Page 43 - Pra U STPM 2022 Penggal 1 - Mathematics (T)

Mathematics Term 1 STPM Chapter 1 Functions

Equating the coefficients of x : 3 = A + B
2
3 = 2 + B
B = 1
1
2
Therefore, 3x + 6x – 1 ≡ 2 + 1 + 2
(x – 1)(x + 1) 2 x – 1 x + 1 (x + 1) 2
In the same way, if the denominator of a proper rational function contains a repeated linear factor such as
3
3
(x + 1) , the denominator of the partial fractions concerned are (x + 1), (x + 1) and (x + 1) respectively.
2
Example 36

2
Express x + 1 in partial fractions.
(x – 1)(x + 1) 3
3
Solution: Since the denominator contains the repeated linear factor (x + 1) , the partial fractions
consist of terms with denominators (x +1), (x + 1) and (x + 1) .
3
2
2
Hence, x + 1 ≡ A + B + C + D
(x – 1)(x + 1) 3 x – 1 x + 1 (x + 1) 2 (x + 1) 3
Multiply both sides of the identity with (x – 1)(x + 1) ,
3
2
3
x + 1 ≡ A(x + 1) + B(x – 1)(x + 1) + C(x – 1)(x + 1) + D(x – 1)
2
When x = 1, 2 = 8A
A = 1
4
When x = –1, 2 = –2D
D = –1

Equating the coefficients of x , 0 = A + B
3
0 = 1 + B
4
B = – 1
4
When x = 0, 1 = A – B – C – D
= 1 + 1 – C + 1
4 4
C = 1
2

2
Hence, x + 1 ≡ 1 – 1 + 1 – 1 .
(x – 1)(x + 1) 3 4(x – 1) 4(x + 1) 2(x + 1) 2 (x + 1) 3

Degree of denominator less than or equal to numerator

The rational function f(x) , where the degrees of f(x) and g(x) are m and n respectively, with m  n, is
g(x)
considered to be an improper rational function. In this case, f(x) must first be divided by g(x) before expressing
in terms of partial fractions.

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01a STPM Math T T1.indd 40 3/28/18 4:20 PM