Page 25 - mathsvol1ch1to3ans

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(x + 3)(x − 2)
Interval Sign of (x + 3) Sign of (x − 2) Sign of (x + 4) Sign of ≥ 0
5(x + 4)

(−∞, −4) − − − −

(−4, −3) − − + +

(−3, 2) + − + −

(2, ∞) + + + +
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So the inequality is satisfied in [−4, 3] ∪ [2, ∞].

Exercise - 2.14

Resolve the following rational expressions into partial fractions.
1 3x + 1 x x 1
1. 2. 3. 4. 5.
2
4
2
x − a 2 (x − 2)(x + 1) (x + 1)(x − 1)(x + 2) (x − 1) 3 x − 1
2
3
1
2
(x − 1) 2 x + x + 1 x + 2x + 1 x + 12 6x − x + 1
6. 7. 8. 9. 10.
3
2
2
3
2
2
x + x x − 5x + 6 x + 5x + 6 (x + 1) (x − 2) x + x + x + 1
2
3
2x + 5x − 11 7 + x x − 1
11. 12. 13. .
2
2
2
x + 2x − 3 (1 + x)(1 + x ) x + x + 1
1
1.
2
x − a 2
1 A B
Let = +
x − a 2 x − a x + a
2
Then 1 = A(x + a) + B(x − a)
1 1
When x = a, A = When x = −a, B = −
2a 2a
1 1 1
Hence = −
x − a 2 (2a)(x − a) (2a)(x + a)
2
3x + 1
2.
(x − 2)(x + 1)
3x + 1 A B
Let = +
(x − 2)(x + 1) x − 2 x + 1
Then 3x + 1 = A(x + 1) + B(x − 2)
7 2
When x = 2, A = When x = −1, B =
3 3
3x + 1 7 2
Hence = +
(x − 2)(x + 1) 3(x − 2) 3(x + 1)
x
3.
(x + 1)(x − 1)(x + 2)
2
x Ax + B C D
Let = + +
2
2
(x + 1)(x − 1)(x + 2) x + 1 x − 1 x + 2
2
2
Then x = (Ax + B)(x − 1)(x + 2) + (C)(x + 1)(x + 2) + (D)(x + 1)(x − 1)