Page 44 - Pra U STPM 2022 Penggal 1 - Mathematics (T)

Mathematics Term 1 STPM Chapter 1 Functions

Example 37

2
Express x + 2x + 3 in partial fractions. 1
(x + 2)(x – 1)
2
x + 2x + 3 x + 2x + 3
2
Solution: ≡
2
(x + 2)(x – 1) x + x – 2
2
≡ (x + x – 2) + (x + 5)
x + x – 2
2
= 1 + x + 5
2
x + x – 2
= 1 + x + 5
(x + 2)(x – 1)
14243 x + 5
–––––––––– can be expressed
(x +2)(x – 1)
Let x + 5 ≡ A + B as partial fractions.
(x + 2)(x – 1) x + 2 x – 1
≡ A(x – 1) + B(x + 2)
(x + 2)(x – 1)
x + 5 ≡ A(x – 1) + B(x + 2)
When x = 1, 6 = 0 + 3B
B = 2
When x = –2, –2 + 5 = –3A + 0
3 = –3A
A = –1
x + 5 ≡ – 1 + 2
(x + 2)(x – 1) x + 2 x – 1
2
Hence, x + 2x + 3 ≡ 1 – 1 + 2
(x + 2)(x – 1) x + 2 x – 1

Example 38

3
Express x + 1 in partial fractions.
(x – 1)(x + 2)
3
3
Solution: We have x + 1 ≡ x + 1
2
(x – 1)(x + 2) x + x – 2
By using long division,
x – 1
2
x + x – 2 x + 0x + 0x + 1 Write x + 1 as x + 0x + 0x + 1.
2
3
3
2
3
x + x – 2x
2
3
– x 2 + 2x + 1
– x 2 – x + 2
3x – 1
3
Hence, x + 1 ≡ (x – 1) + 3x – 1
(x – 1)(x + 2) (x – 1)(x + 2)
≡ (x – 1) + A + B
x – 1 x + 2
x + 1 ≡ (x – 1) (x + 2) + A(x + 2) + B(x –1)
3
2
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01a STPM Math T T1.indd 41 3/28/18 4:20 PM