Re: [考古] 台聯大 96 搶先報~
※ 引述《EricDampier (蛋皮)》之銘言：
: 監試人員一時糊塗...竟然說考卷給你們帶回去當禮物
: 後來第二節他去問才知道試卷也要收回
: 所以就被拿回來當禮物了...
: 甲.填充
: 1. if f is a continuous function such that
: x x
: ∫f(t)dt = x．exp(2x) + ∫exp(-t)f(t)dt for all x,
: 0 0
: find an explicit formula for f(x)
: 2. in what direction is the derivative of
: (x^2+y^2)
: f(x,y) = ───── at P(1,1) equal to zero?
: (x^2-y^2)
: 3. find the maximum value of x^2 + y^2 subject to the constraint
: x^2 - 2x + y^2 - 4y = 0
: 4. suppose that f(0) = -3 and f'(x) <= 5 for all values of x
: how large can f(2) possibly be?
: 5. find the tangent plane of the surface
: cos(πx) - x^2．y + exp(xz) + y．z = 4
: x-2y
: 6. evaluate ∫∫ ─── dA , R is the parallelogram enclosed by the lines
: R 3x-y
: x-2y = 0 , x-2y = 4 , 3x-y = 1 , 3x-y = 8
: 7. find the area of surface cut from parabloid x^2 + y^2 - z = 0 by the
: plane z = 2
: 8. evaluate ∮(6y+x)dx+(y+2x)dy , C : (x-2)^2 + (y-3)^2 = 4
: C
: 乙.計算,證明
: 1. evaluate the following limits
: tan(2x) n √(n^2 - j^2)
: (a) lim （tan x） (b) lim （Σ ───────）
: x→(π/4)- n→∞ j=1 n^2
1.(a)
tan(2x)ln（tan x）
lim e
x→(π/4)-
lim tan(2x)ln（tan x）
x→(π/4)-
= e
ln（tan x）
lim -------------- .................(∞／∞) 羅必達
x→(π/4)- cot(2x)
= e
sec^2(x) 1
lim ------------- --------------
x→(π/4)- tan x -2csc^2(x)
= e
1 cos(x) sin^2(x)
lim (----------) * (-----------) * (------------)
x→(π/4)- cos^2(x) sin(x) -2
= e
-1 sin(x)
lim (---) * (----------)
x→(π/4)- 2 cos(x)
= e
-1
(-----)
2
= e
: ∞ n ln(n)
: 2. (a) test the series Σ (-1) ──── for convergence or divergence
: n=1 n-ln(n)
: ∞ x^n
: (b) let f(x) = Σ ── find the intervals of convergence for f' & f''
: n=1 n^2
: 3. evalute
: a/√2 √(a^2-y^2)
: (a) ∫ ∫ exp(x^2+y^2) dxdy
: 0 y
: 8 2 dydx
: (b) ∫ ∫ ────
: 0 x^(1/3) y^4 + 1
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 140.113.138.35
※ 編輯: ROGER2004070 來自: 140.113.138.35 (07/14 21:02)
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07/14 21:12, , 1F
07/14 21:12, 1F
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07/05 18:15, , 2F
07/05 18:15, 2F
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