Page 51 - Hybrid PBD 2022 Tg 5 - Matematik Tambahan

Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan
18. Cari nilai h dan nilai k bagi yang berikut. SP 2.4.1 TP4
Find the value of h and of k for the following.
h
Kecerunan lengkung y = hx – 2kx pada titik (3, 1) (a) Kecerunan lengkung y = + kx pada (1, –1)
2
x
ialah –4. ialah -5.
h
The gradient for the curve y = hx – 2kx at the point (3, 1) is –4. The gradient for the curve y = + kx at (1, –1) is –5.
2
x
Pada (3,1) , 1 = h(3) – 2k(3)
2
1 = 3h – 18k …… Pada/At (1, –1) , –1 = h + k(1)
d y = h – 4kx –1 = h + k ……. 
dx d y = –h + k
dx x 2
Maka/hence h – 4k(3) = –4
h = 12k – 4 …… Maka/Hence –h + k = – 5 ………
2 Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
Gantikan  dalam , 1 = 3(12k – 4) – 18k  + , 2k = –6
Substitute  into , 18k = 13 k = –3
13 h = –1+3
k = , = 2
18
13
h = 12   – 4 = 14
3
18

(b) Kecerunan lengkung y = hx + kx pada (2, 0) (c) Kecerunan lengkung y = hx + kx – 3 pada (–2, 9)
2
3
ialah –8. ialah –7.
The gradient for the curve y = hx + kx at (2, 0) is –8. The gradient for the curve y = hx + kx – 3 at (–2 ,9) is –7.
3
2
Pada/At (2, 0) , 0 = h(8) + k(2) Pada/At (–2, 9), 9 = 4h + k(–2) – 3
0 = 4h + k ……. 6 = 2h – k ……. 
d y = 3hx + k d y = 2hx + k
2
dx dx
maka/hence 3h(2) + k = –8 maka/hence 2h(–2) + k = –7
2
12h + k = –8 ……. -4h + k = –7 ……. 
 – , 8h = –8  + , –2h = –1
h = –1 h =
1
k = 4 2
k = –5

19. Cari persamaan tangen dan normal kepada lengkung pada titik yang diberi berikut. SP 2.4.2 TP4
Find the equation of tangent and normal to the following curves at the given points.

f(x) = x + 3x – 6 pada titik (–1, 5).
2
3
f(x) = x + 3x – 6 at the point (–1, 5).
3
2
f’(x) = 3x + 6x
Apabila/When x = –1, f’ (–1) = 3(–1) + 6(–1)
2
= –3
1
Kecerunan tangen pada titik (–1, 5) ialah –3. Kecerunan normal pada titik (–1, 5) ialah .
Gradient of the tangent at point (–1, 5) is –3. Gradient of the normal at point (–1, 5) is . 3
1
Persamaan tangen ialah 3
Equation of the tangent is Persamaan normal ialah / Equation of the normal is
1
y – 5 = –3(x – (–1)) y – 5 = (x – (–1))
3
y – 5 = –3x – 3 3y – 15 = x + 1
y = –3x + 2
3y – x = 16

© Penerbitan Pelangi Sdn. Bhd. 30

02 Hybrid PBD Mate Tamb Tg5.indd 30 09/11/2021 9:24 AM