JJC H2 MATHS P2 Solution

1 2014 H2 Mathematics Prelim P2 Worked Solutions Qn Solution Notes 1(a) (i)  ec o s ln cos Differentiate wrt d tan 1d yx x y xx x y xx      2 2 2 Differentiate wrt d sec d x y x x  2 2 0 0 d 1d d 1 d x y y x y x     2 ...2 xyx   (a) (ii) h( ) 0.2xy h( ) 0.2 0xy  From GC, 1.12 1.12x  (to 3 s.f)

2 1(b) 133 nn nx xaa a      2 2 2 (1 )1. .32 3 (1 )1. .3 18 n n xn n n xa aa nn nax x a a           2 4( 1 ) 3 18 31 2 nn n a a an   31 2 1 4 3Sub 1 2 1 4 n a a an a     From GC, 2 or 0.16086 (to 5 s.f) when 2, 2 when 0.161 (to 3 s.f), 0.759 (to 3 s.f) a an an    

3 2(a) 2 2 2 2 d ed de d2 x x y ax ya cx     2e 4 xay cx d    2(b) d d x kx pt  Given that d 0d x t  when x = 12. 12 0kp  12 pk  d d1 2 d 12 (Shown)d1 2 xp x pt xp xt     (i) 1 d d12 12 px tx  ln 12 12 px tc  12 12 12 12 , where p tc p t c xe x Ae A e       When t = 0, x = 10 2A  12 12 2 p t x e  (ii) When x = 0, t = T 1212 2 0 ln 612 12 ln 6 p T e p T T p     (c) x t 10 T 1212 2 p t x e

4 3 (i) Since A lies in the plane 1 , 10 03 0 50                . 10 30  3 Perpendicular distance from O to 1 6 1 1 6 OA  n n 

1 1 22 2 30 6 5 5    n n 4 (ii) Acute angle between OA and 2 21 2 sin OA OA  n n 

 1 2 10 1 01 52sin 10 1 01 52                           

1 20sin 125 6 46.9   (iii) 10 1 5 01 1 5 52 1 0                   A normal to plane 3 is 1 3 2   . A cartesian equation is 320xyz .

5 4(i) 22 2 2 2 1i , 1 3 i , i 1( 1 ) 2 , 1( 3 ) 2 , 1 11arg( ) , arg( )43 pq r k pq r k pq                 2 3 2 2 3 1 2 1 2 21 1 22 s pr q k     2 2 12 14 3 k k k       2arg( ) 3 22 arg( ) arg( ) 3arg( ) 3 s pr q       21 1 5arg( ) 2 3 34 3 6r                If 13, then arg( ) 6kr  If 53, then arg( ) 6kr   3k 

6 4(ii) 3 1i23 3 16i 3 4(1 3 i) 8e 8e k k z z          16i 92e , 1,0,1 k zk   57ii i 99 92e , 2e , 2e                      3 3 3 332 3 43 i 4i 3 i 1 4i 1 3 i i 4 1 3i since i 1 41 3 ii w w w w w          Comparing with 3 4(1 3 i)z  i i w z wz   i 2 57ii ii ii292 9 2 9 11 1 1 3ii i 18 18 18 e e2 e , e2 e , e2 e 2e , 2e , 2e wz                                              

7 5(i) This is quota sampling. A disadvantage of quota sampling is that the sample is not a good representation of the population (residents of the town). (ii) Select the following number of students randomly from each stratum below. 18 - 25 year old 26 - 35 year old Male 75 100 30250  36 100 14250  Female 99 100 40250  40 100 16250  6 Given X ~ N(,  2),   12 12 P2 0 . 8 P2 0 0 . 8 XX XX   Now,   2 122~ , 5XX N   P0 . 8 5 Z      0.84162 5    1.8819 0.53138         12 12 P2 0 . 8 P2 0 . 2 XX a XX a     Now,  2 12 ~2 , 2XX N   22P0 . 2 2 aZ     22 0.84162 2 a      0.59512 0.59512 0.53138 0.75974 0.760 a a a a            

8 7(i)   P ' 0.75 P' 0.75P( ') BA BA A    P ' 0.75 1 0.6 0.3 BA   P' '1 P ( ) P ( ' )AB A BA   10 . 60 . 3 0.1    (ii)    P0 . 4 P 0.4P( ) P0 . 4 ( ) AB AB B AB P B       P( ) P ' P 0.3 0.4P( ) B BA AB B    0.6P( ) 0.3B  P( ) 0.5B    P( ) 0.6 P0 . 4 PP ( ) A AB AB A      A and B are not independent. A B

9 8 1A, 1B, 1S, 2E, 1N, 1C No. of code words formed with 2’E’s = 5 2 4! 2!C  120 No. of code words formed with 1 or no ’E’ = 6 4 4!C  360 Total no. of code words 120 360 480 (shown) (i) P(four-letter code words contain distinct letters) = No. of code words formed with 1 or no 'E' No. of code words formed without restrictions = 360 480 =0.75 (ii) code words do not contain code words containP any vowels distinct letters    = 4 4 4! 480 0.75 C   = 1 15

10 9(i) 0.175 3.57 0.175 3.57 yx yx     22.94 38.8 0.175 3.5799 k    2.40 shownk  (ii) Using GC:  0.943 3 s.f.r  Even though r is close to – 1, from the scatter diagram, the data points do not follow a straight line. Therefore, a linear model may not be suitable. (iii) C is the appropriate model as the data points in the scatter diagram follow the graph of fye x . (iv) Equation of regression line 2.7480 2.0651y x 2.75 2.07y x ( to 3 s.f. ) When x = 8.0, 2.7480 2.0651 2.418.0y (to 2 d.p.) Since x = 8.0 falls outside the data range of x, the estimation of y is unreliable. 7.5 x 2.5 y O 3.40 2.40