1101模模模組組組13-16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. Find the following limits.
(a) (5%) lim
x→∞
−x32 +2x 3x32 +3x − 5
(b) (6%) lim
x→0+( tan 2x
x +
1
ln x) (c) (6%) lim
x→0(1 + 3x)arcsin 2x1
(d) (5%) lim
x→0( 1 2x−
1 1 − e−2x)
Solution:
(a) lim
x→∞
−x32 +2x
3x32 +3x − 5 = lim
x→∞
−1 + 2x−12
3 + 3x−12−5x−32 (3 pts) = −1 + 2 ⋅ 0 3 + 3 ⋅ 0 − 5 ⋅ 0= −
1
3 (2 pts) (b) lim
x→0+( 1
ln x) =0 (∵ lim
x→0+ln x = −∞) (1 pt)
x→0lim+ tan 2x
x
L’H
Ô Ô
0 0
x→0lim+
2 sec22x
1 (2 pts for (tan x)′=sec2x. 1 pt for the chain rule.)
=2(1 pt) Hence lim
x→0+( tan 2x
x +
1
ln x) = lim
x→0+( tan 2x
x ) + lim
x→0+
1 ln x =2 Another solution for computing lim
x→0+
tan 2x x .
x→0lim+ tan 2x
x = lim
x→0+
sin 2x x ⋅
1 cos 2x
= lim
x→0+
sin 2x
2x ⋅2 ⋅ 1
cos 2x (2 pts)
= (lim
x→0+
sin 2x
2x ) ⋅ (lim
x→0+2 ⋅ 1 cos 2x)
=1 ⋅ 2 = 2 (2 pts) (Students can use the limit lim
x→0+
sin 2x x =2) (c) ln ((1 + 3x)arcsin 2x1 ) =
ln(1 + 3x) arcsin(2x) (1 pt)
xlim→0+
ln(1 + 3x) arcsin(2x)
L’H
Ô Ô0 0
xlim→0+ 1+3x3
√ 2 1−4x2
(1 pt for (ln(1 + 3x))′, 2 pts for (arcsin(2x))′)
= 3
2 (1 pt) Hence lim
x→0(1 + 3x)arcsin 2x1 =e32 (1 pt) (d)
limx→0( 1
2x− 1
1 − e−2x) =lim
x→0
1 − e−2x−2x 2x((1 − e−2x)
L’H
Ô Ô
0 0
2e−2x−2
2(1 − e−2x) +4xe−2x (2 pts)
=lim
x→0
e−2x−1 1 − e−2x+2xe−2x
L’H
Ô Ô
0 0
x→0lim
−2e−2x
4e−2x−4xe−2x (2 pts)
= − 1
2. (1 pt)
Page 1 of 9
2. Let f (x) =
⎧⎪
⎪
⎨
⎪⎪
⎩
∣x∣ cos (1
x) if x ≠ 0, 0 if x = 0.
(a) (5%) Determine whether f (x) is continuous at x = 0. Explain your answer.
(b) (5%) Determine whether f (x) is differentiable at x = 0. Explain your answer.
Solution:
(a) Since −1 ≤ cos(1
x) ≤1 for all x ≠ 0, −∣x∣ ≤ ∣x∣ cos(1
x) ≤ ∣x∣ for all x ≠ 0.(2 points) By Squeeze Theorem, lim
x→0∣x∣ cos(1
x) =0.(2 points) Since lim
x→0∣x∣ cos(1
x) =f (0), f (x) is continuous at x = 0.(1 points) (b)
xlim→0+
f (x) − f (0)
x − 0 )(1 points)
= lim
x→0+
∣x∣ cos(1x)
x )(1 points)
= lim
x→0+cos(1
x)(1 points) Since lim
x→0+cos(1
x)does not exist(1 points), Hence f (x) is not differentiable at x = 0(1 points).
3. Find f′(x).
(a) (7%) f (x) = arctan (x 2) +ln
√ x − 2
x + 2, for x > 2.
(b) (8%) f (x) = xln(x3)+2x, for x > 0.
Solution:
(a)
d
dx(arctan(x 2)) =
1 1 + (x/2)2⋅
d dx(
x
2)(2 points)
= 2
4 + x2(1 points) d
dx
⎛
⎝ ln
√ x − 2 x + 2
⎞
⎠
= 1
√ x − 2 x + 2
⋅ d dx
⎛
⎝
√ x − 2 x + 2
⎞
⎠
(1 points)
= 1
√ x − 2 x + 2
1
√ x − 2 x + 2
⋅ 2
(x + 2)2(1 points)
= 2
x2−4(1 points) Hence dy
dx = 4x2
x4−16 (1 points).
(b) Let a = xln(x3). Thus
ln a = ln(x3)ln x = 3(ln x)2(2 points)
⇒ 1 a
da
dx =6 ln x ⋅ 1
x(2 points)
⇒ da dx =
6xln(x3)ln x
x (1 points) d
dx(2x) = ln 2 ⋅ 2x(2 points) Thus dy
dx =
6xln(x3)ln x
x +ln 2 ⋅ 2x(1 points).
Page 3 of 9
4. tan(x − y) = x2+sin(2y) defines y as an implicit function of x near (0, 0) which is denoted by y = y(x).
(a) (7%) Compute dy
dx at (0, 0).
(b) (5%) Write down the linearization of y(x) at x = 0. Use the linear approximation to estimate y(0.01).
Solution:
(a) tan(x − y(x)) = x2+sin(2y(x))
d dx
ÔÔ⇒sec2(x − y) ⋅ (1 − y′) =2x + 2 cos(2y) ⋅ y′
(4 pts total. 1 pt for (tan x)′=sec2x. 1 pt for (sin x)′=cos x. 2 pts for the chain rule.) At (x, y) = (0, 0), sec2(0)(1 − y′(0)) = 0 + 2 ⋅ cos 0 ⋅ y′(0)
Hence y′(0) = 1 3.
(3 pts total. 2 pt for plugging in (x, y) = (0, 0). 1 pts for solving y′(0).) (b) The linearization of y(x) at x = 0 is
L(x) = y(0) + y′(0)(x − 0) (1 pt)
=0 +1 3x = 1
3x (2 pts) ← (a)算錯的話，這裡扣 1 分
Hence we can approximate y(0.01) by L(0.01) which is L(0.01) = 0.01
3 . (2 pts) ((a)算錯的話，這裡再扣 1 分)
5. Dominant 7 is an acappella (阿卡貝拉) group formed by seven students at NTU. They have performed and won several international competitions. Currently they are planning to organise their first concert. When x tickets are demanded for their concert, the price of the tickets can be described by the function
p(x) = 96
√x− x
9 where 1 ≤ x ≤ 90.
The cost in organising the concert is given by the function C(x) = 0.001x3−0.105x2+300.
(a) (6%) Find the maximum revenue generated by the ticket sales. (Revenue = Price × Quantity demanded) (b) (5%) Discuss when the economy of scale occurs. i.e. the marginal cost C′(x) is decreasing.
(c) (3%) Write down the profit function Π(x). (Profit = Revenue − Cost).
(d) (4%) Student A claims that when the revenue is maximized, the profit is also maximized. Do you agree with Student A? Explain.
Solution:
(a)
Marking Scheme.
(1M) Writing down the correct revenue function (2M) Find R′(x) correctly (1M for each term) (1M) Find the correct critical number x = 36
(2M) Any argument that x = 36 indeed gives a maximum value Sample solution.
The revenue function is R(x) = x ⋅ p(x) = 96√ x −x2
9
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
(1M)
.
Differentiate : R′(x) = 48
√x− 2x
9
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
(2M)
. Set R′(x) = 0, we obtain x = 36
´¹¹¹¹¹¸¹¹¹¹¶
(1M)
.
(**) Since R′′(x) = −24x−32 − 2
9 and hence R′′(36) < 0, the second derivative test implies that R(x) at- tains a maximum at x = 36.
Alternative for (**). (Using first derivative test)
x ⋯ 36 ⋯
R′(x) + 0 −
Therefore, the first derivative test implies that R(x) attains a maximum at x = 36.
Another alternative for (**). (Using Extreme Value Theorem) Compare the critical value and values at boundaries : R(1) = 96 −1
9, R(36) = 96 ⋅ 6 − 144 and R(90) = 96√
90 − 900. We conclude that R(x) attains a maximum at x = 36.
(b)
Marking Scheme.
(1M) Writing down C′′(x) < 0 or C′′(x) ≤ 0 (2M) Find C′′(x) correctly
(2M) Solving the inequality correctly.
Remark 1. Accept both < and ≤.
Remark 2. No deductions for omitting 1 ≤ x in the final answer.
Sample solution.
The economy occurs when
C′′(x) < 0
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
1M
⇔ 0.006x − 0.21
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
2M
<0 ⇔ x < 35
´¹¹¹¹¹¸¹¹¹¹¶
2M
.
Hence, the economy occurs when 1 ≤ x < 35.
Page 5 of 9
(c)
Marking Scheme.
Almost all or nothing. Minor deduction for obvious typo.
Sample solution.
Π(x) = x (96
√x− x
9) − (0.0001x3−0.105x2+300) ⋯ (3M) (d)
Marking Scheme.
The key argument is to verify that x = 36 is not a critical number for the profit/cost function.
(2M) Correct approach
(2M) Verifying Π′(36) = C′(36) ≠ 0
Remark. Just disagreeing student A or writing ‘No’ only without any elaborations receive no credits.
Remark. A student who gets (a) or (c) incorrect can get at most 2M here.
Sample solution 1. (Arguing by Profit)
If Π(x) attains a maximum at x = x0, Fermat’s Theorem implies that Π′(x0) =0. ⋯ (2M) However, Π′(36) = R′(36) + C′(36) = 0 + C′(36) < 0.⋯ (2M)
Therefore, Π(x) does not attain a maximum at x = 36. This disproves Student A’s claim.
Sample solution 2. (Arguing by Cost)
If Π(x) attains a maximum at x = x0, then marginal cost and marginal revenue are equal.⋯ (2M) i.e. R′(x0) =C′(x0).
However, R′(36) = 0 by (a) whereas C′(36) < 0.⋯ (2M)
Therefore, Π(x) does not attain a maximum at x = 36. This disproves Student A’s claim.
6. f (x) = (x2+x)12.
(a) (10%) Write down the domain of f (x). Find all asymptotes of y = f (x).
(b) (5%) Compute f′(x). Find interval(s) of increase and interval(s) of decrease of f (x).
(c) (5%) Compute f′′(x). Determine concavity of y = f (x).
(d) (3%) Sketch the curve y = f (x).
Solution:
(a)
Marking Scheme. Domain : 3 % + Slant asymptote : 3.5 % each Domain :
1% - Solving x(x + 1) = 0 ⇒ x = 0 or x = −1
2% - The correct domain Asymptotes : (at ±∞ - 3.5% each)
1% - for computing a = limx→±∞f (x)
x correctly
1.5% - for computing limx→±∞(f (x) − ax) correctly
1% - for writing down the equation of the asymptote
Remark. Also okay if a student manages to guess an asymptote y = ax + b and check lim
x→±∞(f (x) − ax − b) = 0 directly.
Sample Solution.
(Domain)
The domain of f = {x∣x2+x ≥ 0}.
Let h(x) = x2+x = x(x + 1). Now we want to determine the sign graph of h(x) We first solve x2+x = 0, i.e.
x(x + 1) = 0 ⇒ x = 0 or x = −1. (1 point)
These two points divide the real line into three subintervals (−∞, −1) ∪ (−1, 0) ∪ (0, ∞). Evaluate g(−2) = (−2)(−2 + 1) > 0, g(−0.5) = (−0.5)(−0.5 + 1) < 0 and g(1) = 1(1 + 1) > 0. So g(x) = x2+x > 0 on (−∞, −1) ∪ (0, ∞).
So the domain of f is (−∞, −1] ∪ [0, ∞). (2 points)
Note that g is continuous on (−∞, −1] ∪ [0, ∞). There is no vertical asymptote.
To find slant asymptotes, we first compute
xlim→∞
f (x) x =lim
x→∞
(x2+x)12
x = lim
x→∞
(x2(1 +x1))
1 2
x
=lim
x→∞
x((1 +x1))
1 2
x = lim
x→∞((1 +1 x))
1 2 =1.
(1 point)
Page 7 of 9
Then we compute
xlim→∞(x2+x)12 −x
=lim
x→∞x(1 +1 x)
1
2−x = lim
x→∞x[(1 +1 x)
1 2−1]
=lim
x→∞
[(1 +1x)
1 2−1]
1 x
= lim
h→0+
[(1 + h)12−1]
h
= lim
h→0+
[(1 + h)12−1]′
h′ = lim
h→0+ 1
2(1 + h)−12 1
= 1 2.
(1.5 point) Thus lim
x→∞(x2+x)12 − (x +1
2) =0 or equivalently y = x +1
2 is a slant asymptote. (1 point).
On the other hand,
x→−∞lim f (x)
x = lim
x→−∞
(x2+x)12
x = lim
x→−∞
(x2(1 +x1))
1 2
x
= lim
x→−∞
−x((1 +1x))
1 2
x = lim
x→∞−((1 + 1 x))
1 2 = −1.
(1 point) Now we compute
x→−∞lim (x2+x)12 +x
= lim
x→−∞−x(1 +1 x)
1
2+x = lim
x→−∞x[−(1 +1 x)
1 2 +1]
= lim
x→−∞
[−(1 +1x)
1 2+1]
1 x
= lim
h→0−
[−(1 + h)12+1]
h
= lim
h→0−
[−(1 + h)12+1]′
h′ = lim
h→0−
−12(1 + h)−12 1
= − 1 2
(1.5 point) Thus lim
x→−∞(x2+x)12 − (−x −1
2) =0 and y = −x −1
2 is also also slant asymptote. (1 point).
(b)
Marking Scheme.
2% - correct f′(x)
1% - for correctly determining the signs of each sub-intervals 1% - for the correct interval of increase
1% - for the correct interval of decrease Sample solution.
f′(x) = d
dx(x2+x)12 = 1
2(x2+x)−12(2x + 1). (2 point)
Recall the domain of f is (−∞, −1]∪[0, ∞). f′(x) = 0 has no solution in its domain. f′(−2) =1
2(4−2)−12(−4+
1) < 0 and f′(1) = 1
2(1 + 1)−12(2 + 1) > 0.
. (1 point for signs)
So f′(x) < 0 and f is decreasing on (−∞, −1). (1 point)
(c)
Marking Scheme.
3% - correct f′′(x)
2% - for the correct interval/explaining why the curve is always concaving downward Sample Solution.
Using f′(x) = 1
2(x2+x)−12(2x + 1), we have f′′(x) =1
2([(x2+x)−12]′(2x + 1) + (x2+x)−12(2x + 1)′)
= 1 2(−
1
2(x2+x)−32(2x + 1)2+ (x2+x)−122)
= − 1
4(x2+x)−32((2x + 1)2−4(x2+x))
= − 1
4(x2+x)−32(4x2+4x + 1 − 4(x2+x))
= − 1
4(x2+x)−32.
(3 point) Recall the domain of f is (−∞, −1] ∪ [0, ∞) and x2+x > 0 on (−∞, −1) ∪ (0, ∞). So f′′(x) < 0 and f is concave down on (−∞, −1) ∪ (0, ∞). (2 points)
(d)
Marking Scheme.
1% - appropriate ‘ends’ (or x-intercepts or one-sided vertical tangents) 1% - the two slant asymptotes
1% - the shape (increase, decrease, always concaving downward)
Remark. It’s okay that the students do not demonstrate vertical-ness of the one-sided tan- gents at the points x = 0 and x = −1.
Sample solution.
Note that f (−1) = f (0) = 0. (1 point)
f is decreasing and concave down on (−∞, −1). f is increasing and concave down on (0, ∞).
Also, note that lim
x→−1−f′(x) = lim
x→−1−
1
2(x2+x)−12(2x+1) = −∞ and. lim
x→0+f′(x) = lim
x→0+
1
2(x2+x)−12(2x+1) = ∞.
f has a one sided vertical tangent at x = −1 and x = 0
(Slant asymptotes - 1 point) (Correct shape - 1 point)
Page 9 of 9
