JPJC 2024 J2 Math Prelim P1 Solutions for students

2024 JPJC J2 H2 Prelims Paper 1 Solutions 1 3 2Let nv an bn cn d    1 2 3 4 9 : 9 (1) 7 :8 4 2 7 (2) 47 : 27 9 3 47 (3) 141: 64 16 4 141 (4) v a b c d v a b c d v a b c d v a b c d                         3 2 Use GC: 5, 18, 35, 31 5 18 35 31n a b c d v n n n         2 2 11 19 22 8 x x x    2 11 19 2 02 8 x x x     2 2 11 19 2 4 16 02 8 x x x x x       2 2 2 7 3 02 8 x x x x       2 1 ( 3) 0( 4)( 2) x x x x     + 14 or 2 or 32x x x    Replace x by e x 1e 4 or e 2 or e 32 x x x      (no solution) , 1ln ln 2 or ln 32 x x    ln 2 ln 2 or ln 3x x    3 d5sin 5cos d yy     0, 0 5sin 0 5 5 33, 3 5sin sin2 2 2 3 y y                5 3 22 0 5 3 22 0 Volume = d 25 d x y y y      y x O R – 4 3 + + 21 2   +

2   23 0 25 25sin 5cos d       23 0 5 25(1 sin ) cos d       23 0 23 0 25 cos cos d 25 cos d              3 0 25 1 cos 2 d2      3 0 25 sin 2 2 2          2 25 1 2 sin2 3 2 3 25 3 2 3 4 25 25 3 exact6 8                   4(i) (ii)    3f ( ) 2 5 xy x x x         2 3f (2 ) 2 2 2 5 xy x x x            2 32 f (2 ) 2 2 2 2 5 xy x x x        2 3 1 2 5 xy x x    Scaling parallel to the x-axis by scale factor 1 2 . Scaling parallel to the y-axis by scale factor 2. x x = 2 O (0.172,0.299) (3,0) y 5x (5.83,0.0682) (0,0.3) y = 0

3 5(i) 1f  exists for x c k c (ii)     2 2 e ln( ) ln( ) x c y x c y x c y       ln( )x c y  Since , ln( )x c x c y   1f ( ) ln , 1x c x x    (iii) Graphs of f and 1f  are reflections of each other about the line y = x (iv) Range of f = [1,  and Domain of g = (0,  Range of f  Domain of g Hence gf exists gf(x)= ln   2 e x c = 2( )x c , x c 6(i)   ( 1) AB b a AC ma nb a m a nb                Area of triangle ABC     1 2 1 ( 1)2 AB AC b a m a nb              O y x O x y= x (1, c ) y ( c , 1) y=f(x) y= 1f ( ) x y=f( x) ( c ,1)

4             1 ( 1) ( 1)2 1 ( 1) 0 02 1 ( 1)2 m b a n b b m a a n a b m b a n b a m n b a                                        1 ( 1) sin 302 1 1( 1) 42 2 m n b a m n b b                2 ( 1)m n b    (ii) 1 3 2 5OD a OE b      (iii)   1 1 :BDl r b BD b d b                1 1 1 2 2 , where 2 b a b b a b                    1 2a b     (shown) 1:AEl r a AE            1 1 1 3 5 3 5 , where 5 1 5 3 a e a a b a a b a a b                                 Since lines BD and AE meet,  1 2a b     1 5 3 a b    Comparing coefficients of a , 1 5 5 1        --- (1) Comparing coefficients of b , 1 2 3 2 3 1 