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作者: 拉麵仔 時間: 17/2/2009 09:56 PM
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作者: sapphire 時間: 17/2/2009 11:52 PM
Note that sinθ lies between 1 and -1.

x² + 2x + k = 0 has root.
2² - 4k ≥ 0
k ≤ 1

Since sinθ is a root of the equation, (sinθ)²+ 2sinθ + k = 0 holds for some real values k.
(sinθ)² + 2sinθ + k = 0
-sinθ(sinθ + 2) = k

-1 ≤ sinθ ≤ 1
1 ≤ sinθ + 2 ≤ 3
-1 ≤ sinθ(sinθ + 2)          or             sinθ(sinθ + 2) ≤ 3
1 ≥ -sinθ(sinθ + 2)          or             -sinθ(sinθ + 2) ≥ -3
1 ≥ k                                     or             k ≥ -3
Combine the two conditions, the range is -3 ≤ k ≤ 1.

[ 本帖最後由 sapphire 於 17/2/2009 11:53 PM 編輯 ]

作者: 超夢夢的夢境 時間: 18/2/2009 12:56 AM
Note that sinθ lies between 1 and -1.

x² + 2x + k = 0 has root.
2² - 4k ≥ 0
k ≤ 1

Since sinθ is a root of the equation, (sinθ)²+ 2sinθ + k = 0 holds for some real values k.
(sinθ)² + 2sinθ + k = 0
-sinθ(sinθ + 2) = k

-1 ≤ sinθ ≤ 1
1 ≤ sinθ + 2 ≤ 3
以上正確
答案也正確
不過看不到怎樣 conclude 以下結果,因為 sinθ 跟 sinθ+2 不是 independent
-1 ≤ sinθ(sinθ + 2) and sinθ(sinθ + 2) ≤ 3

(註:如果是 independent 的話,算式如下)
(-1 ≤ x ≤ 1
1 ≤ y ≤ 3
For -1 ≤ x ≤ 0,
-3 ≤ xy ≤ 0

For 0 ≤ x ≤ 1,
0 ≤ xy ≤ 3,

=> -3 ≤ xy ≤ 3)

回正題
(sinθ)² + 2sinθ + k = 0
k= 1 - (sinθ+1)²
0 ≤ sinθ+1 ≤ 2
0 ≤ (sinθ+1)² ≤ 4
-3 ≤ 1 - (sinθ+1)² ≤ 1
-3 ≤ k ≤ 1
Combine the two conditions, the range is -3 ≤ k ≤ 1.

作者: sapphire 時間: 18/2/2009 07:18 AM
啊, 竟然把AND打錯了OR

sinθ和sinθ + 2不是independent, 但這兩個term也能同時去到自己的max或是min, 而且也是同時increase/decrease, 兩者同時去到min/max的時候也是sinθ(sinθ + 2)去到min/max的時候
不會出現 sinθ = -1 AND sinθ + 2 = 3 的情況

其實這個還要用differentiation prove min = -1的, 正常情況下用回你的completing square比較好

[ 本帖最後由 sapphire 於 18/2/2009 07:40 AM 編輯 ]

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