[理工] [微積分]-台大91-考古題核對答案
題目連結http://www.lib.ntu.edu.tw/exam/graduate/91/91049.htm
懇請諸位好手幫忙核對看看...特別是第10題和11題，感恩!
1. √5/5
2. n!/x
3. 4/3
4. 4/3
5. (1/5)ln6
6. -9/14
7. 1/3
8. 52/9
9. x-y-2√2=0
10.
(1) 令f(x)=tanx a,b為[π/4,π/2)任意兩點，且b>a
由MVT知(tanb-tana)/(b-a)=f'(c)=(secc)^2 c為[π/4,π/2)內之一點
故tanb-tana>b-a>0 => tanb-b>tana-a>0 =>tanx-x>0
所以tanx>x 當x在[π/4,π/2)
(2) f'(x)=(xcosx-sinx)/x^2=[cosx/x^2]*(x-tanx)
由(1)知x-tanx<0，又cosx≧0 當x在[π/4,π/2]
所以f'(x)<0，當x在[π/4,π/2]時
=>f(x)=sinx/x在此區間為遞減函數
(3) 依據推文高手們給的提示...我試著寫了如下，請看這樣對不對!!
method I
令f(x)=sinx/x，由2得
f(π/2)≦f(x)≦f(π/4) 當x在[π/4,π/2]
a a a
2/π∫dx≦∫f(x)dx≦2√2/π∫dx a=π/2, b=π/4
b b b
=> 1/2≦∫f(x)dx≦√2/2
method II 由積分均值定裡
a
∫f(x)dx=f(c)*(π/2-π/4) ,c為[π/4,π/2]上一點
b
又f(π/2)(π/2-π/4)≦f(c)*(π/2-π/4)≦f(π/4)(π/2-π/4)
得證1/2≦∫f(x)dx≦√2/2
11. max √6/18, min-√6/18
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 61.230.113.150
推
02/23 19:05, , 1F
02/23 19:05, 1F
→
02/23 19:09, , 2F
02/23 19:09, 2F
推
02/23 19:32, , 3F
02/23 19:32, 3F
推
02/23 20:08, , 4F
02/23 20:08, 4F
→
02/23 20:08, , 5F
02/23 20:08, 5F
推
02/23 20:11, , 6F
02/23 20:11, 6F
→
02/23 20:11, , 7F
02/23 20:11, 7F
※ 編輯: TreeMan 來自: 61.230.112.70 (02/23 21:29)
推
02/23 21:35, , 8F
02/23 21:35, 8F
→
02/23 21:35, , 9F
02/23 21:35, 9F
→
02/23 21:36, , 10F
02/23 21:36, 10F
→
02/23 21:39, , 11F
02/23 21:39, 11F
→
02/23 21:42, , 12F
02/23 21:42, 12F
→
02/23 21:44, , 13F
02/23 21:44, 13F
→
02/23 21:45, , 14F
02/23 21:45, 14F
推
02/23 21:46, , 15F
02/23 21:46, 15F
推
02/23 22:44, , 16F
02/23 22:44, 16F
→
02/23 22:46, , 17F
02/23 22:46, 17F
推
02/23 23:00, , 18F
02/23 23:00, 18F
推
02/23 23:35, , 19F
02/23 23:35, 19F
→
02/24 07:25, , 20F
02/24 07:25, 20F
推
01/30 16:08, , 21F
01/30 16:08, 21F