 40 Using GC, 2 or 1 Point 6,8 2 20 2 2 10 2 or 1 tt tt t tt ttt tt − += = =− − −− = −−+ = = =− Alt At 1t =− , 3 and 1.xy= =− The coordinates of point P are ( )3, 1 .− (d) At ,tm= the normal to the curve is ( ) 3 3 22 23 224i.e., 3 33 ym xm m myx m mm −= − −− = − ++ + When 0,x= 324 33 mym m=++ (Point R) When 0,y= 4 23 22 mxm + += (Point Q) x y 2

The mid-point F is 42 332 ,.4 2 33 2 1mm m m m  + ++ + Qn Suggested Solutions 8(a) ( ) ( )( ) ( ) 1 1 3 ( 2) 3 ( 2) 3 1 ( 1) 63 6 36 13 63 63 6 constant n n nn nn S nn uSS nn n n n uu n n nn − − = + = − = +− − + = + − = +− −+ = +− + = Since the difference between two consecutive terms is a constant, the series is an arithmetic progression. The common difference is 6. (b) ( ) ( ) 12 27 6 2 3 15 6 7 3 45 45common ratio, 3 15 vu vu r = = += = = += = = ( ) ( ) 2 3 th 15 3 135 The term of the series in (i), 135 6 3 135 3 226 v m m m = = = + −= = Since 3r = does not lie within 1 1,r−< < the sum to infinity of nv does not exist. (c) ( ) ( ) ( ) ( ) 1 5 ( 1) 5 1 ( 1) 5 ( 1) 1 ( 1)5 ( 1) 1 ( 1) 1 common ratio e e ee ee e e n n nx x n xx nx x n xx n xx n xx xx w w − ++ +− + + −+ +−− + + = = = = = For the series to converge, ( 1)e1xx + < , ( 1) 0xx +< The range of values of x is 10 x−< < .

(d) Sum of first n terms of nv , nvS ( ) ( ) 15 3 1 15 3131 2 n n− = = −− nvnSw> using 0.5,x=− ( ) ( )5 0.5 (0.5)15 31e2 nn +−−> From the graph, the least value of n is 3. Alternative (table method) x y 2.28