ACJC EJC NJC RVHS 9649 2024 Prelim P1 Solutions

2024 H2 FM 9649 Prelim Paper 1 (ACJC_EJC_NJC_RVHS) Solution P1 Q1 (a) A is a 3 4 matrix. Hence rank( ) nullity( ) 4+=AA . As 3rank( ) A , 1nullity( ) A . As the dimension of the null space is at least 1, it is not the zero vector space. (b) 21 1 1 2 1 2 2 1 3 3 3 3 2 2 33 2 3 4 8 1 1 1 3 1 1 1 3 2 3 4 8 3 6 9 15 3 6 9 15 1 1 1 3 1 0 1 1 0 1 2 2 0 1 2 2 0 3 6 6 0 0 0 0 RR R R R R R R R R R R R R  → − → − → + → − −−       = − ⎯⎯⎯→ −       −−    − − −       ⎯⎯⎯⎯⎯ → − ⎯⎯⎯⎯⎯ → −       −    A 1 0 1 1 0 0 1 2 2 2 2 0 0 0 0 0 0 0 x x z wy y z wz w − − − −            − = + − =                   11 22 10 01 x y zwz w            −     =+                  11 22ker( ) , , 10 01 T a b a b        −   = +          

2024 H2 FM 9649 Prelim Paper 1 (ACJC_EJC_NJC_RVHS) Solution P1 Q2 (a) Using GC, 3 2 1 e sin d 100.96xI x x== (2 d.p) (b) Using Simpson’s Rule,  0 1 2 1 43I h y y y= + + ( )1 3 1 6.217676 4 49.645957 56.93187532 −=  + +   87.24446008 87.24== (2 d.p.) Percentage error 100.96 87.245 100% 13.6%100.96 −  = (3 s.f.) (c) From diagram, we observe that there is a large discrepancy in terms of area between the actual curve 2e sinxyx= and the approximated quadratic curve for the interval 23 x , which is caused by the existence of both a turning point and a point of inflexion . Hence Simpson’s Rule using 3 ordinates will not provide a good approximation in this case. Increasing the number of intervals/ordinates will give a better approximation. x O y 1 2 3 (2.68, 94.7) (3, 56.9) (2, 49.6) (1, 6.22)

2024 H2 FM 9649 Prelim Paper 1 (ACJC_EJC_NJC_RVHS) Solution P1 Q3 (a) From the GC, k 1 2 … 50 51 52 21kM − 1 2 … 2.0288 2.0288 2.0288 2kM 2 2.25 … 4.1686 4.1686 4.1686 Since 2 4.1686kM → and 21 2.0288kM − → as k→ , The sequence  nM alternates between the 2 values of 2.0288 and 4.1686 as n→ . (b) Observe that n 1 2 3 … nM 2 2 2.25 … So the sequence in part (b) is the same as moving the terms in the sequence in part (a) up by 1 term, and it will have the same alternating behaviour but odd terms will approach 4.1686 and even terms will approach 2.0288 as n→ . (c) For a constant sequence, 12n n nM M M ++= = = for all n + . Then we have 2 2 1 2 1 2 2 22 9 4 nn n nn MMM MM    + + +  = +  = +  =

2024 H2 FM 9649 Prelim Paper 1 (ACJC_EJC_NJC_RVHS) Solution P1 Q4 (a) f ( ) 8sin 2 5 2 xxx = − + . f (4.5) 2.2246 0= and f (5) 0.21222 0=−  Since f (4.5) f (5) 0 and f is continuous in the interval  4,5 , the equation f ( ) 0x = has a root  in the interval  4.5,5 . 4.5 5   (Shown) (b) Note that 54sin 22 xx =+  is equivalent to 8sin 2 5 02 x x − + = . Let 5g( ) 4sin 22 xx =+  , then g'( ) 2cos 2 xx =  Since 4.5 5 , we have 1.6023 2cos 1.25632 −  −  g'( ) 1 The sequence defined by the recurrence relation 1 54sin 22 n n xx + =+  will not converge to  . Hence it is not suitable to find a good approximation to  . Alternative: Let ( )g( ) 4sin 0.5 2.5xx=+ From the sketch, the iterative method will spiral outwards away from the intersection. Hence it is not suitable to find a good approximation to  . (c) Newton-Raphson iterative formula: 1 8sin 2 52 4cos 2 2 n n nn n x x xx x +  −+=−  − Starting with 0 4.5x = and using GC, we have y = x y = g(x)

2024 H2 FM 9649 Prelim Paper 1 (ACJC_EJC_NJC_RVHS) 0x 4.5 1x 4.99296 2x 4.95916 3x 4.95903 4x 4.95903 Check: f (4.9585) 0.00273 0= and f (4.9595) 0.00243 0=−  Hence 4.959 = (3 d.p.) (d) 1f '( ) 4cos 2 0 cos2 2 2 xxx    = − =  =       2π 2.09443x =  is the x-coordinates of a stationary point. Since 0 2x = is very close to the stationary point, the tangent to C at 2x= is almost a horizontal line and thus will cut the x-axis far away from the initial approximation. So 0 2x = is not a suitable starting value.

2024 H2 FM 9649 Prelim Paper 1 (ACJC_EJC_NJC_RVHS) Solution P1 Q5 (a) ( ) ( )( ) ( )( ) 4 4 2 4 2 2 4 2 22 2 2 4 2 2 2 (1 ) 22 2 2 cos sin 2cos sin cos sin 2 cos (shown) x y x y y x x y y x y x y x y r r r r       + = − + + = += = == Alternative method ( )( ) ( ) ( ) ( )( ) 4 4 2 4 4 4 4 2 2 24 2 2 2 2 3 2 2 2 2 2 2 2 (1 ) cos sin 2 cos sin (1 sin ) sin cos 2sin cos 2 cos sin (1 sin ) 1 2sin cos 2cos sin 2 cos sin 2cos sin cos sin 2 cos (shown) x y x y y r r r r r r r r rr r                       + = − + = −  + − = − − = − = = (b) Line of symmetry is π 2 = or 0x= . (c) d 2cos2 cos sin 2 sind r     =− Length of curve C 2 π 2 0 d dd rr  =+  ( ) ( ) π 222 0 sin 2 cos 2cos2 cos sin 2 sin d      = + − 3.58= units (3 s.f.)

2024 H2 FM 9649 Prelim Paper 1 (ACJC_EJC_NJC_RVHS) Solution P1 Q6 (a) 1 111 1 1 1 11 1 1 1 d d d d 1 d 1 d dd p( ) q( ) p( ) q( )d 1 d 1dDividing both sides of the DE by , p( ) q( )1d d (1 )p( ) (1 )q( ) (shown)d n nn n n n n nn n n n n y u u u uu y y u x n x n x y u u x y x y x u x ux n x uu x u x nx u n x u n xx −− − − −− − − −=  =  = = −− + =  + = − +=− + − = − Alternative: 1 dd (1 )dd nn uyu y n y xx −−=  = − 1 d p( ) q( )d Multiplying (1 ) to both sides of the DE, d(1 ) p( )(1 ) (1 )q( )d d (1 )p( ) (1 )q( ) (shown)d n n nn y x y x yx ny yn y x n y n xx u n x u n xx − −− += − − + − = − + − = − (b) 22d 1 d 1 0, ( )d 2 d 2 yy xy xy x y x yxx − + =  + − = −  . Substituting p( ) ,xx=− 1q( ) 2xx=− and 2n= in part (a), d1 ( 1)( ) ( 1)d2 d d2 u x u xx ux uxx + − − = − −  += Integrating factor 2 d 2ee xxx== 2 2 2 2 2 2 2 1 12 22 2 2 2 2 2 2 2 22 de e ed2 1e e d e 22 1 e2 1 1 2 , where 2 . 1 e 1 e2 x x x x x x x xx ux xux xu x C uC y u A C u CA − − −− += = = + =+ = = = = = ++   Sub 0, 0.1:xy== 20.1 1 20 191 AAA=  + =  =+ . 2 2 2 1 19e x y − = + .

2024 H2 FM 9649 Prelim Paper 1 (ACJC_EJC_NJC_RVHS) When 0.1,x= 20.1 2 2 0.100476 0.1005 (to 4 d.p.) 1 19e y − = = = + When 0.2,x= 20.2 2 2 0.101917 0.1019 (to 4 d.p.) 1 19e y − = = = + (c) ( )1 0.1 0 0.1ln1.1 0.1y = + = ( )( )1 0.10.1 0 ln1. 071 0.1ln1.12 0.1 047y = + + = When 0.1005 (to 4 d.p.)0.1, xy= . ( )2 0.1 0.1ln1 0.1014340.100477 .100477y = + = ( )2 0.100477 0.1004 10.1 0.1ln 0.2ln1.101434 0.102 7 17 92y = + + = When 0.2, 0.1019 (to 4 d.p.)xy= (d) The first two terms of the Maclaurin series of ( )ln 1 y+ is 21 ,2yy− which is close in value to ( )ln 1 y+ for small values of y. Thus, the values found in part (c) are close to the values found in part (b).

2024 H2 FM 9649 Prelim Paper 1 (ACJC_EJC_NJC_RVHS) Solution P1 Q7 (a) ( ) ( ) 2 2 22 22 d1 1,d 20 100 100 1 dd 20100 1 1 1 d100 20 1 1 10 1ln 0 1020 10 20 10 20ln , where 20 .10 xx xt xt xx x t cxx x t c xxx xt A A c xx =−   = − + = +− +− + = +   − += − + =− −     Sub 0t = , 2x= : 12 200 ln 82 310 ln . 2 A A = − + =− Thus, ( ) ( ) 10 20 3ln 10 ln10 2 2 10 20ln 10.3 10 xt xx xt xx += − + − −  += − + − (b) The point of inflexion corresponds to the instant when the rate of infection is the highest. (c) Possible answers: • Recovery rate of infected individuals • Rate of infected individuals leaving the city • Fatality rate of infected individuals x t O x = 10 (8.53, 7.07) (0, 2) ( ) ( ) 2 10 20ln 103 10 xt xx  += − + −

2024 H2 FM 9649 Prelim Paper 1 (ACJC_EJC_NJC_RVHS) Solution P1 Q8 (a) The curve is symmetrical about the x-axis. When 0y= , πsin 2 0 0, 2tt=  =  Observe that π 2t=− and π 2t = correspo