Page 64 - Hybrid PBD 2022 Tg 5 - Matematik Tambahan
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Matematik Tambahan  Tingkatan 5  Bab 2 Pembezaan
                     Jawapan / Answer :                                 2.                     16 cm
                                                                                           y
                     x + y = 5    z  = xy 2                                               y
                                                 2
                                       2
                                     = y (5 – y) = 5y  – y 3
                      dz   = 10y – 3y 2
                       dy   = y[10 – 3y] = 0
                          y = 0   y  =   10
                                      3
                          x = 5   x  =  5                                 Sekeping  logam  yang  berbentuk  segi  empat
                                     3                                    sama dengan bersempadan 16 cm, dilipat  untuk
                      2
                      d z   = –6y                                         membentuk sebuah kotak terbuka setelah
                     dy 2                                                 dikeluarkan 4 segi empat sama yang bersempadan
                                                 10
                                          2

                                        ,
                     Apabila/When  y =  10 d z  = –6    = –20 , 0       y cm daripada setiap pepenjurunya.
                                      3  dy 2     3                       A square piece of metal has a side of 16 cm, is to be folded to
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                        5 10
                       
                     ∴  ,      ialah maksimum                            make an open box by cutting off 4 squares each with sides y cm
                                                                          from each of the  tips of the square.
                        3 3
                                                                          (a)  Apakah nilai  y yang menjadikan kotak itu
                                                                               mempunyai isi padu maksimum?
                                                                               What is the value of y which will make the volume
                                                                               maximum?
                 Kertas 2                                                                             [5 markah / 5 marks]
                                                                          (b)  Seterusnya, cari nilai isi padu maksimum itu.
                                                                               Then, find the maximum volume.
                  1.  Diberi bahawa persamaan suatu lengkung ialah                                    [5 markah / 5 marks]
                        3
                     y =  .
                        x 3
                     Given that the equation of a curve is y =   x 3 3  .  Jawapan/Answer:
                                                                                          2
                     (a)  Cari nilai  dy  apabila x = 3.                           (a)  V = (16 – 2y) y
                                  dx                                           dV
                         Find the value of   dy   when x = 3.                     = 2(16 – 2y)(–2)y + (16 – 2y) 2
                                     dx                                        dy  = (16 – 2y)[–4y + 16– 2y]
                                                [3 markah / 3 marks]
                     (b)  Seterusnya, anggarkan nilai bagi   3  .                  = (16 – 2y)(16 – 6y)
                                               3       (2.98) 3                                             dV
                         Then, estimate the value of   .                       Apabila maksimum / minimum      = 0
                                             (2.98) 3                                                       dy
                                                [4 markah / 4 marks]           When maximum / minimum
                                                                               y = 8 atau  8
                     Jawapan/Answer:                                                    3
                     (a)  y =  x 3 3                                           d V
                                                                                2
                         dy   = –   9                                     (b)   dy 2  = (16 – 2y)(–6) + (16 – 6y)(–2)
                         dx     x 4                                                 = –2[48 – 6y + 16 – 2y]
                                                                                   = –2[64 – 8y]
                         Apabila/When  x = 3 ,  dy  =  –9  =  –1
                                                                                                   2
                                            dx   3 4  9                        Apabila/When  y = 8 ,  d V  = 0
                              dy                                                                  dy 2
                     (b)  δy =   .δx
                                                                                                 
                                                                                      8 d V
                                                                                                        8
                              dx                                                  y =   ,   2    = –2 64 – 8  
                               1
                            = –  [–0.02]                                              3 dy 2            3
                               9                                                                  , 0   (Maksimum)
                                     1                                              8
                           =      =                                            ∴ y =   cm.
                              900   450                                             3
                                                                                  
                                                                                             2  8
                                                                                         8
                                    3    1                                     V =  16 – 2      = 303 11   cm 2
                         ∴ y + δy  =    +                                                3     3        27
                                   27   450
                                    1
                                   =   +   1   =   17   = 0.113
                                    9  450   150







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