2023 TJC NYJC VJC FM Prelim Paper 2 (Suggested Solution)

1 2023 TJC H2 FM Prelim Paper 2 (Suggested Solution) Question 1 (a) 2d ( 1)d y yx  Let 2f ( , ) ( 1)x y y Using the Euler Method with step size 0.5, nx ny d d y x 0 1.6 6.76 0.5 1.6 0.5 6.76 1.78    0.6084 1 1.78 0.5 0.6084 2.0842   Using Euler method, 2.8042 2.80y (3 s.f.) when 1x (b) The particular solution is 15 1 13xy   As seen in the diagram below, the tangent to the curve at 0.5x is steep which results in the first approximation to be already exceed the asymptotic value of y. (c) Using the improved Euler method, 0 0x  , 0 1.6y  , 0.5h , 2f ( , ) ( 1)x y y 0 0 0 0 0 0 10 f ( , ) f ( , f ( , )) 2 6.76 0.60841.6 0.5 2 0.2421 x y x h y h x yy y h         1 1 1 1 1 1 21 f ( , ) f ( , f ( , )) 2 x y x h y h x yy y h     0.441 (3 s.f.) The accuracy of the estimate can be improved by reducing the step size.

2 Question 2 2 1 1 1 1 (dividing throughout by ( 1)) 1 nn nn nn u n au n b n n uu a b n n nn v av b             1 1 11 1 1 1 1 1 ( l a)(i) 2 General solution, (2 ) Note that Thus 0 2 2 1000 1 1, (10000 )2 Particu ar solution, (1000 )2 (10000 )2 (1 0 2 0000 )2 n n nn n n nn n n vuv B b A b v b b u bb n u b b n v v b v A B v A B u A B B B b                              (a)(ii) This will eventually go to zero w hen 10000. The population of virus will first increase to a maximum value and then decrease to zero. b 1( 22b) nnvv  19998(2 ) 2n nun  Using GC: 1000000 0 when 7 i.e. after 1 week nun   Question 3 (i) (a) Characteristic equation: 0ab cd     2 ( )( ) 0 ( ) 0 a d bc a d ad bc            2( ) 4( ) 2 a d a d ad bc      2( ) 4 4(1 )(1 ) 2 a d a d ad d a       2( ) 4(1 ) 2 a d a d a d      2( ) 4( ) 4 2 a d a d a d     

3 2( 2) 2 a d a d    ( 2) 2 a d a d    2( ) 2 2 or 22 a d a d a d      1 or 1 (shown)ad   Or 1 ,1 a b a d c d a d            M Characteristic equation: 0MI  1 01 ad ad          1 1 0a d d a          2 10a d ad a d ad          2 10a d a d          1 1 0 ad     1 or 1 (shown)ad   1: a b x x c d y y               ( 1) 0 ( 1) ( 1) ( 1)( 1) 0 1// 1 a x by by a x d y a xcx d y xd ya                         1: ( 1)a b x xa d a d c d y y                   (1 ) 0 ( 1)(1 ) 0 1// 1 by d x by d x by bxcx a y x y                      The eigenvectors corresponding to 1 and 111 are and respectively.11 dad a           

4 Question 4 (a) a = 2, b + c = 6 Equation of top ellipse: 2 2 2 2 2 2 2 1 4 12 y x y xbb       Equation of bottom ellipse: 2 2 2 2 2 2 2 1 4 12 y x y xcc       Volume of capsule 220 22 0 4 1 d 4 1 d b c yy yycb               033 22 0 33 22 4 33 4 33 b c yyyy cb cbcb cb                              2( )4 3 bc   24 (6)3  16 (b) 2 2 2 2 2 2 2 2 2 2 11(2 ) 4 4 y x y y x a aa a a        d 2 d2 d 4 d 4 x y x yx y y x    2 22 2 2 2 2 2 2 2 2 16 4d 16 16 31 d 16 16 16 yay x x y a y y x x x        Surface area, S = 2 220 0 0 22 222 d 16 32 1 d 2 d 16 3 dd 16 2c a a x a yx y x y a y yyx            Consider 0 22 2 16 3 d a a y y  

5 00 0 2 2 2 2 22222 20 22 222 2 2 20 2 222 200 2 2 2 2222 116 3 d 16 3 ( 6 ) d 2 16 3 32 16 3(4 ) d 16 3 (16 3 ) 164d 16 3 164 16 3 d d 16 3 aaa a a aa a y y y a y y y y ay ya a a y ay a y aay ay aa a y y y ay                            020 2 2 2 1 2 2 220 2 2 2 1 2 2 16 32 16 3 d 4 sin 43 8 3 816 3 d 2 sin 2 2333 a a a aya y y a a aaa y y a a                            Hence surface area 2 2 22 22 822 33 4 cm 33 aa aa      Alternative method of finding 0 22 2 16 3 d a a y y   using substitution 3 3 3 3 0 22 2 0 22 2 0 2 2 0 02 2 2 2 16 3 d 416 (1 sin ) cos d 3 16 cos d 3 8 (cos2 1) d 3 8 sin 2 23 83 433 82 33 a a y y aa a a a a aa                                  Let 4 sin 3 ay  d4 cosd 3 ya   When 2, 3ya   When 0, 0y 

6 Question 5 (a) dd dd diffu y uy xy u x yx x x      22 22 22 22 d d d d d d d d d 1 d 2 d d d d diff y y y ux x x x x y u y x x x x x        2 2 2 2 2 2 2 2 2 3 1 d 2 1 d dd 1 d 2 d 2 dd 1 d 2 d 2 dd uu yx x x x x u u u x x x x x x u u u x x x x x          Substituting the above into the given DE gives 2 2 2 3 1 d 2 d 2 2 1 d 25 0d d d u u u u u u x x x x x x x x x x          2 2 d 25 0d u ux   AE: 2 25 0 5kk    GS:   5 5 5 5 1e e e ex x x xu A B y A B x      . (b)                 2 2 sin 2 cos 2 d 2 cos 2 sin 2 2 sin 2 cos 2d 2 sin 2 2 cos 2 d 2 2 cos 2 2 sin 2d 2 2 sin 2 2 cos 2 4 sin 2 4 cos 2 y px x qx x y p x x x q x x xx p qx x px q x y p qx x q xx px q x p x px q x p qx x                       Substitute the above into the given DE gives       4 sin 2 4 cos 2 4 sin 2 cos 2 sin 2 4 sin 2 4 cos 2 sin 2 4 1, 4 0 1 ,04 px q x p qx x px x qx x x q x p x x qp qp                 PI is 1 cos 24y x x

7 AE: 2 4 0 2ikk    CF: cos 2 sin 2y A x B x GS: 1cos 2 sin 2 cos 2 4y A x B x x x   Substitute , x n n  into above GS gives 1cos 2 sin 2 cos 2 4 4 for large 44 y A n B n n n nyA yA nnn              Question 6 0H : gender of customers and colour preference of Jpad pro tablet are independent. 1H : gender of customers and colour preference of Jpad pro tablet are not independent. Level of significance: 5% Computation: Under 0H , Expected frequencies, ij ij rc E N   (N = grand total) are shown below: Colour preference White Black Grey Total Gender Male 10.8 21.6 21.6 54 Female 19.2 38.4 38.4 96 Total 30 60 60 150 Test statistic:   2 2 ij ij ij ij E E O    Calculation of contribution to test statistic: Colour preference White Black Grey Gender Male   2 0.8 10.8    2 20.6 21.6 k   2 21.4 21.6 k Female   2 0.8 19.2   2 20.6 38.4 k   2 21.4 38.4 k

8 Therefore         2 2 2 2 2 38.4 20.6 21.6 20.6 38.4 21.4 21.6 21.45 54 829.44 829.44 k k k k             22 60 20.6 60 21.45 54 829.44 829.44 kk       225 125 20.6 21.454 1728 kk         225 125 20.6 21.454 1728 kk     Degrees of freedom:   2 1 3 1 2     Rejection region: 2 5.991  Given that there is no association at 5% level between gender and colour preference of Jpad pro tablet,  0H is not rejected at 5% level.     225 125 20.6 21.4 5.99154 1728 kk      Using GC, 14.6 27.3k Therefore, possible values of k are 15 27k where k When 30k      222 5 125 30 20.6 30 21.454 1728        2 2 11.8344 ) 8 11.8344value P 0.0026926216 0.00269 (3.s8 fp       value 0.0