※ 引述《EricDampier (蛋皮)》之铭言： : 监试人员一时糊涂...竟然说考卷给你们带回去当礼物 : 後来第二节他去问才知道试卷也要收回 : 所以就被拿回来当礼物了... : 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.计算,证明 : 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 -------------- .................(∞／∞) 罗必达 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 --

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