NYJC H2 Maths P2 Solution

1 NYJC 2014 Preliminary Examination 9740/02 NANYANG JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATION SOLUTIONS Higher 2 MATHEMATICS 9740/02 Paper 2 17th September 2014 3 Hours Additional Materials: Cover Sheet Answer Paper List of Formulae (MF15) READ THESE INSTRUCTIONS FIRST Write your name and class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.

1 By definition, nl m nl and nm . Therefore, .0ln . Let the foot of perpendicular from C to plane OAB be C’. //CC' ab    CC' t ab  for some scalar t () OC' OC CC' t    ab ca b      Therefore, there is a scalar such that ()t     ab c a b Multiplying both sides by a, () ( )t     aa b a c a b 0    ac aa ba (using result from first part) ()()   aa ab ac  C b O   B  A  C’ a c

2 Since hgR 0, , D  hgRD , gh exists.    ln( 2) gh g ln 2 e gh : 2, , 3 x xx xx x x      ghR1 , fg hx ax a    22xaaxa xa          22 0 21 0 xxa xa xa x       1 or 2x xa  Since ghD3 , , 3xa  3x a , hence range of values of x is  3, a  . Alternative solution:  fg hx x   22x ax x         22 0 21 0 xax x xx a        1 or 2xa x  Replacing by x xa , 1+a or 2xa xa  1 or 2x xa  Since ghD3 , , hence range of values of x is  3, . Range of values of x for (ii) is 3, a  . Maximum point  2 24 ,2 2 a a      1 fy x x O y

3 ln 1 1 xy x   1l n ( 1 ) 21 xy x   2(1 ) ln(1 )x yx dy 12(1 ) 2 dx 1xy x   2 22 dy d y 12(1 ) 4 dx dx (1 )x x    32 32 3 dy dy 22(1 ) 6 dx dx (1 )x x   (i) When 0x , 0y , 1 2 dy dx  , 2 2 3 2 dy dx  , 3 3 11 2 dy dx  23ln 1 1 3 11 0. . .12 2 2 6 2 xx x xx                    2313 1 1 ...24 1 2 xxx  (ii) 1ln 1 1 [ln(1 )](1 )12 x x xx    23 231 [ ...](1 ...)22 3 xxxx x x      233 231 [ ...]22 2 3 xxxxx x      2313 1 1 ...24 1 2 xxx  (iii) 11 23 44 00 13 1 1 d ... d 24 1 2 0.120 xxyx x x         Approximation is not good as 4x  is not close to 0. (iv) 1 2yx

4 (a)(i) The first three triangular numbers are 1, 3, 6. (ii) () () 11 11 11 11 nn rr rr rr tt r tt r -- ++ == -= + Þ - = + åå 21 32 1 nn tt tt tt - Þ- +- +-  = () 1 1 1 n r r - = +å () 1 1 1 1 n n r tt r - = Þ- = + å ()1123 1 2 ntn n n Þ= + + + + = + (iii) Sum of the first N triangular numbers = () 2 11 1 1 11 122 NN N N n nn n n tn n n n == = = æö ÷çéù=+ = + ÷ç ÷ëû ç ÷èøåå å å = () ( ) ()1 12 1 126 2 NNNN Né ùê ú++ + +ê úë û = () ( )126 N NN++ (i) na > 2014  3 1 2014n n . n 31n n 6 734 7 2193 8 6568 By GC, least n = 7. So the population of the bacteria first exceeds 2014 thousands 7 days after the start of the experiment. (ii) Rate of growth of H2 Bacteria 2 days after the start of the experiment =  22 d 3l n 3 1 9 l n 3 1d nn nn a n     thousands/day. or by GC, 2 d 10.9d n n a n   thousands/day. Let nb denote the number of H3 Bacteria (in thousands) exactly n days after the start of the experiment. Then {}nb is an arithmetic sequence with first term 900 and common difference 800. Therefore () ( )900 1 800 800 100nbn n=+ - = + . We want least n such that  0 3 1 800 100 0 3 799 101 0nn nnab n n n         .

n 37 9 91 0 1n n 7 –3507 8 68 9 12391 By GC, the population of H2 Bacteria will exceed that of H3 Bacteria by the 8th day. 5 Numbered all 880 students names from 1 to 880. Chose a random starting number from 1 to 11 th (as 880/80 = 11). Subsequently take every 11 th name after the starting name. The process continues until 80 students have been chosen. The 80 students chosen may not have the correcti on proportion of male and female students as the list of students may not have an even spread of male and female students in the list. Stratified Sampling 6 probability that A wins the match = 0.6 0.6(1 )(0.3) 0.4(0.3)pp p  = 0.18 + 0.54 p P(B won the first set | Player B loses the match) =  0.4 0.3 0.18 0.54 p p  0.4 0.3 0.150.18 0.54 p p  Solving, p = 0.692 or 9 13 When p = 0.65, P(A wins) = 0.531 P(3rd win in nth match) =     231 2 0.531 1 0.531 0.531 0.10866 nn C        3312 0.531 0.469 0.108662 nnn   From GC, n = 7 7 P(Wise misses the bus) = P(X – Y < 0)    ~2 5 1 5 , 9 4 i.e. ~ 10,13 XY N XY N    Thus, probability required = 0.0027728  0.00277 (3 s.f.) T = 45 –（W + X）

E(T) = 45 – 30 – 25 = – 10 Var(T) = 3 + 9 = 12 P(the bus arriving after 8.30 a.m.) = P(T < 0 ) = 0.99805  0.998 (3 s.f.) 8 (A) (B) (ii) (iii) From the diagram in (ii), as x increases, y increases in decreasing amount.  (B) is the appropriate for modelling these values. 0.972 1935 91.6 r y x    0 10 20 30 40 50 60 70 80 0 1 02 03 04 05 0 y x y Field Code Changed Field Code Changed Field Code Changed Field Code Changed Field Code Changed Field Code Changed Field Code Changed Field Code Changed Field Code Changed Field Code Changed Field Code Changed Field Code Changed Field Code Changed

(iv) when 64 33.8yx  As 64y falls within the given data range, the estimated value is valid. Neither regression line is suitable to estimate x since x is the independent variable which can be controlled. 9 Number of ways = 7! + 7! – 6! = 9360 Number of ways = (11-1)! x 2 / 4! x 12 = 3628800 Number of ways = (10–1)!/4! x 10 x 12 = 1814400 10 (a)(i) 0H: 4 0  1H: 4 0  at 5% level of significance necessary assumption: the length of steel pins follows a normal distribution Apply t-test reject  0H  if   value 0 05p .    2 2 3.740 40.46258 1( 3 . 7 )[7.27 ] 0.79410778 0.09278 x s p value     Since p value >0.05, we do not reject 0H . There is insufficient evidence at 5% level of significance to conclud