A Level 2015 P2 Solutions for 9780 CWY 2018

2015 H2 Maths Paper 2 (edited for new H2 Syllabus 9758) Qn Suggested Solutions Section A (Pure Math) 1(i) Max height is when d 0d h t  . 11 16 010 2 1 162 32 m h h h    (ii) Separating variables and integrating w.r.t. t,  1 21 1dd 10116 2 2 ht h     112 16 22 10h t c    114 16 2 10h t c    When t = 0, h = 0: 16c 140 16 1602th   When the tree is 16m, 140 16 (16) 1602 46.9 years t t     2(i) Let the angle between L and the x-axis be . cos  = 21 3 . 0 60 | 49 1                   = 2 7   = 73.4 Acute angle between L and the x-axis = 73.4 (ii) 12 : 2 3 , 46 L                      r . Let B be a point on L such that 33PB  In general, 1 2 1 2 1 2 1 2 f '( ) d f '( )[ f ( )] d f ( ) [ f ( )] 2[ f ( )] x x x a x x ax ax C a x C         In particular, 1()( ) d ( )( 1) n n a bxa bx x C bn     P B B’ 33 L N

and let N be the point on L closest to P. 1 2 2 2 3 5 for some 4 6 6 12 73 26 PB OB OP                                            2 2 2( 1 2 ) ( 7 3 ) (2 6 )PB           = 254 70 49 254 70 49 33    Square both sides and solve for  , 2 2 2 54 70 49 33 0 49 70 21 0 7 10 3 0 (7 3)( 1) 0 31 or 7                    1 2 3 2 1 3 = 1 4 6 10 OB                                or 1 2 13 312 3 = 5774 6 46 OB                                 The points are (3, 1, −10) and 1 (13, 5, 46)7  . Hence, as N is the midpoint of BB’, 3 13 17 1 1 1 1 5 = 12 7 7 10 46 58 ON                                (iii) Normal to the new plane: 1 2 36 7 3 = 2 2 6 11                              36 1 36 . 2 2 2 4 11 4 11                                 r Cartesian equation: 36 2 11 4x y z   .

3(ai) 2 1f ( ) 1x x  , x > 1 Any horizontal line y = k intersects the graph of f ( )yx at most once.  f is a one-one function and so its inverse exists. (ii) 2 1f ( ) , 11xx x Let 2 1 1y x  Make x the subject: 2(1 ) 1yx  2 2 1 1 1 y yx yx y yx y    As 1x , 11 1yx yy    1 1f ( ) 1 , , 0x x x x      1 ( ,0).ffDR    (b) Let 2 2g( ) 1 xyx x   . 2 2 (1 ) 2 (2 ) 0 (1) y x x yx x y         The range of g corresponds to the values of y for which (1) has real root x i.e. D  0. 2 2 1 4 (2 ) 0 1 8 4 0 yy yy       Solve 21 8 4 0yy   : 8 64 4(4) 1132(4) 2y    −6 −4 −2 2 4 6 −4 −3 −2 −1 1 2 3 4 x y x = 1 y = 0

33R : 1 or 1 22 g y y y       4(bi)     2 2 4 8 3 2 1 2 3 2 2 3 2 1 AB r r r r A r B r         Let 33: 2 2 1 122r B B         Let 11: 2 2 3 122r A A         2 2 1 1 4 8 3 2 1 2 3r r r r      (ii) 2 11 2 1 1 4 8 3 2 1 2 3 nn n rr S r r r r         1 11 2 1 2 3 11 35 11 57 11 2( 1) 1 2( 1) 3 11 2 1 2 3 11 3 2 3 n r rr nn nn n               (iii) Let S be the sum to infinity. From (ii), as 1, 3 nnS  . 1 3S [Note: Since 1 1 1 3 2 3 3 nS n    , nSS .] For nS to be within 310 of the sum to infinity,

3 3 3 10 1 1 1 103 3 2 3 1 1023 2 3 1000 498.5 nSS n n n n            Smallest n = 499. Section B (Statistics) 6(i) Let X be the random variable number of red sweets in a small pack of 10. X~B(10, 0.25) P( 4X  ) =1 – P( 3X  ) = 0.224 (iii) Y~B(100, 0.25) P( 30Y  )=1 – P( 29X  )  0.14954 Let S be the random variable number of large packs that contain at least 30 red sweets out of 15 large packs. S~B(15, 0.14954) P( 3S  ) = 0.824 8 An unbiased estimate of population mean is 0.8825x  An unbiased estimate of population variance is 22 0.074785 0.00559s  9(i) P(B|A) = P(B) since A and B are independent = 0.4 (ii) Let P( )BC = 0.1+x. P( ) P( ) (0.22 ) (0.165 ) 0.45 0.385 0.835 A B C A x x x x x             P( ' ' ') 1 (0.835 ) 0.165 (1) A B C x x         When B and C are also independent, P( ) P( ) P( ) 0.1 0.4 0.3 0.02 B C B C x x        Sub into (1): P( ' ' ') 0.185A B C  

(iii) When B and C are not independent, Min x = 0 Max x = 0.165 (as 0.165 x cannot be negative) Sub into (1): min P( ' ' ') 0.165 0 0.165 max P( ' ' ') 0.165 0.165 0.33 A B C A B C           10 (i) (ii) (a) 0.9807r (b) 0.9748r (c) 0.9986 (all to 4 d.p.)r  (iii) Most appropriate case is (c) since r is closest to 1. Equation is 0.147 34.8 (3 s.f.)Ph  (iv) (Changing from feet to metres is a scaling by factor 1 3.28 parallel to the x-axis.) The equation changes from f ( )Ph to f (3.28 )Ph : 0.14687 3.28 34.789 0.266 34.8 (3 s.f.) Ph Ph     11(i) No. of different arrangements 8! 100802!2! since there are 2 As and 2 Bs (ii) No. of different arrangements = 10079 (iii) No. of different arrangements = 6! = 720 AA BB C G E S (iv) No. of different arrangements = Total no. of ways – No. of ways with As or Bs together or both = 7! 7!10080 6! 57602! 2!     Note that: No of different arrangements with 2 As or 2 Bs together: AA B B C G E S = 7! 2! . 2000 45000 4.28 27.8 h/feet P/inches

No of different arrangements with 2As and 2Bs together = 6!. 12(i) Let X be the mass of an apple in grams and Y be mass of a pear in grams. X ~ N(300, 202) Y ~ N(200, 152) 2 15 ... ~ (5 300,5 20 ) (1500,2000)X X N N     P( 15 ... 1600) 0.0127XX    . (ii) Let 2 15 ... ~ N(5 300,5 20 ) N(1500,2000)A X X      and 2 18 ~ N(8 200,8 15 ) N(1600,1800)B Y Y      ~ N(1500 1600, 2000 1800) N( 100,3800)AB     P( 0) 0.0524AB   . (iii) The total mass of 5 apples and 8 pears after preparation is the random variable 220.85 0.9 ~ (0.85 1500 0.9 1600,0.85 2000 0.9 1800 ) ~ N(2715, 2903) T A B N T         P( 2750) 0.742T  .