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        --- (2) Using GC, 2 1,7 7   2 2 1 27 7 2 3 7 7 OF a b a b                 

5 7(a) 4 3 : 1 5 , 0 0 l r t t                     ---(1) For plane p, 1 3 5 5 4 1 1 15 a n a a                                   Since l and p do not meet in a unique point, 5 3 4 5 0 15 0 a a                      3 5 20 0 3 35 35 3 a a a        (b)(i) Given a = 7, 5 7 3 4 4 1 7 15 2 n                      p: 3 3 3 1 0 1 1 2 4 2 r                                --- (2) Subst. (1) into (2): 4 3 3 1 5 1 1 0 0 2 t                                   13 14 1 1 t t    Position vector of the point of intersection, B = 4 3 1 1 5 4 0 0 0                               1, 4,0B 

6 (ii) To find F, the foot of perpendicular from A to p: 4 3 : 1 1 , 0 2 AFl r s s                     4 3 3 1 1 1 1 0 2 2 13 14 1 1 s s s                                      4 3 1 1 1 0 0 2 2 OF                               Since F is the mid-point of AA’, ' 2 ' 2 1 4 2 2 0 1 1 2 0 4 OA OAOF OA OF OA                                        2 1 3 ' 1 4 3 4 0 4 BA                                   1 3 ': 4 3 , 0 4 1 4 3 3 4 l r x y z                           l l’ p

7 8 (i) Month Amount owed at beginning of the month Amount owed at the end of the month 1 200000(1.005) 200000(1.005) –x 2 200000(1.005)2 –(1.005)x 200000(1.005)2 –(1.005)x –x 3 200000(1.005)3 – (1.005)2x– (1.005)x 200000(1.005)3 – (1.005)2x – (1.005)x – x Amount owed at the end of n months 1 2200000(1.005) (1.005) (1.005) ... 1.005n n n x x x x       2 1200000(1.005) 1 1.005 (1.005) (1.005)n n n x         … 1 1.005 200000(1.005) 1 1.005 200000(1.005) 200 1.005 1 n n n n x x           (ii) 200000(1.005) 200 1.005 1 0n n x      200000(1.005) 200 1500 1.005 1 0 300000 100000(1.005) 0 (1.005) 3 ln 3 ln1.005 220.27 n n n n n n           Alternatively, Use GC table n 200000(1.005) 200(1500) 1.005 1n n     219 1896.19 220 405.67 > 0 221 −1092.30 < 0 n = 221 At the end of 220 months, Selena owed 220 220200000(1.005) 200(1500) 1.005 1 $405.67     Last repayment amount to be repaid on the 221st month = $405.67 1.005  $407.70 (2d.p.) 221 months = 18 years 5 months Full repayment on: 31 May 2043 (iii) 120 120200000(1.005) 200 1.005 1 0 363879.3468 163.8793468 0 2220.410039 x x x         $2220.42 (2d.p.) [$2220.41 not accepted]

8 9(a) Sub. 1 2iz  into 4 3 2 9 0z z z sz t             4 3 2 1 2i 1 2i 9 1 2i 1 2i 0 s t                     7 24i 11 2i 9 3 4i 1 2i 0 31 2 58 i 0 s t s t s                  Comparing imaginary parts, 2 58 0 29 s s    Comparing real parts, 31 0 31 60 s t t s       Now 4 3 2 9 29 60 0z z z z     Using GC the other roots are 1 2i, 3, 4  . Alternative solution Since 4 3 2 9 0z z z sz t     is a polynomial equation with real coefficients and 1 2i is a root, 1 2i is another root. Quadratic factor =    1 2i 1 2iz z                2 2 2 1 2i 1 2i 1 2i 2 5 z z z z z                Let    4 3 2 2 2 9 2 5z z z sz t z z z az b         . By comparing coefficients, 3 : 1 2 1z a a      2: 9 2 5 12z b a b      : 2 5 29 z s b a   constant term : 5 60 t b  Now 4 3 2 9 29 60 0z z z z     Using GC (polyroot finder), the other roots are 1 2i, 3, 4  . (b)     3 3arg arg arg i *i * 3arg arg arg * 3arg arg 2 4arg 2 w w ww w i w w w w                  

9 For 3 i * w w to be purely imaginary, 3 3 5arg , , ,...i * 2 2 2 w w               3 54arg , , ,... 2 2 2 2 4arg 0, , 2 , , 3 , 2 3arg ,0, , , ,4 2 4 4 2 w w w                        Since iw a b  and a and b are positive real numbers,  0 arg 2w   .  arg 4w   1tan 4 b a      1b a b a   iw a a  10(i) 0.1 0.1 d 2ed 2e d t t v t v t     0.1 0.1 e2 0.1 20e t t v c v c          When t = 0, v = 0 00 20e 20 c c    0.120 20e tv   Subt 10v 0.1 0.1 0.1 10 20 20e 20e 10 1e 2 t t t        10.1 ln 2 110ln 10ln 2 (exact)2 t t     (ii) As 0.1,e 0, 20tt v    Eventually, the speed increases and tend to 20 ms-1

10 (iii) d2 ( 3)( 2)d w w wt    1 d 1 ( 3)( 2) d 2 1 1 d d( 3)( 2) 2 w w w t w tw w       Method 1: Partial fractions 1 ( 3)( 2)