Page 63 - Hybrid PBD 2022 Tg 5 - Matematik Tambahan

Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan
Pentaksiran Sumatif
SPM

PRAKTIS PdPR Jawapan
Bab 2 Praktis PdPR Bab 2
3
Kertas 1 3. Isi padu V cm bagi suatu gas di bawah tekanan P
unit memenuhi persamaan PV = 220. Apabila P =
40, P bertambah dengan kadar 4 unit/s. Tentukan
1. Cari nilai bagi / Find the value of kadar perubahan isi padu pada ketika itu.
(a) had/lim (4 – x ) The volume V cm of a gas under the pressure P unit is given
2
3
x → 1
[2 markah / 2 marks] by the equation PV = 220. When P = 40, P increases at a rate
2
(b) f‘ (–2) jika/if f(x) = 4 – 2x + x of 4 unit/s. Determine the rate of change of the volume at that
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instant.
[3 markah / 3 marks]
[3 markah / 3 marks]
Jawapan / Answer : Jawapan / Answer :
(a) had/lim (4 – x ) = 3
2
x → 1
(b) f(x) = 4 – 2x + x PV = 220 –1
2
f‘ (x) = –2 + 2x V = 220P
f‘ (–2) = –2 + (2)(–2) = –6 dV = –220
dP P 2
dV = –220 dP
.
dt P 2 dt
2. Diberi bahawa A = 5t + 2 dan x = 2t – 6t. = –200 [4] = –0.5 cm /s
2
3
Given that A = 5t + 2 and x = 2t – 6t. 40 2
2
(a) Ungkapkan dA dalam sebutan t.
dA dx
Express in terms of t. 4. Persamaan lengkung diberi oleh y = 2x + 3.
2
dx
[2 markah / 2 marks] Tentukan perubahan hampir dalam y, dalam
(b) Cari perubahan kecil bagi x apabila A sebutan β apabila x berubah daripada β kepada
berubah daripada 4 kepada 3.98 pada ketika β + δβ.
t = 2. The equation of the curve is given by y = 2x + 3. Determine the
2
Find the small change in x when A changes from 4 to 3.98 small change in y, in terms of β when x changes from β to β +
when t = 2. δβ.
[3 markah / 3 marks] [3 markah / 3 marks]

Jawapan / Answer : Jawapan / Answer :
(a) A = 5t + 2 x = 2t – 6t y = 2x + 3 . δx = β + δβ – β
2
2
dA = 5 dx = 4t – 6 dy
dt dt dx = 4x = δβ
dA = 5 δy = dy .δx
dx 4t – 6 dx
= 4x.δβ
(b) δA = 3.98 – 4 = –0.02 = 4β.δβ
δA = dA .δt
dt
–0.02 = 5 δt
δt = –0.004 5. Dua kuantiti, x dan y berubah supaya x + y = 5.
2.
δx = dx .δt t = 2 Satu kuantiti lain z ditakrifkan oleh z = xy Cari
dt nilai-nilai x dan y yang menjadikan z maksimum.
= (4t – 6).δt Two quantities, x and y changes such that x + y = 5. Another
2
= (8 – 6)(–0.004) quantity z is defined by z = xy . Find the values of x and y which
will make z maximum.
= 2(–0.004) [7 markah / 7 marks]
= –0.008

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