A Level 2012 P2 Solutions for new syllabus YAN

GCE A Level H2 Mathematics Paper Two Section A: Pure Mathematics [40 marks] a) 2 2 2 d 16 9d y xx   Integrating both sides w.r.t. x, 3d 16 3dy x x Cx    , where C is an arbitrary constant Integrating both sides w.r.t. x, 2 4 38 4y x x Cx D    The general solution is 2 4 38 4y x x Cx D    , where C and D are arbitrary constants. b) 2d 16 9du ut   2 1 d 116 9 d uu t      Integrating both sides w.r.t. t, 2 1 d 1 d16 9 u tu   2 2 1 d 119 4 3 du t u       Using MF15 formula, 1 4 3ln , where is an arbitrary constant24 4 3 u t E Eu   Given 1u when 0t  ,

1 7 10 ln ln 724 1 24 E E    1 4 3 1 ln ln 724 4 3 24 ut u      1 4 3ln24 7 4 3 ut u 

Q2 is out of syllabus i) ii) 3 2 2 4 4x x x    3 2 2 8 0x x x       22 3 4 0x x x    2x or 2 3 4 0x x   Since 2 2 2 2 3 3 3 73 4 4 0 for all 2 2 2 4x x x x x                            , 2 3 4 0x x   has no real solutions. Thus, the integer solution is 2x and there are no other real solutions. (proved) iii) Let 3y x  in       3 2 3 3 2 3 4 4x x x       3 2 2 4 4y y y     From (ii), this has 1 real solution, 2y . Hence 3 2x  i.e. 1x O 1.659 (0.548, -4.631) (-1.215, -1.887) -4 y x O  fy x

iv) v)  f 4x      3 2 3 2 f 4 or f 4 2 or 4 4 2 4 4. x x x x x x x x              For the former, 2x . For the latter,      3 2 3 2 2 2 4 4 2 0 2 0 2 1 0 0, 1 or 2 x x x x x x x x x x x x x                 The roots are 2, 0, 1, and 2 . y x O  fy x 4 (-1.215, 1.887) (0.548, -4.631) 1.659

i) nth month Amount put in ($) 1 100 2 100 10 3  100 2 10 4  100 3 10 n   100 1 10n       2 5000 2(100) 1 10 50002 19 1000 0 42.52 rej. as 0 or 23.52 nS n n n n n n n             Least 24n The value of Mrs A’s account became greater than $5000 on 1 Dec 2002. ii) nth month Amount put in ($) Amount at end of the month ($) 1 100  100 1.005 2 100        2100 1.005 100 1.005 100 1.005 100 1.005      3 100             2 3 2 100 1.005 100 1.005 100 1.005 100 1.005 100 1.005 100 1.005         n 100          1 100 1.005 100 1.005 100 1.005 100 1.005 1.005 1 1.005 1 n n n         -42.52 23.52

Amt in account on nth month =   100 1.005 1.005 1 20100(1.005 1)1.005 1 n n    20100(1.005 1) 5000n   Hence the value of Mr B’s account first became greater than $5000 in September 2004. iii) Let %r be the interest rate per month. Let 1 100rv  nth month Amount put in ($) Amount on 2nd day of the month ($) Amount on last day of the month ($) 1 100 100 100v 2 100 100 100v    2 100 100 100 100 v v v v     3 100 2100 100 100v v   2 3 2 100 100 100 100 100 100 v v v v v v      n 100   1 2100 100 100 100 100 1 1 n n n v v v v v          When 36n ,   36100 1 50001 v v       36 36 100 1 5000 1 50 49 0 v v v v       Using GC, 1 (rejected since 0 1)v r v    or 1.0179576v 1.0179576 1 100r   1.79576 1.80r   (to 3 s.f.). Hence the interest rate per month is 1.80%. Algebraic method: 501 1.005 201n  501.005 1 201n       50ln 1 201 ln 1.005n   44.541n least 45n  GC method: least 45n 

Section B: Statistics [60 marks] i) Let D represent the event that the patient has the disease. Let T represent the event that the test gives a positive result. a)  test gives positive result     T 0.001 0.999 1 0.999 0.998 0.00599 p p p        b)  D|T     D T T 0.001 0.00599 0.166 (to 3 s.f.) p      ii)  D|T 0.75          D T 0.75T 0.001 0.750.001 0.999 1 0.001 0.75 0.999 0.998 0.7495 0.74925 0.999666 (to 6 d.p.) p p p p p p p          D D’ T T’ T T’

i) Let X be random variable “tail length of a randomly chosen squirrel in cm”. Let  be the population mean of the tail length. The appropriate hypotheses are: 0 1 : 14.0 : 14.0 H H     ii) Under 0H , 2 3.8~ 14.0, 20X N     The test statistic is  14.0 ~ 0,13.8 20 XZ N  For 0H to not be rejected, 1.96 1.96 14.01.96 1.963.8 20 12.335 15.665 z x x         Solution set is { :12.3 15.7}x x     (to 3 s.f.) Alternatively For 0H to be rejected, 1 1 2 2 2 1 2 2 ( ) 0.025 12.335 ( ) 0.025 ( ) 0.975 15.665 Or : By symmetry, 14 2 15.665 P X C C P X C P X C C C C C                -1.96 1.96 0.025 Z~N(0,1) C1 C2 0.025 2 3.8~ 14.0, 20X N    

For 0H to not be rejected, 1 2Solution set for is { : } { :12.3 15.7}x x C x C x x          iii) When 15.8 15.7x   , z falls in the rejection region, so 0H is rejected. Thus there is sufficient evidence, at the 5% level, that the squirrels on the island do not have the same mean tail length as the species known to the zoologist.

i) Let S represent the event that the sisters are next to each other in the line. Let B represent the event that the brothers are next to one another in the line.  sisters are next to each other  S no. of arrangements with sisters together total no. of arrangements 14! 2! 15! 2 15     ii)   13! 3! 1B 15! 35    brothers not all next to one another     B' 1 B 11 35 34 35       iii)  sisters are next to each other and broth ers are next to one another  S B 12! 2! 3! 15! 2 455      Note: 13! = (2+10+1)!, where (2+10+1) comes from counting the two sisters separately, the ten other people and the brothers as one item. 3! comes from counting the number of ways of arranging the three brothers within that single item. Note: 12! = (10+1+1)!, where (10+1+1) comes from counting the ten other people, the sisters as one item and the brothers as one item. Note: 14! = (3+10+1)!, where (3+10+1) comes from counting the three brothers separately, the ten other people and the sisters as one item. 2! comes from counting the number of ways of arranging the two sisters within that single item.