數學寒假作業三問

Q.1
(Please draw a diagram if you cannot understand below.)
Side BC = 4
Let P be a point on BC such that AP┴BC.
Let BP = x, then CP = 4 - x.
Also, let AP = height = h.
Then:
　tan B = h / x
　tan C = h / (4 - x)
　tan B + tan C = [h / x] + [h / (4 - x)]
　　　　　　　 = 4h / [x (4 - x)]
Now, tan B + tan C = 3.
So,
　4h / [x (4 - x)] = 3
　h = (3/4) x (4 - x)
　　= (-3/4) [(x - 2)² - 4]
　　= (-3/4) (x - 2)² + 3
h(max) = 3 , as -(x - 2)≤0 for any x.
So,
maximum area = ½ (4) (3) = 6.

======================
Q.2 and Q.3 follow below...

2007-02-20 21:49:24 補充：
Q.2y = cos² x－cos x＋2　= (cos x－½)²＋7/4As x increases from (2nπ) to (2nπ＋π/3),　(cos x) decreases from 1 to ½　and, y decreases from 2 to 7/4.As x increases from (2nπ＋π/3) to (2nπ＋π),　(cos x) decreases from ½ to -1　and, y increases from 7/4 to 4.

2007-02-20 21:49:46 補充：
Q.2 contd:As x increases from (2nπ＋π) to [2(n＋1)π－π/3],　(cos x) increases from -1 to ½　and, y decreases from 4 to 7/4.As x increases from (2(n＋1)π－π/3] to [2(n＋1)π],　(cos x) increases from ½ to 1　and, y increases from 7/4 to 2.

2007-02-20 21:53:47 補充：
Q.2 contd:In other words,y is increasing for　(2nπ＋π/3)＜x＜(2nπ＋π)　and　(2(n＋1)π－π/3]＜x＜[2(n＋1)π],y is decreasing for　(2nπ)＜x＜(2nπ＋π/3)　and　(2nπ＋π)＜x＜[2(n＋1)π－π/3],for any integer n.

2007-02-20 22:05:39 補充：
Sorry, no idea yet for Q.3.Would you mind opening a new thread for Q.3?I believe someone here can answer you there.

瞭解一次、多一次的保障。尋找八大行業工作。必須謹慎小心一點。
小巴常說:保險跟冒險只差一個字，意義卻是大不同!
一通電話、一次詢問。都是為自己的著想。