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 28 1N N      As N  , 7 1N  0 and 2 N 0,   3 9 4 2 1 N n n n n n    8 which is finite. Hence   3 9 4 2 1 N n n n n n    is convergent and the sum to infinity is 8.

4 (iv)   3 9 4 2 1 N n n n n n    = 23 32 41 9 4 .........(1)(2)(3) (2)(3)(4) (3)(4)(5) ( 2)( 1) N N N N        2 9 14 ( 1)( 2) N n n n n n    = 32 41 9 4 ....(2)(3)(4) (3)(4)(5) ( 2)( 1)( ) 9 5 9 14 ( 1)( )( 1) ( 1)( 2) N N N N N N N N N N N N              =    2 4 9 4 2 1 N n n n n n      Note : replace n by n  2 =    2 3 9 4 2 1 N n n n n n      23 (1)(2)(3) = 7 28 2 1 2N N        23 6 = 25 7 2 6 1 2 N N       5(i) 5 0 0 OA         , 4 3 0 OB          , 5 0 6 OC         Since ABCD is a parallelogram,  AD =  BC    OD OA =    OC OB = 5 4 1 0 3 3 6 0 6                                 OD = 1 5 3 0 6 0                  = 4 3 6        

5 5 4 9 0 3 3 0 0 0 9 1 3 3 0 0 6 BA BA BC                                                    i i Since ABCD is a parallelogram and AB BC , ABCD is a rectangle. (ii) 9 1 3 1 3 3 3 3 1 3 6 9 0 6 0 6 5 BA BC                                                         Normal to the plane ABC 3 9 5 n           . 3 9 5 OA         i = 5 3 0 9 15 0 5                   i Vector equation of plane ABC is 3 9 15 5 r         i Cartesian equation of plane ABC is 3 9 5 15x y z   . (iii) Normal vector of the base is parallel to OE  . Hence, normal vector of the base = 0 0 1         Acute angle between plane ABC and the base = 1 3 0 9 0 5 1cos 3 0 9 0 5 1                                    i = 1 2 2 2 5cos 3 9 5 1    = 1 5cos 115  = 62.2 (1 dp)

6 (iv) Since : 1: 5AF AE  , 0 5 1 1 1 0 0 05 5 10 0 2 AF AE                                       Length of projection of AF onto plane ABC 1 3 0 9 2 5 3 9 5 AF n n                                 22 2 18 11 9 3 9 5            526 115 2.1387 2.14    6(a)(i) Since there is only one way the letters are in alphabetical order, Total number of ways = 11! 3!2!2! –1 = 1 663 199 (a)(ii) A A O I E R R T PP P Total number of ways = 5 2 4! 5! P2! 2!  = 14 400 (b) Method 1: Probability = 6 5 4 3! 5 4 3 11 10 9 2! 11 10 9      = 14 33 (or 0.424) Method 2: Probability = 6 5 5 1 2 3 11 3 C C C C  Method 3: Probability = 6 5 6 3 1 2 11 11 3 3 C C C 141 C C 33  

7 7(i) (ii) (a) Correlation coefficient between x and y is r = 0.9483 (4 d.p.) (b) Correlation coefficient between ln x and y is r = 0.9849 (4 d.p.) (iii) From the scatter diagram, as x increases, y increases at a decreasing rate. In addition, the product moment correlation coefficient between ln x and y, 0.9849, is closer to +1 as compared to that between x and y, 0.9483. Hence y = cln x + d is the better model. (iv) Since x is the independent variable, neither the regression line of x on y nor the regression line of ln x on y should be used to estimate the value of x when y = 200. (v) Equation of regression line of y on ln x is y = 106.5611 ln x – 31.2643 y = 107 ln x – 31.3 when y = 200, 200 = 106.5611 ln x – 31.2643 x = 8.76 (3 s.f.) Since y = 200 is within the given range of data, which is an interpolation, and r is close to +1, indicating a strong positive linear correlation, the estimate is reliable. 8(i) P(packet is unsatisfactory) = 1 – (0.99)(0.98)2 (0.96) = 0.087236  0.0872 (3 s.f.) (ii) X ~ B(180, 0.087236) P(5  X < 10) = P(X 9) – P(X 4) = 0.042669  0.0427 (iii) P(X > r)  0.12 1 – P(X r)  0.12 P(X r)  0.88 From GC, P( X 19) = 0.8429 (< 0.88) P(X 20) = 0.8947 (> 0.88) P(X 21) = 0.9323 (> 0.88)  least r = 20 (iv) Each packet has an equal chance of being selected and the selection of the packets is independent of one another. This method of selection is done so that a random sample will be obtained which is free from bias and will be representative of the population. y x 6 18 155 265

8 (v) Required probability =       2 6 8!0.087236 0.912764 0.087236     2!6!   = 0.010749  0.0107 Alternative Method Let Y be the the number of packets (out of 8) that are unsatisfactory. Y ~ B(8, 0.087236) Required probability = P(Y = 2)  (0.087236) = 0.010749  0.0107 9(i) Box A Box B (B) 4  1 3 (R) 5  2 3 (B) 1  1 5 5 5 (R) 2  2 5 8 7 (R) 3  2 5 12 8 P(X = 8) = P(2 from Box A and 4 from