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台灣大學 · 物理學系 · 轉學考考古題 · 民國108年（2019年）

108 年度 微積分(B)

台灣大學 · 物理學系 · 轉學考

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第 1 題8 分
Compute the following limits
(•)4 分
lim⁡x→0(cos⁡x+12x2)3x=\lim\limits_{x \to 0} \left(\cos x + \frac{1}{2}x^2\right)^{3x} = __(1)__
(•)4 分
lim⁡x→03x−sin⁡3x5x−tan⁡5x=\lim\limits_{x \to 0} \frac{3x - \sin 3x}{5x - \tan 5x} = __(2)__
第 2 題4 分
The graph of f(x)=(x+1)2/3(x−2)1/3f(x) = (x+1)^{2/3}(x-2)^{1/3} has an inflection point at x=x = __(3)__
第 3 題4 分
If x5+y5=33x^5 + y^5 = 33, then d2ydx2∣x=1=\frac{d^2y}{dx^2}\Big|_{x=1} = __(4)__
第 4 題4 分
If ∫12x+1f(t)ttdt=tan⁡−1x\int_1^{2x+1} \frac{f(t)}{t^t} dt = \tan^{-1} x, then f(3)=f(3) = __(5)__
第 5 題12 分
Calculate the following integrals:
(•)4 分
∫02x3(x2+4)3dx=\int_0^2 \frac{x^3}{(x^2+4)^3} dx = __(6)__
(•)4 分
∫0ln⁡2ex−1dx=\int_0^{\ln 2} \sqrt{e^x-1} dx = __(7)__ (Hint: Use u=ex−1u = \sqrt{e^x-1})
(•)4 分
∫12x−2x2−1dx=\int_1^2 \frac{x-2}{\sqrt{x^2-1}} dx = __(8)__
第 6 題4 分
The integral ∫0∞(tan⁡−1x)4xadx\int_0^{\infty} \frac{(\tan^{-1} x)^4}{x^a} dx converges if and only if aa is in the interval __(9)__
第 7 題4 分
Suppose that y′=xy−xy' = xy - x with y(0)=2y(0) = 2, then y(2)=y(2) = __(10)__
第 8 題4 分
Suppose that y′+y=2cos⁡xy' + y = 2\cos x with y(0)=2y(0) = 2, then y(π6)=y\left(\frac{\pi}{6}\right) = __(11)__
第 9 題4 分
Let RR be the region below the curve y=sin⁡2xy = \sin^2 x when 0≤x≤π0 \leq x \leq \pi and VV be the volume of the solid obtained by rotating RR about the yy-axis. Then V=V = __(12)__
第 10 題8 分
Find the sum
(•)4 分
∑n=2∞(n−2)!+2nn!=\sum_{n=2}^{\infty} \frac{(n-2)!+2^n}{n!} = __(13)__
(•)4 分
∑n=1∞12n−1(13)n=\sum_{n=1}^{\infty} \frac{1}{2n-1}\left(\frac{1}{\sqrt{3}}\right)^n = __(14)__
第 11 題4 分
The 3th nonzero term in the Maclaurin series of ln⁡(2x3+5)\ln(2x^3 + 5) is __(15)__
第 12 題4 分
The 3rd nonzero term of the Mclaurin series of the function f(x)={csc⁡x−cot⁡x,x≠0,0,x=0,f(x) = \begin{cases} \csc x - \cot x, & x \neq 0, \\ 0, & x = 0, \end{cases} is __(16)__
第 13 題8 分
Let r⃗(t)=(etcos⁡t,etsin⁡t,et)\vec{r}(t) = (e^t \cos t, e^t \sin t, e^t), −1≤t≤1-1 \leq t \leq 1.
(•)4 分
The length of the curve is __(17)__
(•)4 分
The curvature at the point t=0t = 0 is __(18)__
第 14 題4 分
The shortest distance from the origin to the paraboloid z=x2+2y2−364z = \frac{x^2 + 2y^2 - 36}{4} is __(19)__
第 15 題4 分
∫01∫02∫y/21yzcos⁡(x3−1)dx dy dz=\int_0^1 \int_0^2 \int_{y/2}^1 yz\cos(x^3-1) dx \, dy \, dz = __(20)__
第 16 題4 分
The volume of the solid described by x2+y2≤1x^2 + y^2 \leq 1 and x2+y2+z2≤4x^2 + y^2 + z^2 \leq 4 is __(21)__
第 17 題4 分
The area of the surface x2+y2+z2=4x^2 + y^2 + z^2 = 4, (x−1)2+y2≤1(x-1)^2 + y^2 \leq 1, is __(22)__
第 18 題4 分
Let F⃗(x,y,z)=(sin⁡y,xcos⁡y+cos⁡z,−ysin⁡z)\vec{F}(x,y,z) = (\sin y, x\cos y + \cos z, -y\sin z), C:r⃗(t)=(sin⁡t,cos⁡t,2t)C: \vec{r}(t) = (\sin t, \cos t, 2t), 0≤t≤π0 \leq t \leq \pi. ∫CF⃗⋅dr⃗=\int_C \vec{F} \cdot d\vec{r} = __(23)__
第 19 題4 分
Let CC be the curve consisting of line segments from (0,0)(0,0) to (2,1)(2,1) to (1,2)(1,2) to (0,0)(0,0). ∫C(x−y)dx+(2x+y)dy=\int_C (x-y) dx + (2x+y) dy = __(24)__
第 20 題4 分
The flux of the vector field F⃗=(sin⁡y+x2,3y2+ex,3zy2)\vec{F} = (\sin y + x^2, 3y^2 + e^x, 3zy^2) through the surface x2+y2+z2=1x^2 + y^2 + z^2 = 1, z≥0z \geq 0, where the surface is equipped with the upward normal, is __(25)__
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