2017 年 4 月 22 日
姓名： 學號： 學系：
說明:
1. 本試題含封面共 7 頁，8 大題。
2. 考試時間 100 分鐘。
3. 請在每個試題所屬的頁面作答。如欲使用試題背面，請標示清楚。
4. 如果題目附有答案欄，請將答案寫在答案欄上。
5. 清楚地寫出計算及證明的過程，沒有過程的答案將不予記分。
題號 配分 分數 1 20
2 15
3 10
4 10
5 10
6 10
7 15
8 10
總分 100
1. 選擇題 (每題選出一個正確答案) (a) (5 points) 若 ex/y = x + y, 求 dy
dx =?
(A) xy
x2+ xy + y2 (B) xex/y+ y2
yex/y− y2 (C) xy
xex/y− y2 (D) y
x − 2
xy2ex/y (E) yex/y+ y2
xex/y− y2
(b) (5 points) 令 f (x) = 2x3− 9x2+ 12x. 選出一個正確選項
(A) f (x) 有一個局部極大值在 x = 1, 一個絕對極大值在 x = 2, 和一個反曲 點在 x = 3/2;
(B) f (x) 有局部極大值在 x = 1 和 x = 3/2, 及一個局部極小值在 x = 2;
(C) f (x) 有局部極小值在 x = 1 和 x = 2, 及一個反曲點在 x = 3/2;
(D) f (x) 有一個局部極大值在 x = 1, 一個局部極小值在 x = 2, 和一個反曲 點在 x = 3/2.
(E) f (x) 有反曲點在 x = 1, x = 2 和 x = 3/2;
(c) (5 points) 令 f (x) 連續且可微分在 [−1, 2] 區間上. 已知 f(−1) = −5 和 f(2) = 7.
下列哪一個敘述是「錯」的?
(A) f (x) 有一個絕對極大值在 [−1, 2]���;
(B) 存在一個點 c 介於-1 和 2 之間, 使得 f (c) = 0;
(C) 存在一個點 c 介於-1 和 2 之間, 使得 f′(c) = 4;
(D) 存在一個點 c 介於-1 和 2 之間, 使得 f′(c) > 0 ;
(E) 上述 (A), (B), (C), (D) 選項皆有可能因 f (x) 的其他性質不同而產生 錯誤.
(d) (5 points) 級數
∑∞ n=1
(−1)n (n2+ 1)p 是 (A) 絕對收斂當 p≥ 12;
(B) 絕對收斂當 p > 1 和條件收斂當 0 < p≤ 1;
(C) 絕對收斂當 0 < p≤ 12 和條件收斂當 p > 12; (D) 發散的對於任意 p;
(E) 絕對收斂當 p > 12 和條件收斂當 0 < p≤ 12.
Solution: (a) A (b) D (c) E (d) E
2. 計算
(a) (5 points) lim
x→−∞
( x x + 2
)x
(a)
Solution: Let y =( x x + 2
)x
. Then ln y = (
lnx+2x
)
1 x
. By L’Hopital’s Law,
xlim→∞ln y = lim
x→∞
(
lnx+2x )
1 x
= lim
x→∞
x+2
x · (x+2)2 2
−x12 = lim
x→∞− 2x
x + 2 =−2.
Hence, lim
x→∞y = lim
x→∞eln y = e−2. (b) (5 points)
∫ π
4
0
x cos x dx.
(b) Solution: Let u = x and dv = cos xdx. Using the integration by parts,
∫ π
4
0
x cos x dx = x sin x π4
0 −
∫ π
4
0
sin x dx = x sin x π4
0 + cos x π4
0
= π 4 sinπ
4 − 0 + cosπ
4 − cos 0 = π√ 2 8 +
√2 2 − 1
(c) (5 points) 寫出函數圖形 y = xcos(πx) 於 x = 3 的切線方程式.
(c) Solution: Let y = f (x) = xcos(πx). Then ln y = cos(πx)· ln x. Hence
d
dx(ln y) = 1 y
dy
dx = π· [− sin(πx) · ln s] + cos(πx) · 1 x We have
dy
dx = xcos(πx)[cos(πx)
x − π ln x · sin(πx)]
Also, f (3) = 3−1 = 13 and the slope of the tangent line at x = 3 is dydx
x=3 =−19. Therefore, the equation of the tangent lineis
y =−1 9x + 2
3.
3. (10 points) 令 f (x) =
∫x2
x et2 dt
x− 1 當 x̸= 1. 試給出一個值 f(1) 使得 f 連續在 x = 1.
3.
Solution: Let g(x) = ∫x2
x et2 dt, then g(1) = 0. By the Fundamental Theorem of Calculus,
g′(x) = d dx
( ∫ x2
1
et2 dt−
∫ x 1
et2 dt )
= 2xe(x2)2− ex2
= 2xex4 − ex2 Consider lim
x→1f (x) = lim
x→1
g(x)− g(1)
x− 1 = g′(1) = e. Hence, if we define f (1) = e, then f is continuous at x = 1.
4. (10 points) 已知 f (0) = 0, f′(0) = 1和 |f′′(x)| ≤ 2 對於所有 x ∈ [0, 3]. 求 f(3) 最大可 能的值.
4.
Solution: By Mean Value Theorem, for every t ∈ [0, 3], there is c ∈ (0, t) such that
|f′(t)− f′(0)| = |f′′(c)||t − 0| ≤ 2t.
Hence, f′(t)≤ f′(0) + 2t = 1 + 2t. By the Fundamental Theorem of Calculus (Type II),
f (3) = f (0) +
∫ 3
0
f′(t) dt≤ 0 +
∫ 3
0
1 + 2t dt = 12 Hence, the largest possible value of f (3) is 12.
(In fact, if f (x) = x + x2, then f (3) = 12.)
5. (10 points) 試證明 1− x
1 + x < e−2x 對於所有 x∈ (0, 1).
Solution: Let f (x) = (1 + x)− (1 − x)e2x. Then f′(x) = 1−[
− e2x+ 2(1− x)e2x]
= 1− (1 − 2x)e2x and
f′′(x) =−[
2e2x(1− 2x) − 2e2x]
= 2(1 + 2x)e2x.
Hence, for every x∈ (0, 1), f′′(x) > 0 and this implies that f′(x) is strictly increasing on (0, 1). We have f′(x) > 0 for every x∈ (0, 1) since f′(0) = 0.
Again, f (x) is strictly increasing on (0, 1) and f (0) = 0. Then f (x) > 0 on (0, 1).
Therefore,
1− x
1 + x < e−2x for every x∈ (0, 1)
6. (10 points) 已知以 r 為半徑的三維球體體積為 V = 43πr3. 若以每分鐘 3 立方英吋的 速度吹入空氣於一個氣球中，試求當氣球體積為 36π 立方英吋時，其半徑 r 的改變速 度為何?
6.
Solution: When V = 4
3πr3 = 36π, the radius r is equal to 3. By chain rule, d
dtV = d dt(4
3πr3) = 4πr2dr dt. Hence,
dr dt =
dV dt
4πr2 We have
dr dt
r=3 = 3
4π· 9 = 1
12π (inch/min)
7. 已知一曲線的參數方程式為
x = cos3t y = sin3t t∈ [0,π 2].
(a) (5 points) 求出所有時間 t∈ [0,π2], 使得當一質點延此曲線軌跡運動時，此質點在 該時間有最大速度.
(a)
Solution: x′(t) =−3 cos2t sin t and y′(t) = 3 sin2t cos t. The speed function of the point tracing the curve is
v(t) =√[
x′(t)]2
+[ y′(t)]2
= 3 sin t cos t.
To find the maximum of v(t) on [0,π2], we consider
v′(t) =−3 sin2t + 3 cos2t = 3 cos(2t), v′′(t) =−6 sin(2t).
Hence, v′(π4) = 0 and v′′(π4) = −6 < 0. By the second derivative test, the point has a maximal speed at t = π
4. (b) (5 points) 計算此曲線的孤長
(b) Solution: The arc-length of the curve is
∫ π
2
0
√[x′(t)]2
+[ y′(t)]2
dt =
∫ π
2
0
3 sin t cos t dt = 3 2
∫ π
2
0
sin(2t) dt = 3 2.
(c) (5 points) 計算此曲線繞 x 軸旋轉後的旋轉曲面之表面積
(c) Solution: The area of the surface rotated about x-axis is
S =
∫ π
2
0
2πy(t)√[
x′(t)]2
+[ y′(t)]2
dt
= 6π
∫ π
2
0
sin4t cos t dt (u = sin t) = 6π
∫ 1 0
u4 du
= 6π 5
8. (10 points) 求級數
∑∞ n=0
2n
n!(n + 2) 的值. (提示: 對函數 f (x) = xex 的泰勒級數做積分) 8.
Solution: Consider the Taylor expansion ex =
∑∞ n=0
xn n!
Hence,
xex =
∑∞ n=0
xn+1 n!
We have ∫ x
0
tet dt =
∑∞ n=0
xn+2 n!(n + 2) On the other hand,
∫ x 0
tet dt = tet x
0−
∫ x 0
et dt = xex− ex+ 1 Set x = 2 to obtain
∑∞ n=0
4· 2n
n!(n + 2) = e2+ 1 and hence
∑∞ n=0
2n
n!(n + 2) = e2+ 1 4
