Page 27 - mathsvol1ch1to3ans

27

(x − 1) 2
6.
3
x + x
(x − 1) 2 A Bx + C
Let = +
x + x x (x + 1)
3
2
2
2
Then (x − 1) = A(x + 1) + (Bx + C)(x)
When x = 0 A = 1
2
Comparing x coefficients, we have B = 0
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Comparing x coefficients, we have C = −2

(x − 1) 2 1 2
Hence = −
2
x + x x (x + 1)
3
2
x + x + 1
7.
2
x − 5x + 6
2
x + x + 1 B C
Let = A + +
2
x − 5x + 6 x − 3 x − 2
2
Then x + x + 1 = A(x − 3)(x − 2) + B(x − 2) + C(x − 3)
When x = 2, C = −7

When x = 3, B = 13

When x = 4, A = 1
2
x + x + 1 13 7
Hence = 1 + −
2
x − 5x + 6 x − 3 x − 2
3
x + 2x + 1
8.
2
x + 5x + 6
3
x + 2x + 1 C D
Let = Ax + B + +
2
x + 5x + 6 x + 3 x + 2
3
2
Then x + 2x + 1 = (Ax + B)(x + 5x + 6) + C(x + 2) + D(x + 3)
When x = −2, D = −11
When x = −3, C = 32

1 − 3D − 2C
When x = 0, B = = −5
6
3
By comparing the coefficients of x we have A = 1.
3
x + 2x + 1 32 11
Hence = x − 5 + −
x + 5x + 6 x + 3 x + 2
2
1
x + 12
9.
2
(x + 1) (x − 2)
x + 12 A B C
Let = + +
2
(x + 1) (x − 2) x + 1 (x + 1) 2 x − 2