NYJC H2 MATH P2 Answer

2015 NYJC JC2 Prelim Exam 9740/2 Solutions Page 1 of 10 Qn Solution 1 Let nP be the statement     1 sin 2 1 scos 2 i i 2s n nn r nr    for n  When n = 1, LHS   1 1 cos 2 cos 2 r r   RHS sinsin3 2sin    2cos2 cos22sin ins  1P is true. Assume that kP true for some , 1kk    . i.e.     1 sin 2 1 scos 2 i i 2s n nk r kr    For n = k +1, we want to prove    1 1 sinsin 2 3cos 2 2sin k r kr      LHS       1 11 cos 2 cos 2 cos 2 2 kk rr r r k            sin 2 1 cos 2 22sin sink k         sin 2 1 cos 2 sin 2 in in s2 2s kk              sin 2 1 sin 2 3 sin 2 1 2sin sink k k           sin 2 3 2s n in sik    kP is true 1kP is true Since P1 is true and Pk is true  Pk+1 is true, by the principle of Mathematical Induction, Pn is true for all n  +.     11 sinsin 2 1 cos cos 2 2sin ( 1) 122·22 2 nn rr nn rr          

2015 NYJC JC2 Prelim Exam 9740/2 Solutions Page 2 of 10 Qn Solution 2(a) (b) Let y = 21 22 x x   (2 2) 2 1 (2 2) 2 1 21 22 y x x x y y yx y          1 21f : , 1 22 xxx x    f(x) = f –1(x)  f 2(x) = x f 1965 (0.5) = 0 (i), (ii) (iii) For equation 11g g( ) g ( )x g x  , 26 x . x y O y = x g( )yx 1 )g (y x (0,2) (16,6) x (2,0) (6,16) x

2015 NYJC JC2 Prelim Exam 9740/2 Solutions Page 3 of 10 Qn Solution 3(i) (ii) Since 1 52iz   and 1 103iz   , thus 1 4 2iz  satisfies the equation 52iz  and i 03 1z  . (iii) Note that triangle OBC is a right angle triangle. Further B lies on the locus of 52iz  . Thus       2 2 21 1 1 5 10 102 4 2 55 522 5( 1) area              (iv) 1 5sin 5  Thus 2   and 2   . Hence the required range is 2.03 arg( 1.12) 14iz  . A(2,-1) O Re Im C(4,-2) D(2,4)    B(3,1)

2015 NYJC JC2 Prelim Exam 9740/2 Solutions Page 4 of 10 Qn Solution 4 (i) Let 1tanyx  . Then tan yx Differentiate with respect to x, 2 dsec 1 d yy x    2 d1 tan 1 d yy x 2 d1 d1 y xx  (shown) (ii) From graph, area of shaded rectangle < area under curve for 1n x n   .  22 1 11 1d11 n n n n xnx        1 2 1 1 tan1 n n xn        11 2 1 tan tan 11 nnn    (show) (iii) Comparing area of rectangles with area under curve for 0 xn , 2 2 2 2 2 0 1 1 1 1 1 d1 1 1 2 1 3 1 1 n xnx            1 2 0 1 1 1 1 tan2 5 10 1 n xn         1 2 1 1 1 1 tan2 5 10 1 nn       Alternative: Use Method of Difference using (i) result   2 1 13 y x     2 131x y   131x y   (rejected) or 131x y   131x y   Required volume 1 2 2 1 10 dxy 21 2 1 10 13 1 d 2.5998 yy       2.60 units3 or 0.828 

2015 NYJC JC2 Prelim Exam 9740/2 Solutions Page 5 of 10 Qn Solution 5 i) A random sample refers to the sample is obtained by selecting 50 students from 500 students in such a way that each of the 500 students will have an equal chance of being selected. Each selection is independent. ii) To obtain a sample of 50 students usin g stratified random sampling, we will determine the sample size of the method of transport such that the sample size of each method of travel is proportional to the size of each method of travel in the school. Car Public transport On foot Sample Size 30% of 50 (=15 ) 50% of 50 (= 25) 20% of 50 (= 10) Then conduct simple random sampling from each method of travel. These simple random samples from the different method of travel groups are combined to form the overall stratified random sample of 50 students. 6 Let X be the weight of the chicken sold in the supermarket. 2~ N( , )X  P( 1) 0.2X  1P = 0.20 1P = 0.20 X Z         Using GC, 1    =-0.84162 -----(1) 1.8P = 0.15 1.81-P = 0.15 1.8P = 0.85 X Z Z             Using GC, 1.8    =1.0364 ------(2) Solving equation (1) and (2), 1.36 and =0.426 7 (i) The calls may not be independent as people may be calling in to lodge a complaint on bad service rendered by the company. (ii) Since 60 50n and  60 0.08 4.8 5np   ,  ~ Po 4.8C .    P 6 1 P 5 0.349CC     (3 s.f.)

2015 NYJC JC2 Prelim Exam 9740/2 Solutions Page 6 of 10 Qn Solution (iii)  ~ 60,0.08CB Since 80n is large, 4.416~ 4.8, 80CN   approximately by CLT.  5 0.197PC  (3 s.f.) 8 (i) required probability = 45 9 140 28 (ii) P(father born in Asia | mother born in UK) P(father born in Asia mother born in UK ) P(mother born in UK) 8 2140 60 15 140   (iii) P(parent born in either UK or Europe) 1 P( both parents born in Asia) 45 19 1 140 28     OR Let A be the event at least one parent in UK Let B be the event at least one parent in Europe Req d = P( )+P( ) P( ) 80 38 23 19 140 pro 140 140 2 b. 8 A B A B      9(a) Let  be the mean of X. To test 0 1 : 455 : 455 H H     Level of Significance: 100  Perform Z test Calculation: p-value = 0.096875 For sufficient evidence to reject 0H , p – value < 100  . Thus 0.096875100 9.6875     Thus the least value of level of significance is 9.69%.

2015 NYJC JC2 Prelim Exam 9740/2 Solutions Page 7 of 10 Qn Solution 9(b)   22 10 0.569219 0.360s  0 1 : 455 : 455 H H     Level of Significance: 0.05 Perform t test Reject 0H if p-value < 0.05 If 455x    455 2 0.05 0.025 / 10 Xx xT s P P         Since ( 2.2622) 0.025PT  , thus 455 2.2622 / 10 455.4292    x s x If 455x    455 2 0.05 0.025 / 10 Xx xT s P P         Since ( 2.2622) 0.025PT   , thus 455 2.2622 / 10 454.5707    x s x Reject 0H if calc 0.025 (9) 2.2622Tt  Since 0H is rejected, 455 10 455 2.2622 10 or 455 2.2622 10 454.5707 or 4 2 55.4292 .2622 / x sx sx xx s         Thus the required set of values of x is (0,454.57] [455.43, ) .

2015 NYJC JC2 Prelim Exam 9740/2 Solutions Page 8 of 10 Qn Solution 10(i) 10(ii) The scatter diagram shows that S is decreasing at a decreasing rate as t increase. Further, if a linear model is used, there will be a negative amount of sleep after a certain age, which is impossible. Thus a linear model is inappropriate. 10(iii) Since S is decreasing at a decreasing rate as t increase and approaching a value, the proposed model may be appropriate. 10(iv) Using GC, ˆˆ 7.51, 14.8ab 10(v) Since t =50 is out of the given data range, it will be inappropriate to use the model due to extrapolation.

2015 NYJC JC2 Prelim Exam 9740/2 Solutions Page 9 of 10 Qn Solution 11 (i) No of ways = 610 1000000 (ii) No of possible 3 identical digits on the left = 10 No of possible 3 identical digits on the right and is different from the left = 9 Total no of ways = 10 x 9 = 90 (iii)No of possible ways in the other 2 digits = 6 152   Total no of ways = 15 x 6! = 10800 (iv) Case 1: LHS – all 3 identical No of ways = 10 x 9 x 9 x 9 = 7290 Case 2 : LHS – 1pair identical No of ways = 310 3! 2! 8 1382402 2! 