Page 40 - Hybrid PBD 2022 Tg 5 - Matematik Tambahan

Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan
2. Cari nilai bagi setiap yang berikut. SP 2.1.1 TP3
Find the value for each of the following.

x + 4x – 12  x– 4 x
2
(i) had/ lim (ii) had/ lim (iii) had/ lim 3 – 

x → –6 x + 6 x → 16 x – 16 x → 0 x+ 9

(x + 6)(x – 2) ( x– 4) ( x+ 4) x
= had/ lim = had/ lim = had/ lim 3 – 
x+ 9

x → –6 x + 6 x → 16 (x – 16) ( x+ 4) x → 0

x+ 9)

= –6 – 2 = had/ lim (x– 16) = had/ lim x(3 + 
x+ 9) (3 + 

= –8 x → 16 (x – 16) ( x+ 4) x → 0 (3 –  x+ 9)
= had/ lim 1 x(3 + 
x+ 9)

x → 16  x+ 4 = had/ lim

1 x → 0 9 – (x+ 9)
= x(3 + 

x+ 9)
= had/ lim Bhd. All Rights Reserved.
8 = had/ lim
x → 0 –x
= –(3 +  9)
= –6
Tip Penting
(−6) + 4(−6) − 12
2
Dengan penggantian secara langsung, didapati nilai
−6 + 6
tidak tertakrif, kita perlu melakukan pemfaktoran fungsi dahulu.
2
(−6) + 4(−6) −12
By direct substitution, −6 + 6 is undefined, hence we need to factorise
first.
x – 25 2x – 17x + 8 (3 + x) – 9
2
2
2
(a) had/ lim (b) had/ lim (c) had/ lim
2
x → 5 x + 2x – 15 x → 8 8 – x x → 0 x
Penerbitan Pelangi Sdn
(x+ 5) (x– 5) (2x – 1)(x – 8) 9 + 6x + x – 9

2

= had/ lim = had/ lim

x → 5 (x + 5) (x– 3) x → 8 –(x – 8) x → 0 x
= 0 = –15 = had/ lim 6 + x
x → 0
= 6
 x– 2  x
2x+ 22 – 4

(d) had/ lim (e) had/ lim (f) had/ lim
x + 16
x → 4 x – 4 x → –3 x + 3 x → 4 4 – 
2x+ 22 – 4][ 2x+ 22 + 4]
( x– 2) ( x+ 2) [   x(4 + 
x+ 16)

= had/ lim = had/lim = had/ lim
x+ 16)(4 + 
2x+ 22 + 4)

x → 4 (x – 4) ( x+ 2) x → –3 (x + 3)( x → 4 (4 – x+ 16)
x– 4

= had/ lim = had/lim [2x+ 22 – 16]

x+ 16)

x → 4 (x – 4) ( x+ 2) x → –3 (x + 3)( x(4 + 
2x+ 22 + 4)
1 2(x + 3) = had/ lim
= had/ lim = had/ lim x → 4 16 – (x+ 16)

2x+ 22 + 4)
x → 4  x+ 2 x → –3 (x + 3)( = –(4 + 
20)
1 1
= = = –(4 + 2 5)
4 4
19 © Penerbitan Pelangi Sdn. Bhd.
02 Hybrid PBD Mate Tamb Tg5.indd 19 09/11/2021 9:24 AM