Page 102 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities

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Proof 2: a + b + c =1 ⇒ (a + b + c) =1 ⇒ a + b + c =1 − 2(ab + bc + ca) (i).
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a + b ≥ 2ab, b + c ≥ 2bc, c + a ≥ 2ca , add them up to obtain 2(a + b + c ) ≥ 2(ab + bc + ca)
2(a + b + c ) ≥ 2(ab + bc + ca)(ii).
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(i)+(ii)⇒ 3(a + b + c ) ≥ 1 ⇒ a + b + c ≥ .
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3.44 The function f(x) is defined on [0, 1], and f(0) = f(1) . For any distinct
x 1 ,x 2 ∈ [0, 1], we have |f(x 2 ) − f(x 1 )| < |x 2 − x 1 | . Show |f(x 2 ) − f(x 1 )| < .
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Proof: Let 0 ≤ x 1 <x 2 ≤ 1. We consider two cases:
1) If x 2 − x 1 ≤ , then |f(x 2 ) − f(x 1 )| < |x 2 − x 1 |≤ .
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2) If x 2 − x 1 > , then from f(0) = f(1) we obtain
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|f(x 2 ) − f(x 1 )| = |f(x 2 ) − f(1) + f(0) − f(x 1 )|≤|f(x 2 ) − f(1)| + |f(0) − f(x 1 )|
|f(x 2 ) − f(x 1 )| = |f(x 2 ) − f(1) + f(0) − f(x 1 )|≤|f(x 2 ) − f(1)| + |f(0) − f(x 1 )|. Hence, (1 − x 2 ) + (x 1 − 0) = 1 − (x 2 − x 1 ) < 1 2

(1 − x 2 ) + (x 1 − 0) = 1 − (x 2 − x 1 ) < .
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3.45 The equation |x| = ax +1 has one negative root but no positive root, find the
range of the parameter a .
Solution 1: Let x be the negative root of the equation |x| = ax +1 , then
−x = ax +1 ⇒ x = −1 < 0, thus a +1 > 0. Equivalently, when a> −1, the equation has
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a negative root.
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Suppose the equation has a positive root x, then x = ax +1 ⇒ x = 1−a > 0 , thus a< 1.
As a conclusion, the condition that the equation has one negative root but no positive root
is equivalent to a> −1 holds but a< 1 fails, that is, a ≥ 1.

Solution 2: Another approach is to plot the functions y = |x| and y = ax +1 on the
Cartesian plane. This will directly give us the same conclusion.

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3.46 Solve the inequality 3x −4x−23 > 2.
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x −9
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Solution: 3x −4x−23 > 2 ⇔ x −4x−5 > 0 ⇔ (x+1)(x−5) > 0. This inequality is equivalent to
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x −9 x −9 (x+3)(x−3)
(x + 3)(x + 1)(x − 3)(x − 5) > 0 whose solution set is (−∞, −3) ∪ (−1, 3) ∪ (5, +∞).

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