Page 93 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities

√
3
3.13 The inequality x > ax + has the solution set {x|4 <x< b} , find the values of
2
a and b .
√ √ √
2
Solution: x > ax + 3 ⇔ a( x) − x + 3 < 0 . The solution set {x|4 <x< b} is
2 2
√ √
equivalent to {x|2 < x< b} . Vieta’s formulas imply
√ 1
2+ b =
a
√ 3
2 b =
2a
a> 0
⇒ a = ,b = 36.
1
8
2
3.14 If the inequality x − ax − 6a ≤ 0 has solutions, and the two roots x 1 ,x 2 of
x − ax − 6a =0 satisfy |x 1 − x 2 |≤ 5. Find the range of the real number a .
2
2
Solution: The inequality has solutions if and only if Δ= a + 24a ≥ 0 ⇔ a ≥ 0 or a ≤−24.
Vieta’s formulas imply

x 1 + x 2 = a
x 1 x 2 = −6a
√
2
2
2
|x 1 − x 2 | = (x 1 − x 2 ) = (x 1 + x 2 ) − 4x 1 x 2 = a + 24a ≤ 5, that is
2
a + 24a − 25 ≤ 0 ⇒ (a + 25)(a − 1) ≤ 0 ⇒−25 ≤ a ≤ 1.
As a conclusion, a has the range: −25 ≤ a ≤−24 or 0 ≤ a ≤ 1.
3.15 If 0 <a < 1, 0 <b< 1, 0 <c < 1 , show it is impossible that (1 − a)b, (1 − b)c, (1 − c)a

are all greater than 1/4.

Proof 1: We prove the conclusion by contraction. Suppose
(1 − a)b> , (1 − b)c> , (1 − c)a> . Multiply them to obtain
1
1
1
4 4 4
abc(1 − a)(1 − b)(1 − c) > 1 . On the other hand, 1−a+a 2 1 , similarly
64 0 < (1 − a)a ≤ [ ] =
2 4
we have 0 < (1 − b)b ≤ , 0 < (1 − c)c ≤ ≤ . Multiply them to obtain
1 1
1 1
, 0 < (1 − c)c
0 < (1 − b)b ≤
4 4 4 4
1
abc(1 − a)(1 − b)(1 − c) ≤ , a contractions.
64
2
2
2
Proof 2: Since 0 <a < 1, 0 <b< 1, 0 <c < 1, we let a = sin α, b = sin β, c = sin γ , then
2
2
2
2
2
2
(1−a)b·(1−b)c·(1−c)a = abc(1−a)(1−b)(1−c) = sin α sin β sin γ cos α cos β cos γ =
2
2
2
2
2
2
1
2
2
1 2 (1−a)b·(1−b)c·(1−c)a = abc(1−a)(1−b)(1−c) = sin α sin β sin γ cos α cos β cos γ =
64 sin 2α sin 2β sin 2γ ≤ 64 1 , thus it is impossible that (1 − a)b, (1 − b)c, (1 − c)a are all
2
2
2
1
64 sin 2α sin 2β sin 2γ ≤ 64
greater than 1/4.
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