Page 31 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I reAl numBers

Solution 2: Obviously S = x + y > 0 (since x, y cannot be both zero due to
2
2
√ √
2
2
4x − 5xy +4y =5). Let x = S cos θ, y = S sin θ , and substitute into
4x − 5xy +4y =5:
2
2
5 8S − 10
2 2 . Since
4S cos θ − 5S cos θ sin θ +4S sin θ =5 ⇒ 4S − sin 2θ =5 ⇒ sin 2θ =
2 5S
8S − 10 10 10
8S − 10 .
| sin 2θ|≤ 1, then ≤ 1 ⇒−1 ≤ ≤ 1 ⇒ ≤ S ≤
5S 5S 13 3

√
1.83 For a positive integer n , find the integer part of ( n +2n + n) .
2
2
Solution: For a positive integer n , we have
√
2 √
2
2
2
2
n <n +2n<n +2n +1 = (n + 1) ⇒ n< n +2n<n +1 ⇒ 0 <
2
2
2
2
2
√ n <n +2n<n +2n +1 = (n + 1) ⇒ n< n +2n<n +1 ⇒ 0 <
2 √
n +2n − n< 1 √ 2 2 √ 2 2
n +2n − n< 1 . Let x =( n +2n + n) ,y =( n +2n − n) , then
2
√ √ √
2
2
2
2
2
2
x + y =( n +2n + n) +( n +2n − n) =4n +4n . Since 0 < n +2n − n< 1,
√
then 0 < ( n +2n − n) < 1, then
2
2
√ √
( n +2n + n) =4n +4n − ( n +2n − n) ∈ (4n +4n − 1, 4n +4n), thus the integer
2
2
2
2
2
2
2
√
part of ( n +2n + n) is 4n +4n − 1.
2
2
2

.

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