PastExamLabPastExamLab

台灣大學 · 生物環境系統工程學系 · 轉學考考古題 · 民國114年（2025年）

114 年度 微積分(B)

台灣大學 · 生物環境系統工程學系 · 轉學考

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Each answer counts for 5 points.
第 1 題5 分
lim⁡x→0+(1−sin⁡xx)1lnx=\lim\limits_{x \to 0^+} \left(1 - \frac{\sin x}{x}\right)^{\frac{1}{lnx}} = __(1)__.
第 2 題15 分
Let f(x)=∫1x−1e−(t−1)2dtf(x) = \int_1^{x-1} e^{-(t-1)^2} dt.
(a)5 分
Find the Taylor series for e−x2e^{-x^2} centered at a=0a = 0. __(2)__
(b)5 分
Find the Taylor series for f(x)f(x) centered at a=2a = 2. __(3)__
(c)5 分
lim⁡x→2f(x)−arctan⁡(x−2)(x−2)4ln⁡(2x−3)=\lim\limits_{x \to 2} \frac{f(x) - \arctan(x - 2)}{(x - 2)^4 \ln(2x - 3)} = __(4)__.
第 3 題10 分
Suppose that near (x,y)=(1,−1)(x, y) = (1, -1) the equation
ex2+y−ln⁡(xy2)=1e^{x^2+y} - \ln\left(\frac{x}{y^2}\right) = 1
defines yy as a twice differentiable function of xx which is denoted by y(x)y(x).
(a)5 分
The linearization of y(x)y(x) at x=1x = 1 is __(5)__.
(b)5 分
y′′(1)=y''(1) = __(6)__.
第 4 題5 分
∫024−xxdx=\int_0^2 \sqrt{\frac{4-x}{x}} dx = __(7)__.
第 5 題15 分
Suppose that f(x,y,z)f(x, y, z) is differentiable near (x,y,z)=(0,1,2)(x, y, z) = (0, 1, 2) and f(0,1,2)=10f(0, 1, 2) = 10. Assume that curves
r1(t)=(3t,e2t,2cos⁡t),\mathbf{r}_1(t) = (3t, e^{2t}, 2\cos t),
r2(t)=(ln⁡t,t2,2t),\mathbf{r}_2(t) = (\ln t, t^2, \frac{2}{t}),
lie on the level surface f(x,y,z)=10f(x, y, z) = 10 and fy(0,1,2)=32f_y(0, 1, 2) = \frac{3}{2}.
(a)5 分
The tangent plane to f(x,y,z)=10f(x, y, z) = 10 at (x,y,z)=(0,1,2)(x, y, z) = (0, 1, 2) is __(8)__.
(b)5 分
Let
g(x,y,z)={∣z−2∣sin⁡(πx)∣x∣+2∣y−1∣+∣z−2∣,for (x,y,z)≠(0,1,2)0,for (x,y,z)=(0,1,2),g(x, y, z) = \begin{cases} \frac{|z - 2|\sin(\pi x)}{|x| + 2|y - 1| + |z - 2|}, & \text{for } (x, y, z) \neq (0, 1, 2) \\ 0, & \text{for } (x, y, z) = (0, 1, 2) \end{cases},
and u=∇f(0,1,2)∣∇f(0,1,2)∣\mathbf{u} = \frac{\nabla f(0, 1, 2)}{|\nabla f(0, 1, 2)|}. Then Dug(0,1,2)=D_{\mathbf{u}}g(0, 1, 2) = __(9)__.
(c)5 分
Assume that f(x,y,z)f(x, y, z) attains the maximum value at (x,y,z)=(0,1,2)(x, y, z) = (0, 1, 2) when restricted to a level surface h(x,y,z)=1h(x, y, z) = 1, where h(x,y,z)h(x, y, z) is differentiable and hz(0,1,2)=6h_z(0, 1, 2) = 6. Using linear approximation, estimate the maximum value of f(x,y,z)f(x, y, z) subject to the nearby level surface h(x,y,z)=0.9h(x, y, z) = 0.9. The estimated value is approximately __(10)__.
第 6 題30 分
Suppose that f(x)f(x) is continuous on [0,6][0, 6] and f(0)=f(4)=0f(0) = f(4) = 0. The graph of f′(x)f'(x) is given as below, but values of f′(1)f'(1), f′(2)f'(2) and f′(4)f'(4) are not determined. It is known that lim⁡x→4−f′(x)=∞\lim\limits_{x \to 4^-} f'(x) = \infty and lim⁡x→4+f′(x)=−∞\lim\limits_{x \to 4^+} f'(x) = -\infty.
第 6 題圖表
(a)10 分
Find lim⁡x→2+f(x)−f(2)x−2\lim\limits_{x \to 2^+} \frac{f(x) - f(2)}{x - 2} and lim⁡x→2−f(x)−f(2)x−2\lim\limits_{x \to 2^-} \frac{f(x) - f(2)}{x - 2} by the Mean Value Theorem. Is f(x)f(x) differentiable at x=2x = 2? Justify your answers.
(b)5 分
List all critical numbers of f(x)f(x) in the interval (0,6)(0, 6). Where does f(x)f(x) attain a local maximum? Where does f(x)f(x) attain a local minimum?
(c)8 分
Find intervals on which y=f(x)y = f(x) is concave upward. Find intervals on which y=f(x)y = f(x) is concave downward. Find the inflection points of f(x)f(x).
(d)6 分
Sketch the graph of f(x)f(x) for x∈[0,6]x \in [0, 6].
(e)6 分
Suppose that f′(x)=32x2+2x−6x+18f'(x) = \frac{32}{x^2 + 2x - 6x + 18} for x>6x > 6. Find lim⁡x→∞f(x)−f(6)\lim\limits_{x \to \infty} f(x) - f(6).
第 7 題20 分
Let S\mathcal{S} be the part of the graph
z=cos⁡xsin⁡y,0≤x≤π2,0≤y≤π,z = \cos x \sin y, \quad 0 \leq x \leq \frac{\pi}{2}, \quad 0 \leq y \leq \pi,
with upward orientation. Consider the vector field F(x,y,z)=zi+yzk\mathbf{F}(x, y, z) = z\mathbf{i} + yz\mathbf{k}.
(a)10 分
Parametrize the surface S\mathcal{S} and evaluate ∬ScurlF⋅dS\iint_{\mathcal{S}} \text{curl}\mathbf{F} \cdot d\mathbf{S} directly.
(b)10 分
Let CC be the boundary curve of S\mathcal{S}. Decompose CC as the union of differentiable parametric curves and evaluate ∮CF⋅dr\oint_C \mathbf{F} \cdot d\mathbf{r} directly. Show that the line integral equals the surface integral in part (a) which is consistent with Stokes' Theorem.
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