1001微乙01-05班期中考解答和評分標準
1. (10%) 計算 lim
x→∞(x− 1 x + 1)x 。 Sol:
xlim→∞
(
1 + −2 x + 1
)x
= lim
x→∞
[
(1 + −2
x + 1)x+1(1 + −2 x + 1)−1
]
= lim
y→∞
[
(1 + −2
y )y(1 + −2 y )−1
]
= e−2 評分標準:
有利用 e = lim
x→∞
( 1 + 1
x )x
的概念得 7 分; 依算式完整度得 7 至10分 。
2. (10%) 令f (x) = xln x 。 求 f0(x) 。 Sol:
f (x) = xln x = e(ln x)2 ⇒ f0(x) = e(ln x)2 · 2 ln x · 1
x = 2 ln x· x(ln x)−1. (There is no partial credit)
3. (10%) 令f (x) = tan−1(√
x)· tan (x2) 。 求 f0(x)。 Sol:
Let f (x) = arctan(√
x)· tan(x2). Try to find f0(x)
f0(x) = arctan(√
x)0 · tan(x2) + arctan(√
x)· tan(x2)0 The Prodcut Rule : 2pts
=
( 1
1 + (√
x)2) ·√ x0
)
· tan(x2) + arctan(√ x)·(
sec2(x2)· x20)
The Chain Rules : 2pts + 2pts
=
( 1
1 + x · 1 2√ x
)
· tan(x2) + arctan(√ x)·(
sec2(x2)· 2x)
The Differentations : 2pts + 2pts
4. (10%) 說明 f (x) = 4x3+ 2x2+ 4x + 1 和g(x) = 2x2+ cos(x) 僅有一個交點 。 Sol:
Let F (x) = f (x)− g(x) = 4x3+ 4x + 1− cos x be the difference of f(x) and g(x).
Since F (0) = 0, f (x) and g(x) has an intersection at x = 0. (3 %)
M1: Suppose there is another intersection at x = a6= 0, i.e. F (a) = 0.
By Mean Value Theorem(1%),∃c lies between a and 0, such that F0(c) = F (a)− F (0)
a− 0 = 0→←
(∵ F0(x) = 12x2+ 4 + sin x > 0). Therefore, f (x) and g(x) only intersect at x = 0 (6%).
M2: Since F0(x) = 12x2+ 4 + sin x > 0, F (x) is strictly increasing. Moreover F (0) = 0, so f (x) and g(x) only intersect at x = 0 (7%).
5. (20%) 假設 y3+ xy− x = 1 。
(a) 求過點(1, 1) 之切線方程式 。 (b) 求 d2y
dx2 在點 (1, 1) 之值 。 Sol:
(a) If the reader can find the clues of chain rules (3 pts).
3y2y0+ y + xy0− 1 = 0 (5 pts)
After that you plug in (0, 0) and find y0 = 0, thus y = 1. (2 pts)
(b) If you have already get (8 pts) in part A, you may get (3 pts) from starting your second order operation.
6y(y0)2 + 3y2y00+ y0+ y0+ xy00 = 0 (5 pts) After that you plug in (0, 0) and find y00= 0. (2 pts)
6. (10%) 利用線性逼近去估計 ln 0.97之值 。 Sol:
設 f (x) = ln x .
因為已知 f (1) = ln 1 = 0 且0.97 與1 很接近,
所以用 f (x) 在x = 1 的切線來估計 f (0.97) = ln 0.97的值. 由公式 f (x)≈ f(a) + f0(a)(x− a)
代入 x = 0.97, a = 1. 則f0(1) = (ln x)0¯¯¯
x=1 = 1 x
¯¯¯
x=1 = 1
ln 0.97 = f (0.97)≈ f(1) + f0(1)(0.97− 1)
= ln 1 + 1∗ (−0.03)
= 0− 0.03 = −0.03 評分標準:
f (x)≈ f(a) + f0(a)(x− a) (5%) (ln x)0|x=1 = 1
x|x=1 = 1 (2%) ln 0.97≈ −0.03 (3%)
7. (20%) 若y = f (x) = x2+ 1 x 。
(a) y = f (x) 在 (區間) 遞增 。
y = f (x) 在 (區間) 遞減 。
(b) y = f (x) 之極大值 (若存在的話) : (座標) 。
y = f (x) 之極小值 (若存在的話) : (座標) 。
(c) y = f (x) 在 (區間) 凹向上 。
y = f (x) 在 (區間) 凹向下 。
(d) y = f (x) 所有的漸近線為
。 (e) 畫出 y = f (x) 之圖形 。
Sol: f (x) := x2+ x−1 defined on x6= 0
f0(x) = 2x− x−2 = x−2(2x3− 1)
< 0, x < 2−1/3, x6= 0
> 0, x > 2−1/3.
Hence f (x) is increasing on [2−1/3,∞). (2%)
f (x) is decreasing on (−∞, 0) and (0, 2−1/3]. (2%)
Critical points: x1, 0 = f0(x1)⇒ x1 = 2−1/3
f00(x) = 2 + 2x−3 = 2x−3(1 + x)(1− x + x2)
> 0, x∈ (−∞, −1),
= 0, x =−1,
< 0, x∈ (−1, 0)
> 0, x∈ (0, ∞).
f00(x1) = 2 + 4 > 0, f (x1) = 2−2/3+ 21/3 = 3 221/3. Hence (x, y) = (2−1/3,3
221/3) is local minimum. (2%)
There are no other critical points, hence no local maximum. (2%) f is concave up when f00> 0, i.e. x ∈ (−∞, −1) and x ∈ (0, ∞). (2%)
f is concave down when f00< 0, i.e. x ∈ (−1, 0). (2%)
lim
x→0±f (x) =±∞, ⇒ ’x = 0’ is a vertical asymptote. Since f is defined and continuous on {x 6= 0}, there is no other vertical asymptote.
Let y = mx + b be an asymptote, then m = lim
x→±∞f (x)/x = lim
x→±∞(x− x−3) =±∞. Hence no such asymptote.
The only asymptote is x = 0. (4%)
The graph is
8. (10%) 若高鐵每月載客量為 40, 000人 , 票價為 1, 500元/人 。 高鐵公司希望調整票價 , 增加收益 。 若票 價每調高 10 元, 則會損失乘客 200 人 。 請問該如何調整票價 ,才能達到最大的收益 ?
Sol:
Let increasing x dollar then it will lost 20x people
f (x) = (40000− 20x)(1500 + x)
=−20x2+ 10000x + 40000∗ 1500 then
f0(x) = −40x + 10000 if f0(x) = 0 then x = 250 and f0(x) =−40 < 0 so it has max value Hence increasing 250 dollar.
(4%) The standard of anwser:
if you get the f (x) =−20x2+ 10000x + 40000∗ 1500 then 6 points.
if you get the x = 250 then 4 points.
