A Level 2016 P2 Solutions for new syllabus (edited CSC)

2016 H2 Maths Paper 2 SECTION A Qn Solution 1 Let h m and V m3 be the depth and volume of water respectively at time t minutes. Given: d 0.1 (1)d V t   and 5.0tan  d d d (2)d d d h h V t V t   , so we find V in terms of h first. tan 0.5 0.5 rh    2311 3 12(0.5 )V h h h 2 21 4 d (3)d4 Vh hh    Sub (1) & (3) into (2): 22 d 4 0.4 0.1d h t h h   When 3V  ,   3 1 3361 12 3hh     , so d 0.0251 (3 s.f.)d h t  The rate of increase of the depth of water is 0.0251 m per minute. 2(a)(i) 2 2 sin 2cos d sin d x nxx nx x x nx x nn 2 cos cossin d (1) d cos 1 cos d cos 1 sin x nx nxx nx x x nn x nx nx xnn x nx nxnn           2 2 2 2 23 sin 2 cos 1cos d sin sin 2 cos 2sin where is an arbitrary constant x nx x nxx nx x nx n n n n x nx x nx nx ccn n n            h r  Integration by parts 2 d d d sin d cos 2 vux x u nx vx x n nx    d d d co sin d 1 s vux x u n nxvx x n    

(a)(ii) 2 3 2 3 22 22 2 2 22 23 224 sin2 4 cos2 2sin2 sin 2 cos 2sin 4 cos2 2 cos 4 (1) 2 (1) cos d sin 2 cos 2sin ( the other terms are 0) (when is even) n n n n n nnn n n n n nn nn nn x nx x x nx x nx nx n n n n                                      22 22 4 (1) 2 ( 1) 26 or (when is odd) or nn nn n    (b) 29 x xxy   22 2 2 )9( )( x xxy            2 2 0 32 20 2 5 29 5 29 5 12 9 51 9 2 2 9 52 5 4 529 Volume of solid d d 9 99 1. d 29 9 d2 9d ln 9 ln 5 ln 9 1 ln yx x x x uu uu u u uu u u u uu                               3 (i) ttx cos , ty cos1 , for 02 t  When D meets the x-axis, 0 cos 1 0 or 2y t t t       1x or 12  x d d d sin d d d 1 sin y y x t x t t t    At the maximum point, d 0 sin 0d y tx    0, , 2t  When 0t or 2 , we get the points on the x-axis. When t , we get the maximum point. cos 1x       , 1 cos 2y    Note:       sin 0 cos 2 1, cos 2 1 1 n nn       29 x xxy  Let 2299u x x u     d 2 2 9d 1d d 29 u xux xu u      When x = 2, u = 5; x = 0, u = 9.

(ii) 1 10 dArea required d d d xa xy x y t t     0 0 0 1 2 1 4 11 44 3 1 44 (1 cos )(1 sin ) d 1 cos sin sin 2 d sin cos cos 2 sin cos cos 2 1 sin cos cos 2 a a a t t t t t t t t t t t a a a a a a a a                                (iii) At P, where 2t  , 2x  , 1y and 2 2 sind1 d 1 sin 2 y x   . Equation of normal at P,  212yx    21yx    At E, 0y , 1 2xa  At F, 0x , 1yb     Area of triangle OEF is      21111 2 2 2 2 1ab      units2 4(b)(ii) 2 2iw 222 ( 2) 8w     , 1 2 24tan    i 48ew          1 ii i 44 44 * 8e 8e 8e n n n n ww         D ( ,0)Ea ( 1, 2)  t  (2 1,0)  2t  ( 1,0) 0t  2( ,1)P  (0, )Fb Note: The extra line is for part (iii).

 43 3 5 9 2 2 2 2 4 4 2 arg( * ) ..., , , , ,... , 1 2(4 3) 1 8 6 78 n mn ww m nm nm m                  The smallest positive whole number value of n is 7 when 0m . Section B 5 Let W be the event that a player wins a game. Let R, B and Y be the events that the spinner comes to rest over the red, blue and yellow sections respectively. (i)           1 1 2 1 4 1 7 7 72 3 6 11 42 P  W    (ii)         21 7 3 4 1111 42 P    P |   P    BWBW W   (iii)               1 1 2 1 4 1 7 7 72 3 6 4 1029 Required probabili 3 t P P 3! ! yP W R W B W Y       6(iii) Let X be the age of employees in the company and µ be the population mean age. Test 0H : 37  against 1H : 37 at 5% level of significance Under 0H , 140 80~ N(37, )X approx. by Central Limit Theorem since 80n is large. Using a one-tail z-test, R B Y 1 7 2 7 4 7 W W’ W’ W’ W W 1 2 1 2 1 3 2 3 1 6 5 6 Do the following to get the generalisation of 59 2 2 2, , ,...   : 1. Observe that    2 2 21 , (5), 9 ,...   2. Factorise 2  , you would have 1, 5, 9, … 3. The sequence follows an AP with first term 1, common difference 4. 4. Find the mth term, 43mTm Note: Do not use n for the generalisation

Critical value = 34.8 Critical region: 34.8x  For H0 to be rejected, x must lie in the critical region. 34.8x Critical value = 1.64485 Critical region: 1.64485z 37 140 80 cal xz  For H0 to be rejected, calz must lie in the critical region. 1.64485 37 1.64485 140 80 calz x     34.8x  The set of values of x is  4.8| 3xx  (iv) Now do a one-tail z-test at the α% level of significance. For H0 not to be rejected, -value = P( 35.2) 100pX  0.086809 100 8.68 (3 s.f.)     The set of values of  is  8 68|0 .  7(i) No. of ways 4 34 3 2 or C 3!    = 24 (ii) No. of ways they are all men 6 36 5 4 or C 3!    = 120 Required no. of ways = Total without restriction – 24 – 120 10 9 8 24 120 576       0 Critical value = 1.64485 0.05 Critical region Z 37 Critical value = 34.834 0.05 Critical region X 0 : 37 : 140 : 35.2 :80 x n   This info suggests we use a p-value method.

(iii) Required probability 7! 3! 1 9! 12  (iv) Required probability   7 36! 3! 5 9! 12 C  8(i) [Do note that it was instructed by the question that the scatter diagram is to be drawn on a graph paper] (ii) The scatter diagram shows a non-linear relationship between x and y. Thus a linear model in the form of y ax b is not appropriate. (iii) From the scatter diagram, as x increases, y increases. Hence c is negative. As x approaches infinity, y approaches d, which is a positive limit (representing the maximum efficiency as power increases). (iv) Product moment correlation coeefficient, 0.980 (3 s.f.) 17.484 17.5, 91.750 91.8 r cd      (v) Using the model 17.484 91.750y x  , 3 85.9 (3.s.f.)xy   The value that Xian copied wrongly is estimated to be 85.9. This estimate is reliable as x = 3 is within the data range for x and 1r  indicates strong linear correlation between 1x and y. 9(a) Given: 2~ N(15, )Xa  10 15P( 10) 0.25 P 0.25 aXZ      5 0.67449 7.41 (3 s.f.) a a    x x x x x x x C S T x x x x x x x y

(b) Given: ~ B(4, )Yp and P( 1) P( 2) 0.5YY    3 2 2 2 3 2 2 2 3 4 2 3 4 42 42 44 (1 ) (1 ) 0.512 4 (1 3 3 ) 6 (1 2 ) 0.5 4 12 12 4 6 12 6 0.5 2 6 4 0.5 4 12 8 1 (shown!) p p p p p p p p p p p p p p p p p p p p p p p p                                  From GC, 2.01, 1.25, 0.166, 0.599p (3 s.f.) As 01 p , 0.166 or 0.599p