SAJC 9758 2023 Prelim P2 Solution

2023 H2 Math Prelims Paper 2 Solutions 1 No. Solutions 1 Substitute i into the equationzkk= + ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 22 22 2 2 2 3 i (5 i) i 0 3 1i ( 5i ) 1i 0 3 1 2i i (5 5i 5i+i ) 0 3 (2i) (4 6i) 0 ----------(*) 4 6 6i0 kk kk kk kk kk k kk α α α α α + −+ + += + − + ++= ++ − ++ += − + += −+ − = Comparing real and imaginary parts: 40kα −= ----(1) ( ) 26 6 0 (2) 6 10 1 or 0 (rejected since is non-zero) kk kk kk k − = −−−− −= = = Substitute 1k = into (1) 4α = Let the other root be w ( )( ) 23 (5 i) 4 3 1 iz z zwz− + + = − −− Comparing constants: ( ) ( ) ( )( ) 4 3( )( 1 ) 4 31 i 41 i 31 i 1 i 22 i33 wi w = − −− = + −= +− = − 2(i) 2 36 6 PQR πππθ π θ −−+  = ∠ − = Using Sine Rule, P Q R 2 3 π 6 π θ+ a

2023 H2 Math Prelims Paper 2 Solutions 2 sin sin3 6 2 PQ a π π θ =  − sin 3 sin cos sin66 2 13cos sin2 2 cos 3 3 2 3 co sis n a PQ a a π ππ θθ θθ θθ − − =   −  = = 2(ii) ( ) 2 12 222 2 2 2 3sin 13 2 13 2 ( 1)( 1) 3 cos 3 3 3 33 1 ( 1) 3 32 2! 2 13 2 713 23 aPQ a a a a a θθ θ θ θθ θθθθ θθθ θθ − − −− −−   −−+ − −−+ −−   ++ = ≈ =   = + = + ++  ≈   ++   2(iii) 2 2 713 2 3s 33 cosf( ) 3 cos inLet 3 sin 713 2 3 sin 3 sin 1 cos 1 cos aa a θθ θθθ θθ θθ θθ θθ ++ − −  −=  −= ++ − − 0f( .) 05θ <∴

2023 H2 Math Prelims Paper 2 Solutions 3 From GC, 0.190( 4 sf )3θ< < 3(a) ( ) ( ) 2 22 2 22 2 2 2 2 2 2 1d d tan , secdt1 1 sec t dt tan tan 1 1 sec t dttan (sec ) sec dttan cos dtsin (cos )(sint) dt 1 , where is an arbitrary constantsin 1 , xx xt t xx tt tt t t t t t CCt x Cx − = = + = + = = = = = −+ += −+ ∫ ∫ ∫ ∫ ∫ ∫ Alternatively, 2 cos dt cosec cot dsin = cosec t t ttt tC = −+ ∫∫ (b)(i) ( ) 3 23d cos 3 sind x xxx =− f( )y θ= 0.05y = 0 0.194 x 1

2023 H2 Math Prelims Paper 2 Solutions 4 (ii) 53 3 23 3 32 3 2 3 3 3 23 3 33 sin d ' sin 1cos cos d ' 3 cos33 1cos 3 cos d33 1cos sin33 x xx ux v x x x x x xx u x v x x x x xx x x xC = = = −+ = = − = −+ = − ++ ∫ ∫ ∫ 4(i) (i) 2 90xy x y+− = Differentiate with respect x, ( ) dd2 2 19 0dd d 29 21d d 21 d 92 yyxy xx y xyx yy xx + +− = −= −− += − (ii) (ii) Let d d yG x= . 21 92G y x + −∴= Diff wrt x, ( ) ( )( ) ( ) 2 d2 92 2d d 12d 92 y xyxG x x = − − +− − When 3x= , 0 39 0 1 29 6yy xy y yx+= +− = ⇒ − ⇒ = d 2(1) 1 1d 9 2(3) y x + −∴ = = Hence, when 3x= ( ) ( ) ( ) 2 d d 2(1) 9 2(3) 2(1) 1 ( 2) 0.02 9 2(3) dd dd 0.022 or sf)67(375 x t GG tx ⋅ − − +−= × − = = Therefore, required rate is 2 75 units/s.

2023 H2 Math Prelims Paper 2 Solutions 5 (or 0.0267 units/s) 5(i) (ii) (iii) ( ) 11L.H.S. = ! ( 1)! 11 ( 1)! = R.H.S. (Shown)( 1)! rr r r r r − + +−= + =+ ( ) 1 1 123 ...2! 3! 4! ( 1)! = (1)( 1)! 11 = (2)! ( 1)! 11 = 1! 2! 11 + 2! 3!