1101模模模組組組17-19班班班 微微微積積積分分分1 期期期考考考解解解答答答和和和評評評分分分標標標準準準 1. (20 pts) 計算以下的極限。(不可以使用 L’Hospital’s rule)
Evaluate the following limits. (Use of L’Hospital’s rule is not allowed.) 你可以直接使用以下的極限 （不用證明）。
You may use, without proof, the following standard limits.
x→0lim sin x
x =1 lim
x→±∞(1 +1 x)
x
=e lim
x→1
ln x x − 1=1 (a) (5 pts) lim
x→∞
3x+ex
1 + 22x (b) (5 pts) lim
x→0
1 − cos(2x)
√
x2+4 − 2 (c) (5 pts) lim
x→π2
ln(sin2x)
cos2x (d) (5 pts) lim
x→∞( x + 1 x − 1)
x
Solution:
(a)
x→∞lim 3x+ex 1 + 22x = lim
x→∞
1x+ (e/3)x
(1/3)x+ (4/3)x (3%)
=0 (2%).
(b)
limx→0
1 − cos(2x)
√
x2+4 − 2=lim
x→0
1 − cos(2x)
√
x2+4 − 2
√
x2+4 + 2
√
x2+4 + 2
=lim
x→0
(1 − cos(2x))(√
x2+4 + 2)
x2 (2%)
=lim
x→0
2 sin2x x2 (
√
x2+4 + 2) (2%)
=2 ⋅ 4 = 8 (1%).
(c)
x→π/2lim
ln(sin2x)
cos2x = − lim
x→π/2
ln(sin2x)
sin2x − 1 (2%) (let t = sin2x) = −lim
t→1
ln t
t − 1 (2%)
= −1 (1%).
(d)
x→∞lim( x + 1 x − 1)
x
= lim
x→∞(1 + 1 (x − 1)/2)
2⋅x−12 +1
(3%)
(let y = (x − 1)/2) = lim
y→∞(1 +1 y)
2y+1
=e2⋅1 = e2 (2%).
Page 1 of 9
2. (20 pts) 計算以下導函數或導數。 Compute the following derivatives.
(a) (8 pts) f (x) = x3⋅e(x2+2),求 f′(x)和 f′′(x)。Find f′(x) and f′′(x).
(b) (6 pts) g(x) = x
tan−1(sin x),求 g′(x)。Find g′(x).
(c) (6 pts) h(x) = xx⋅ (1 + x2)
1
x,求 h′(1)。 Find h′(1).
Solution:
(a) Product rule and chain rule.
f′(x) = (3x2) ⋅ex2+2+ (ex2+2⋅2x) ⋅ x3= (2x4+3x2)ex2+2
f′′(x) = (8x3+6x) ⋅ ex2+2+ (ex2+2⋅2x) ⋅ (2x4+3x2) = (4x5+14x3+6x) ex2+2 (b) Quotient rule and chain rule.
g′(x) =(1) ⋅ tan−1(sin x) − (1+sincos x2x) ⋅x (tan−1(sin x))2
(c) Logarithmic differentiation.
ln h(x) = x ln x +ln(1 + x2) x h′(x)
h(x) =ln x + 1 + 2 1 + x2−
ln(1 + x2) x2 h′(1) = h(1) ⋅ (0 + 1 + 1 − ln 2) = 4 − 2 ln 2.
Grading scheme:
Part (a) is 5+3. Part (b) is 6. Part (c) is 5+1.
0 points if they couldn’t use the derivative rules correctly.
-3 points if they clearly remembered a derivative formula wrong.
-1 point for computational mistakes/miscopy/oversimplify.
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3. (12 pts) 方程 x2+y2= (2x2+2y2−x)2在點 (0,1
2)附近可以描寫成隱函數 y = y(x)。
Near the point (0,1
2), the equation x2+y2= (2x2+2y2−x)2defines implicitly a function y = y(x).
(a) (7 pts) 求導數 dy dx∣
(0,12)。 Find dy dx∣
(0,12).
(b) (5 pts) 使用 y(x)在 x = 0 的線性逼近去估算 y(0.1) 的值。
Use linear approximation of y(x) at x = 0 to approximate the value of y(0.1).
Solution:
(a) Regard y as a function of x and take derivatives on both sides of the equality above we have 2x + 2ydy
dx =2(2x2+2y2−x) ⋅ (4x + 4ydy
dx−1) (4%) Put (x, y) = (0, 1/2) in the above equality, we have
dy
dx∣(0,1/2)=2 ⋅ 2 ⋅1 4(2dy
dx∣(0,1/2)−1) (2%)
⇒ dy
dx∣(0,1/2)=1 (1%).
(b)
y(x) ≈ y(0) + y′(0)x (2%) ⇒ y(0.1) ≈ 1
2+1 ⋅ 0.1 = 0.6 (3%).
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4. (10 pts) 考慮函數 Consider the function f(x) = 4 tan−1(x3) +ex.
(a) (4 pts) 說明 : f (x) 的反函數存在。 Explain briefly why the inverse function of f(x) exists.
(b) (6 pts) 設 g(x)為 f(x) 的反函數， 求 g′(π + e)。 Let g(x) be the inverse function of f(x). Find g′(π + e).
Solution:
(a) (2M) Correct f′(x)
(1M) Writing f′(x) > 0
(1M) Saying f(x) is strictly increasing Sample solution.
f′(x) = 12x2 1 + x6 +ex
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
2M
>0
°
1M
so f is strictly increasing ⋯(1M).
This implies that the inverse function of f (x) exists.
(b) (1M) Discovering f(1) = π + e
(2M) Correct f′(1)
(1M) Correct formula for g′(f (x))
(2M) Correct answer Sample solution.
Note f (1) = π + e ⋯ (1M) and f′(1) = 6 + e ⋯ (2M).
Therefore, g′(π + e) = g′(f (1)) = 1
f′(1)⋯ (1M)
= 1
6 + e⋯ (2M)
Page 4 of 9
5. (6 pts) 說明 : x = 0 為方程式 x + cos−1x = π
2 的唯一解。
Prove that x = 0 is the only solution to the equation : x + cos−1x = π 2.
Solution:
2M - Correct citing of Rolle/MVT/Consequences of MVT
2M - Correct derivative of x + cos−1(x)
2M - Overall correct and complete argument
Sample solution 1.
Let F (x) = x + arccos(x). Suppose x = α is another solution to the equation. Then F (0) = F (α).
Hence, Rolle’s Theorem implies that F′(c) = 0 for some c lying strictly between 0 and α. ⋯ (2M) However, F′(c) = 0 implies 1 − 1
√ 1 − c2
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
2M
=0 and
hence c = 0 which is a contradiction. ⋯ (Complete, correct argument 2M) Sample solution 2.
Let F (x) = x + arccos(x).
Then 1 − 1
√ 1 − x2
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
2M
<0 for −1 < x < 0 and 0 < x < 1.
Therefore, F is strictly decreasing before reaching x = 0 and also after leaving x = 0.⋯ (2M) Hence the function crosses y =π
2 at most (and hence exactly) once. ⋯ (Complete, correct argument 2M)
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6. (20 pts) 考慮函數 Consider the function f(x) = 6x − x2−4 ln x.
(a) (1 pt)寫出函數 f(x) 的定義域。 Write down the domain of f(x).
(b) (4 pts) 求 f′(x)，找出函數 f(x) 遞增、遞減的區間。
Find f′(x). Write down the interval(s) of increase and interval(s) of decrease of f (x).
(c) (4 pts) 求 f′′(x)， 判斷 y = f(x) 的凹性。
Find f′′(x). Write down the interval(s) on which f (x) is concave upward and the interval(s) on which f (x) is concave downward.
(d) (4 pts) 找出所有局部最大/小值和反曲點。 Write down (if any) the local extremas and inflection points.
(e) (2 pts) 找出 y = f(x) 的所有漸近線。 Find all the asymptotes of y = f(x).
(f) (5 pts) 畫出 y = f(x) 的圖形。 Sketch the graph of y = f(x).
Solution:
(a) x > 0.
(b)
f′(x) = 6 − 2x −4 x=
−2(x2−3x + 2)
x =
−2(x − 1)(x − 2) x Increasing: (1, 2)
Decreasing: (0, 1) and (2, ∞) (c)
f′′(x) = −2 + 4 x2 =
−2(x2−2)
x2 =
−2(x −√
2)(x +√ 2) x2
Concave upward: (0,√ 2) Concave downward: (√
2, ∞)
(d) Local minimum at x = 1, y = 5, local maximum at x = 2, y = 8 − 4 ln 2, inflection point (√ 2, 6√
2 − 2 − 2 ln 2).
(e) Vertical asymptote at x = 0 since lim
x→0+f (x) = ∞.
No horizontal asymptote as x goes to infinity since lim
x→∞f (x) = −∞.
No slant asymptote as x goes to infinity since lim
x→∞
f (x) x = −∞.
(f)
Page 6 of 9
Grading scheme:
(a) (1pts) Right 1 or wrong 0.
(b) (5pts) 2 pts for derivative. 3 pts for determining the signs.
(c) (5pts) Follow through. 2 pts for derivative. 3 pts for determining the signs.
(d) (4pts) Follow through. -2 for each mistake.
(e) (2pts) 1 point for horizontal asymptote and 1 point for vertical asymptote.
(f) (5pts) Follow through. Their picture need to match their answers above. -1 for each item not labeled or different from previous answers.
Page 7 of 9
7. (12 pts) 在拋物線 y = x2 內側且水平線 y = 2 下方畫一長方形使得長方形的頂邊與水平線 y = 2 重合 (見圖)， 求此長方形 最大面積。
Find the largest rectangle that fits inside the graph of the parabola y = x2 below the line y = 2, with the top side of the rectangle on the horizontal line y = 2.
Solution:
4M - for writing down the correct function to maximize
2M - correct derivative
2M - correct critical number
3M - any correct argument that verifies maximality of the critical number
1M - correct answer Sample solution.
Let (x, x2)be the coordinates of the right hand corner of the rectangle. We want to maximize the function A(x) = (2x)(2 − x2) =4x − 2x3
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¶
4M
.
Differentiating gives
A′(x) = 4 − 6x2
´¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¶
2M
.
Setting A′(x) = 0 gives x =
√ 2 3
±
2M
(negative rejected).
(**) Since A′′⎛
⎝
√ 2 3
⎞
⎠
= −12 ⋅
√ 2 3 <0,
the second derivative test implies that A(x) attains a maximum at x =
√ 2
3 ⋯(3M) and at which A(
√ 2 3) =
8√ 6 9
²
1M
.
Alternative for (**). (Using first derivative test)
x 0 ⋯
√ 2
3 ⋯
√ 2
A′(x) + + 0 − −
Therefore, the first derivative test implies that A(x) attains a maximum at x =
√ 2 3. and at which A(
√ 2 3) =
8√ 6 9 .
Another alternative for (**). (Using Extreme Value Theorem) Compare the critical value and values at boundaries :
Page 8 of 9
A(
√ 2 3) =
8√ 6 9
²
and A(0) = A(√ 2) = 0.
We conclude that A(x) attains a maximum at x =
√ 2 3.
Page 9 of 9
