JPJC 2024 J2 Math Prelim P2 Solutions for students

2024 JC2 H2 Maths Prelim Paper 2 Solutions 1((a) (b) At point of intersection where 0 < x < a, 2 2 2 2 2 2 2 ( ) 3 3 3 0 3 9 4 2 x a x x a x x x a ax            23 9 4 2 ax    (since x > 0) 2(a) sin cos dpx qx x 1 2sin cos d2 px qx x     1 sin sin d2 p q x p q x x       1 1 1 cos cos2 p q x p q x Cp q p q           Alternatively, sin cos dpx qx x 1 2 cos sin d2 qx px x     1 sin sin d2 q p x q p x x       1 1 1 cos cos2 q p x q p x Cq p q p           (b) 2 1 1sin d cos cos d 1 1 cos sin x mx x x mx mx xm m x mx mx Cm m        d sind d cos1d vu x mx x u mx vx m     2 2y x a  3y x y x aa 2a

2 (c) 0 2 0 2 2 sin d 1 1 cos sin 1 1 1 1cos sin (0) cos 0 sin(0) x mx x x mx mxm m m mm m m m                            1 cos mm  When m is odd, 0 1 1sin d ( 1)x mx x m m      When m is even, 0 1 1sin d (1)x mx x m m     k = 1 or – 1 3(i) 1tan 2 2 1f '( ) e 1 (1 )f '( ) f ( ) xx x x x x         Alternatively, 1tan 1 f ( ) e ln f ( ) tan xx x x     Differentiate with respect to x       2 2 1 1f 'f ( ) 1 (1 )f ' f xx x x x x     (ii) 2(1 )f '( ) f ( )x x x  Differentiate with respect to x 2(1 )f ''( ) 2 f '( ) f '( )x x x x x   Differentiate with respect to x 2 2 (1 )f '''( ) 2 f ''( ) 2 f ''( ) 2f '( ) f ''( ) (1 )f '''( ) 4 f ''( ) 2f '( ) f ''( ) x x x x x x x x x x x x x x          When x = 0, 1tan 0f (0) e 1y     (1 0)f '( ) 1 f '( ) 1x x    (1 0)f ''( ) 0 1 f ''( ) 1x x      (1 0)f '''( ) 0 2 1 1 f '''( ) 1x x      1tan 2 3 1 1f ( ) e 1 ... 2! 3! xy x x x x         2 31 11 2 6x x x    (iii) 0.5 0.5 2 3 0 0 1 1f ( ) d 1 d 0.6432 2 6x x x x x x      (4s.f.) (iv) The series is a good approximation for f( x) if x is close to 0. Since x =1 is not close to 0, it is not suitable to be used to estimate 4e  .

3 4(i)           7 1 5 2 2 2 17 5 2 2 1 2 1 n n n n n n n n n n n n            =    2 2 27 7 5 10 2 6 4 2 1 n n n n n n n n n         Obtain    9 4 2 1 n n n n    (Shown) Alternative : By partial fractions Let    9 4 2 1 n n n n    = 2 1 A B C n n n   9 4n =       1 2 2 1A n n B n n C n n      Subst n = 2 , obtain A = 7 Subst n = 1 , obtain B =  5 Subst n = 0, obtain C =2 Obtain    9 4 2 1 n n n n    (Shown) (ii)   3 9 4 2 1 N n n n n n    =    3 7 5 2 2 1 N n n n n        = 7 5 2 1 2 3 7 5 2 2 3 4 7 5 2 3 4 5 7 5 2 4 5 6 ......... 7 5 2 4 3 2 7 5 2 3 2 1 7 5 2 2 1 N N N N N N N N N                                                      = 7 28 1N N      (iii)   3 9 4 2 1 N n n n n n    = 7