Matematik Tambahan 2 - skema

Document Sample
Matematik Tambahan 2 - skema Powered By Docstoc
					ANSWER TRIAL PAPER 2, 2010                         (b) (i)
SECTION A
                             x 3
1. x  3  3 y or y                    1M        2
                               3
     Substitute x or y into eqn. 2. 1M
     3  3 y  6 y 2  10 or
                     2
                                                                                                2
           x 3                                 –2
     x  6       10 or
           3 
                  equivalent.                           graph cos         1M,
6 y  3 y  7  0 or
     2                                                  Max 2 and min – 2 1M
                                                            1
2 x 2  27x  12  0                                    1     cycle             1M
                                                            2
          3  3  4(6)(7)
                 2
y                                      1M
                2(6)                                             2x
                                                      (ii) y          2         1M,
Or                                                               
          (9)  (9) 2  4(2)(12)                   Draw straight line graph           1M
x
                     2( 2)                             No. of solution = 3                1M
y = 0.859, – 1.359                       1M
x = 5.577 // 5.576,                                              10359 34.5k
                                                       5. (a)                             1M
   – 1.077 // – 1.076                    1M                         242  k
                                                                10359 34.5k
                                                                              41.2       1M
2. (a) y  kx  4k  5                   1M                        242  k

            dA 25 2                                                  k = 58              1M
     (b)           k 8                 1M
            dk    2
           25  2                                           (b)  fx 2  593465           1M
              k 8 0                    1M
           2                                                     593465
                                                                         (41.2) 2        1M
                5         5                                       300
            k , k                     1M
                4         4                                      = 16.76                  1M
            2
           d A
                 25k 3                1M
           dk 2                                                   y x          2
                                                   6. (a)            1 or y  x  8 or
                                    5                             8 12         3
           A minimum, k                1M                     equivalent                1M
                                    4
3. a) 4 + (25 – 1)x = 40                 1M
              x = 1.50                   1M                    3(12)  1(0) 3(0)  1(8) 
                                                       (b)                  ,            
                                                                   1 3         1 3 
     b)
        25
           2(4)  (25  1)(1.50) 1M
        2                                                     Either x or y correct       1M
           = 550                             1M               (– 9, 2)                    1M
     c)
           n
             2(4)  (n  1)(1.5)      1M
           2                                           c) Equation CD or gradient CD           1M
           n
             2(4)  (n  1)(1.5)  30n 1M                        3
                                                             y2   ( x  (9)) or
           2                                                       2
             n  35.67                   1M                         20
                                                             mCD 
            N = 36                           1M                    9 x
                                                                                       20     3
                                                            Find x, when y = 0 or           
       tan                                                                           9 x    2
4. (a)                                   1M
       sec                                                 and solve for x                    1M
           sin  cos    2
                                        1M
           cos    1                                           23 
                                                             C  , 0                         1M
                                                               3 
 Section B                                                          10. a) Solve equations
 7. a) AC  AB  BC or DB  DA  AB                        1M                    y  2 x  3 and x  y 2  1                1M
             AC  8x  y                      1M                                           1
                                                                                  y        , y  1 or
                                                                                           2
                                                                                           5
             DB  6x  4 y                    1M                                  x         , x2                          1M
                                                                                           4
    b) (i) DH  k (6x  4 y)                  1M                                 Q (2, 1)                                   1M
                                                                                                         1    3
         (ii) DH  4 y  m(8x  y)           1M                    Area of trapezium =                    2  (1)        1M
                                                                                                         2    2
                                                                                  1
                                                                                                            y
    4 y  m(8 x  y )  k (6 x  4 y )                                           ( y          1)dy        y
                                                                                           2
                                              1M                                                                            1M
                                                                                                            3
               4  m  4k and                                                   0

                  8m  6k                     1M                                      7  1  
                                                                    Area =                1  0                       1M
     Solve equations                          1M                                      4  3  
               16                     12                                           5
          k             1M, m                   1M                         =       // 0.417                           1M
               19                     19                                          12


                                                                                       y           
 8. a)                                                                                 1             2
log10 y 0.1206
                                                                    (c)     V                2
                                                                                                    1 dx
                     0.301   0.4983       0.699   0.8633    1.061                      0


                                                            1M                                              1
                                                                                 y5 2y3     
   b) log10 y  log10 p  x log10 q                    1M                             y  dx                           1M
       All points plotted                              2M                        5
                                                                                     3      0
                                                                                             
       Line of best fit                                1M                       15 2(1) 3 
 c) (i) log10 p  log10 y  int ercept                 1M                               1  0                              1M
                                                                                5
                                                                                       3     
                                                                                              
          p = 0.832 – 0.871                            1M
                                                                               28                                                1M
                                                                                // 1.867
    (ii)  log10 q  gradient                      1M                          15
          q = 0.013                                1M
   (iii) y = 6.03                                  1M
                                                                    11. (a)
                                                                         p = 0.75 and q = 0.25                               1M
9. a) AR = 12 + 9 = 21                        1M
                                                                          Use    10
                                                                                      C n  0.75 n  0.2510n                1M
 b) SOQ  120       o
                                              1M
        = 2.095 rad                           1M                          (i) P(X = 10) = 0.05631                            1M

                    60
                                                                          (ii) P(X = 9) + P (X = 10)                        1M
 c) arc RQ = 21         or                                                  = 0.244                                       1M
                   180
     arc CQ = 9 2.095             1M
                                                                                 x  174
                            60                                        b) z                                            1M
perimeter = 9 2.095 + 21       +9+3 1M                                          8
                           180
                                                                           (i) 0.6915                                  1M
            = 52.85                           1M
                                                                           (ii) P  1.75  z  0.75                  1M
 d) Area of sector                            1M
    Area of triangle                          1M                             Finding correct area                      1M
Area of shaded region =                                                           0.7333                               1M
1                 1               1
  (21) 2  1.047  (9) 2  2.095  (12)( 12 2  6 2
2                 2               2
                                             1M
= 83.66                                      1M
12. (a) t  0, v  15                   1M                                  1.54
                                                           14. (a)     x          100  110         1M
                                                                             1.4
   (b) 3t 2  18t  15  0 and solve 1M                                y
                                                                          100  115 , y = 2.30       1M
             t =1, t = 5           2M                                  2
                                                                       6.18
                                                                             100  103 , z = 6       1M
                                                                         z
   (c)         a  6t 18               1M
                   =6                   1M                 (b) (i)
   (d)                                                         110(160)  120(45)  115(65)  103(90)
                                                                                                      1M
                                                                                360
         15
                                                                            I  110.4                  1M
               1              5
                                                                     P09
              Shape minimum              1M                  (ii)          100  110.4          1M
                                                                     110
              3 coordinates shown        1M
                                                                      P09 121.44

                        
  (e) Distance = t 3  9t  15t 0  5
                                         1M                  (c) P / 09  80
                                                                  10                             1M
               = 425                     1M
                                                                     P / 09  P09 / 08
                                                                      10                 80  110.4
                   P07  RM 240                                                        =              1M
                                                                          100               100
                                                                                    = 88.32      1M
13. (a) (i) CD = 10                      1M
              CE
     (ii)         cos 70 o              1M                15. (a) I : x  y  80          1M
              10
              CE = 3.42                  1M                        II : x  2 y            1M
                                                                   III : y  x  20        1M
  (b) AB 2  12 2  9 2  2(12)(9) cos 70 o 1M
        AB  12.29                      1M                     (b)

         sin ABC sin 70o
                                         1M
             12    12.29

               sin ABC  0.9175

         ABC  66.56 o // 66.57 o        1M

         1
c) Area = (12)(9) sin 70 or
         2                                                     One straight line drawn            1M
         1                                                     The other two straight lines drawn 1M
   Area = (10)(3.42) sin 70               1M
          2                                                    Region R                           1M

                    1                 1                     (b) (i) 10  x  50              1M
Area ABED =           (12)(9) sin 70 – (10)(3.42) sin 70
                    2                 2                         (ii) p  30x  40 y
                                          1M                      Finding maximum point (30, 50)            1M
                = 34.67                   1M
                                                                     Maximum profit = 30 (30) + 40 (50) 1M
                                                                                   = 2900               1M

				
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
views:202
posted:9/29/2010
language:English
pages:3