Page 28 - mathsvol1ch1to3ans

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Then x + 12 = A(x + 1)(x − 2) + B(x − 2) + C(x + 1) 2
14
When x = 2 C =
9
11
When x = −1 B = −
3
14
2
Comparing x coefficients, we have A = −
9
x + 12 14 14 11
Hence = − −
2
(x + 1) (x − 2) 9(x − 2) 9(x + 1) 3(x + 1) 2
2
6x − x + 1
10.
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3
2
x + x + x + 1
2
6x − x + 1 Ax + B C
Let = +
2
2
3
x + x + x + 1 x + 1 x + 1
2
2
Then 6x − x + 1 = (Ax + B)(x + 1) + (C)(x + 1)
When x = −1 C = 4
When x = 0 B = −3
2
Comparing x coefficients, we have A = 2
2
6x − x + 1 2x - 3 4
Hence = +
x + x + x + 1 x + 1 x + 1
2
2
3
2
2x + 5x − 11
11.
2
x + 2x − 3
2
2x + 5x − 11 B C
Let = A + +
2
x + 2x − 3 x − 1 x + 3
2
Then 2x + 5x − 11 = A(x − 1)(x + 3) + B(x + 3) + C(x − 1)
When x = 1 B = −1
When x = −3 C = 2
Comparing coefficients of x 2 we have A = 2
2
2x + 5x − 11 1 2
Hence = 2 − +
2
x + 2x − 3 x − 1 x + 3
7 + x
12.
2
(1 + x)(1 + x )
7 + x A Bx + C
Let = +
2
(1 + x)(1 + x ) 1 + x 1 + x 2
2
Then 7 + x = A(1 + x ) + (Bx + C)(1 + x)
When x = −1 A = 3
Comparing coefficients of x 2 we have A + B = 0
Comparing coefficients of x we have B + C = 1
Solving the above equations, we have B = −3 and C = 4
7 + x 3 4 − 3x
Hence = +
2
(1 + x)(1 + x ) 1 + x 1 + x 2