2018 年 4 月 28 日
姓名: 學號: 學系:
說明:
1. 本試題含封面共 10 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 15
6 10
7 10
8 10
總分 100
1. 選擇題 (每題選出一個正確答案)
給定函數 f (x),(a)、(b) 兩小題考慮其導函數 f′(x)之圖形如下:
圖形在 0 ≤ x ≤ 2 間為圓弧,在 2 ≤ x ≤ 3 間為直線。另假設 f(0) = 0.
(a) (5 points) f (3) =?
(A) 2π (B) 12 −π2 (C) 34 (D) 0 (E) π2 − 12 (b) (5 points) 以下何者敘述必定錯誤?
(A) f 在 x = 1 有反曲點。
(B) f 在區間 [0, 3] 上為連續函數。
(C) f 在 x = 1 有局部極小值。
(D) f 在 x = 2 有局部極大值。
(E) f (4) = 0。
註記: 原考題中因 y 軸座標之標註有誤 (−1, 0, 1 誤標為 0, 1, 2.),依據該錯誤資 訊,(a) 題無答案,故送分,而 (b) 題 C,D,E 皆為正確答案,選出任何一項皆給 分。
Remark: There is a typo on the original exam, where the numbers on y-axis are mistakenly marked 0, 1, 2 instead of the correct −1, 0, 1. Because of the wrong data, question (a) has no answer, and therefore 5 points are automatically given.
For part (b), options C,D,E are all correct according to the wrong data. Any of those answers results in full 5 points.
(c) (5 points) 以下對瑕積分
∫ ∞
1
f (x) dx
之敘述何者為真?
(A) 若 limx→∞f (x) = 0,則積分收斂。
(B) 若 f (x) 為非零多項式,則積分發散。
(C) 若積分發散,則對所有實數 c,∫∞
1 cf (x) dx 亦為發散。
(D) 若積分收斂,則存在 M > 0 使得對於所有 x > M ,f (x) > 0。
(E) 以上皆對。
(d) (5 points) 假設 P (x) = ∑∞
j=0ajxj 為函數 f (x) 在 x = 0 之泰勒展開式。以下敘 述何者為真?
(A) 若對於所有 j,f(j)(0) 存在。則在 x = 0 附近,f (x) = P (x).
(B) 對於所有 j,aj = f(j)(0).
(C) 對於所有 j,f′(x) =∑∞
j=1jajxj−1。 (D) 存在偶函數 f (x) 使得 a3 = 1。
(E) 以上皆非。
註: f(j) 為 f 的 j 階導函數。
Solution: (a) E (b) C (c) B (d) E
2. 計算
(a) (5 points)
tlim→∞
( 1 + 1
4t )t
=?
(a) e14 Solution: Let u = 4t, then t→ ∞ ⇔ u → ∞.
limt→∞(
1 + 4t1)t
= limu→∞(
1 + u1)u4
=(
limu→∞(
1 + 1u)u)14
= e14.
(b) (5 points) ∫ 1
0
sin√
x dx =?
(b) 2(sin 1− cos 1) Solution: Let u =√
x, we have∫1
0 sin√
x dx = 2∫1
0 u sin u du.. Doing integra- tion by parts, we have
2
∫ 1
0
u sin u du = 2(−u cos u + sin u)|10 = 2(sin 1− cos 1).
(c) (5 points) ∫
x− 2
x2− 4x + 13 dx =?
(c) − log |√ 3
9+(x−2)2| + C Solution: ∫ x−2
x2−4x+13 dx = ∫ x−2
(x−2)2=9 dx. Let x− 2 = 3 tan u, the integral is rewritten into
∫ 3 tan u
32sec2u 3 sec2u du =
∫
tan u du =− log | cos u|+C = − log | 3
√9 + (x− 2)2|+C.
3. (10 points) 描述 a 值範圍,使得曲線
y + y3 − x3 − ax2− x = 0
有水平切線。
3. a≥√
3 or a≤ −√ 3
Solution: Differentiate both sides, we have dy
dx + 3y2dy
dx − 3x2− 2ax − 1 = 0, or
dy
dx = 3x2+ 2ax + 1 1 + 3y2 .
Horizontal tangent lines appear precisely when dxdy = 0, which is true if and only if 3x2+ 2ax + 1 = 0. For the quadratic polynomial to have real root, we have
4a2− 12 ≥ 0, which is true if and only if a≥√
3 or a ≤ −√ 3.
4. (10 points) 將一條一公尺長的鐵絲在某點垂直折成 L 型,並將兩端分別固定在 x 與 y 軸且兩邊平行座標軸 (見下圖)。
將鐵絲繞 y 軸旋轉,求形成之旋轉體積的最大值。
4. 4π27
Solution: Suppose the length of the top is t and the bottom is 1− t. The volume of revolution is V (t) = πt2(1−t) = πt2−t3. To find the maximum value, we differentiate V′(t) = π(2t−3t2) and see that the critical points are 0,23. The corresponding volumes are 0 and 4π27 and the second value is clearly the maximum volume.
5. (15 points) 求
xlim→2
∫x2
−4cos(t4+ t2) sin7t dt
x− 2 .
列出所有充分理由才給滿分。
5. 4 cos 272 sin74
Solution: Let F (x) =∫x2
−4cos(t4+ t2) sin7t dt. Since the integrand is an odd func- tion, we have that F (2) = 0. The limit is rewritten as
xlim→2
∫x2
−4cos(t4+ t2) sin7t dt
x− 2 = lim
x→2
F (x)− F (2)
x− 2 = F′(2).
By the fundamental theorem of calculus, F′(x) = 2x cos(x8+x4) sin7x2and therefore F′(2) = 4 cos 272 sin74.
6. (10 points) 證明不可能存在可微函數 f (x) 滿足 f (0) = 0、f (1) = 4、且 f′(x) < ex 對 於所有 x。
Solution: Suppose such function exist, then by mean value theorem, 4 = f (1)− f (0) = f′(t) for some t∈ (0, 1). But f′(x) < ex < 3 on (0, 1) and therefore f′(t) = 4 is impossible.
7. (10 points) 求所有 p > 0 使得積分
∫ 1 0
cos4px + 1 x2p dx 收斂。列出所有充分理由才給滿分。
7. p < 12
Solution: Note that since 0≤ cos x ≤ 1 on [0, 1], we have
1≤ cos4px + 1≤ 2 for all x ∈ [0, 1]. Then,
0≤ cos4px + 1 x2p ≤ 2
x2p and ∫1
0 2
x2p dx converges when 2p < 1, or p < 12. Therefore, if p < 12, the integral converges.
To ensure that the integral diverges for other values of p, we observe
0≤ 1
x2p ≤ cos4px + 1 x2p and ∫1
0 1
x2p dx = ∞ for 2p ≥ 1, or p ≥ 12. The integral diverges for p ≥ 12 and therefore integral converges if and only if p < 12.
8. (10 points) 給定函數
f (x) = ecos x
√e ,
x∈ (0, π). 證明函數在該區間可逆並求 (f−1)′(1).
8. −√23
Solution: Since f′(x) =− sin xecos x√e < 0 on (0, π). The function is strictly decreas- ing and therefore one-to-one and invertible. To compute the derivative of f′ at 1, we first solve f (x) = 1. That is,
ecos x−12 = e0. Therefore, we must have cos x = 12, or x = π3 and
(f−1)′(1) = 1
f′(π3) =− 2
√3.
註記: 原中文版本考題中,函數定義域誤打為 (0, 2π)(英文版本正確),因此函數並不 可逆。故證明可逆部分送五分 (中英文版本皆送)。因為該筆誤,在中文版本中計算反
函數之微分,有可能算出另一可能性 √2
3,若算出該答案,或是明確指出反函數並不
存在,亦給五分,但英文版本該部分照常批改。
Remark: In the Chinese version of original exam, the domain of the function is mistak- enly given by (0, 2π) (while the English version is correct). Therefore, the 5 points on proving the invertibility of f is automatically given (in both versions). However, the part on computing (f−1)′(1) is well posed for English version and therefore is graded normally.