HCI JC 2 H2 Maths 2012 Paper 2 Solution - for Sharing

1 © Hwa Chong Institution 2012 9740/02/Prelim/12 HCI H2 Math Prelim Paper 2 Solution Qn. Solutions 1 (i) Since y = 0.5 cuts the graph twice, f is not a one-one function. Therefore 1f does not exist. Restrict the domain to 13 22 x so that the range of h is equal to the range of f.  m = 1 2 , n = 3 2 Since h( 5 6 ) = 1 2 , therefore 1 15h ( )26   . (ii) With domain 13 22 x , hR [ 1, 1] , gD [ 1 , )   , Since hgRD , therefore gh exists. Substituting hR as gD into y = g(x) graph,

2 © Hwa Chong Institution 2012 9740/02/Prelim/12 1 ghR [ , ] aa  2 Let w be the width of the paper used and h be the length of the paper used. Area of paper used, A = wh We have ( 4)( 8) 1352wh   1352 4( 8)w h    Therefore 1352 4( 8)A wh h h     1352 4( 8) h hh 

3 © Hwa Chong Institution 2012 9740/02/Prelim/12 2 d ( 8)(1352) (1352 )(1) 4d ( 8) A h h hh   2 10816 4( 8)h   To minimise the paper used, d 0 d A h  2 10816 4( 8)h  2( 8) 2704h   2704 8 60h    First derivative test: h (60)  60 (60)  d d A h -ve 0 +ve OR Second derivative test: 2 23 d 21632 0d ( 8) A hh  Hence at 60h , the amount of paper used is minimised. At 60h , 1352 452w   30w Hence the dimension of the paper is 30 cm by 60 cm. 3 (i) i i i i 11 e e e e cos isin cos isin 2cos (shown) n n n n nnz z n n n n n                   

4 © Hwa Chong Institution 2012 9740/02/Prelim/12 (ii) 3 3 3 3 3 1By taking 1, 2cos 11cos 2 11cos 2 11 8 1 3 1 3 (shown)8 nz z z z z z z z zz zz                        33 3 3 3 1 3 1cos 3 8 1 1 1= 3 8 1= 2cos3 3 2cos8 13= cos3 cos (shown)44 zz zz zz zz                (iii) Method 1 3 13cos 3 d cos9 cos3 d44 11sin 9 sin 336 4 C            Method 2   32 2 2 3 cos 3 d cos3 cos 3 d cos3 1 sin 3 d cos3 cos3 sin 3 d 11 sin 3 sin 339 C                      

5 © Hwa Chong Institution 2012 9740/02/Prelim/12 4 (a)       22 2 2 50 4 5 0 Since we have cos 60 2 2 5 0 2 5 0 2 24 6 1 or 6 1 rej2 61 AB OP                           b a a b a b a b aa b a b a b aa aa a a (b)     3 4 3 2 2 7 7 Let : , then 11 OB OCOC OE AD AB OD OB OA                 a b a ba Since , , are collinear, for some constant O E D OE kOD k  32 17k     ba ba 3 5 : 3: 5AD AB    5 (i) For x > 0,   2 1x > 0 Since a, b and c are positive, for x > 0,   2 1 c x > 0 and ax b > 0    2 1 cy ax b x     > 0 for x > 0 (shown) (ii) Given area of region R = 42   3 2 0 d 1 cax b x x   = 42 3 2 0 1 21 cax bx x    = 42  9 324 ca b c     = 42 93 + 3 + 24a b c = 42 (shown)  (1) (iii)   2 1 cy ax b x    

6 © Hwa Chong Institution 2012 9740/02/Prelim/12 At (0, 5), + bc = 5  (2)   3 d2 d 1 yc ax x   At (0, 5), d 0d y x  2ac = 0  (3) From GC, 8, 1, 4a b c    equation of G is   2 481 1 yx x     . (iv) [Note: x-intercept (1.585171 , 0)] (v) Required volume =   2 3 3 2 2 0 0 48 1 d 8 1 d( 1)x x x x x      = 163.2774895 = 163 units3 6(i) Let X be the time spent by a customer at a supermarket. If 2~ N(35, 30 ), ( 0) 0.122.X P X  i.e. the probability that the time spent being less than zero is significantly big. However, X is a non -negative quantity. Therefore a normal distribution will not provide an adequate model. (ii) Let X be the time spent by a customer at a supermarket. 2~ N(35,10 )X x y 0 y = 8x + 1   2 481 1 yx x     x = 1 (1.59 , 0) (0, 5) x = 3 101 4 R S

7 © Hwa Chong Institution 2012 9740/02/Prelim/12 12 12 3 ~ N( 35, 1100) or 3 ~ N(35, 1100) X X X X X X         12 12 P3 P 3 0 0.146 X X X X X X       7(i) No. of ways 7315 4! 175560   (ii) ( 1)( 2)( 3) 175560n n n n    or 4P 175560n  or 4C 7315n  Using GC or guess and check, 22n . Method 1: Consider the 4 scholarship recipients as 1 unit. No. of ways 7 4! 9!   60963840 Method 2: Consider the 4 scholarship recipients as 1 unit. Case 1: The 4 scholarship recipients are in the row with 6 seats. Case 2: The 4 scholarship recipients are in the row with 7 seats. No. of ways     99 234! 3! 7! 4! 4! 6!CC        60963840 8(i) P( )AB P( ) P( ) AB B  = P(2Curry,1Spicy,1others) P(3Curry,1Spicy) P(exactly 1 Spicy)  7 ways to slot the 4 scholarship recipients. 4! ways to arrange scholarship recipients among themselves. 9! ways to arrange remaining applicants. seat seat seat seat seat 4! ways to arrange scholarship recipients among themselves. 9 2C ways to choose 2 non-scholarship recipients in row with 6 seats. 7! ways to arrange remaining applicants in row with 7 seats. 3! ways to arrange 1 unit of scholarship recipients and 2 non-scholarship recipients.

8 © Hwa Chong Institution 2012 9740/02/Prelim/12 10 9 10 60 4! 10 9 8 10 4! 80 79 78 77 2! 80 79 78 77 3! 10 70 69 68 480 79 78 77                       141 2737 or 0.0515162587 0.0515 (3 s.f.) OR P( )AB P( ) P( ) AB B  10 10 60 10 10 2 1 1 3 1 80 4 10 70 13 80 4 ( ) ( )C C C C C C CC C       141 2737 or 0.0515162587 0.0515 (3 s.f.) (ii) P( ) 1 P(0 curry) P(1 curry) 70 69 68 67 10 70 69 6814 80 79 78 77 80 79 78 77 A                       70 10 70 4 1 3or 1 80 80 44                              0.0741568558 0.0742 (3 s.f.) Or P( ) P(2 curry) P(3 curry) P(4 curry) 10 9 70 69 4! 10 9 8 70 4! 80 79 78 77 2!2! 80 79 78 77 3! 10 9 8 7 80 79 78 77 A                           10 70 10 70 10 2 2 3 1 4or 80 80 80 4 4 4                                             0.0741568558 0.0742 (3 s.f.) Method 1: Show P( ) P( )A B A . Hence A and B are not independent, and  'A and 'B

9 © Hwa Chong Institution 2012 9740/02/Prelim/12 are not independent. From (i), P( ) 0.0515AB  Since 0.0515 P( ) P( ) 0.0742A B A   , A and B are not independent. Hence 'A and 'B are not independent. Method 2: Show P( ) P( ) P( )A B A B   . Hence A and B are not independent, and  'A and 'B are not independent. From (i), P( )AB 10 9 10 60 4! 10 9 8 10 4! 80 79 78 77 2! 80 79 78 77 3!                    10 10 60 10 10 2 1 1 3 1 80 4 ( ) ( )or C C C C C C       0.017830271 0.0178 (3 s.f.) 10 70 69 68P( ) 4 80 79 78 77B      10 70 13 80 4 or CC C    0.3461095866 0.346 (3 s.f.) P( ) P( ) 0.0742 0.346 0.0256663987 0.0257AB     Since 0.0178 P( ) P( ) P( ) 0.0257A B A B     , A and B are not independent. Hence 'A and 'B are not independent. Note: Students can choose to show P( ' ') P( ') P( ')A B A B   or P( ' ') P( ')A B A or P( ' ') P( ')B A B 9(i) Let X be the number of defective SIM cards, out of 10.  ~ B 10,0.15X We assume that the probability of a SIM card being defective remains constant at 0.15. OR We assume that the defective SIM cards produced are independent of one another.

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