1021微
微甲 07-11班
班期 中 考 解 答
1. (9%) Find
d
π
π
(sec x)x , − < x < .
dx
2
2
Solution:
(sec x)x = ex ln sec x
(2%)
d
d x ln sec x
d
(sec x)x =
e
= ex ln sec x (x ln sec x)
(3%)
dx
dx
dx
sec x tan x
d
x
x
(sec x) = (sec x) ln sec x + x
= (sec x)x (ln x + x tan x)
dx
sec x
ˆ
2. (9%) Evaluate
0
1
(2 + x)2
dx.
1 + x2
Solution:
ˆ
ˆ 1
(2 + x)2
4x + 3
(2 + x) = x + 4x + 4 ⇒
dx =
1+
dx
2
1+x
1 + x2
0
0
ˆ 1
ˆ 1
ˆ 1
ˆ 1
x
1
(2 + x)2
1dx + 4
dx =
dx + 3
dx
(3%)
2
2
2
1+x
0 1+x
0
0 1+x
0
ˆ 1
1
1dx = x = 1
(2%)
2
0
0
ˆ
1
4
0
ˆ
3
0
ˆ
0
1
2
1
1
1
x
2 dx
=
2
ln(1
+
x
)
= 2 ln 2
1 + x2
0
(2%)
1
1
3
π
dx
=
4
arctan
x
= 3 arctan 1 = 3 · = π
2
1+x
4
4
0
(2%)
(2 + x)2
3
dx = 1 + 2 ln 2 + π
2
1+x
4
t − ln(1 + t)
t→0
t2
p
t − ln(1 + t)
(b) Use (a) to find lim+
t→0
t
3. (9%) (a) Find lim+
Solution:
(a)5 points
this is indeterminate form of type
lim+
t→0
0
(1 point)
0
1
1 − 1+t
t − ln (1 + t)
1
1
=
lim
=
lim
=
t→0+
t→0+ 2(1 + t)
t2
2t
2
(process 2 points,answer 2 points)
Page 1 of 6
(4%)
(b)4 points
because f (x) =
√
x is continous at x = 1/2 and
lim+
t→0
t − ln (1 + t)
1
=
2
t
2
so
r
r
r
p
t − ln (1 + t)
t − ln (1 + t)
t − ln (1 + t)
1
= lim+
lim+
=
=
lim+
2
2
t→0
t→0
t→0
t
t
t
2
(process 2 points,answer 2 points)
1
4. (9%) Evaluate lim (2x + 3x + 5x ) x .
x→∞
Solution:
解法一
x x 2
3
2 +3 +5 =5 1+
+
(2分)
5
5
x x x1
1
3
2
x
x
x x
+
(3分)
(2 + 3 + 5 ) = 5 1 +
5
5

x x x1 !
ln 1 +

2
3


ln
lim
1
+
+
=
lim


x→∞
x→∞
5
5





x x x1


2
3


⇒ lim 1 +
+
= e0 = 1
x→∞
5
5




或直接說





x x x1



3
2

 lim 1 +
+
= 10 = 1
x→∞
5
5
x
x
x
x
2 x
5
1
x→∞
ln(2x + 3x + 5x )
ln(2 + 3 + 5 ) =
(2分)
x
(ln 2)2x +(ln 3)3x +(ln 5)5x
ln(2x + 3x + 5x ) l0 Hospital
2x +3x +5x
(3分)
lim
======= lim
x→∞
x→∞
x 1
x
x
(ln 2) 52 + (ln 3) 53 + (ln 5)
= lim
(2分)
2 x
3 x
x→∞
+
+
1
5
5
= ln 5 (1分)
x
x
x
1
x
1
lim (2x + 3x + 5x ) x = eln 5 = 5 (1 分)
x→∞
解法三
When x ≥ 1
2x + 3x + 5x < 5x + 5x + 5x = 3 × 5x (2分)
5x < 2x + 3x + 5x (2分)
Page 2 of 6
3 x
5
x
lim (2x + 3x + 5x ) x = 5 × 1 = 5
解法二
+
=0
(3分)
(1 分)
Because
1
lim (5x ) x = 5 (1分)
x→∞
1
1
lim (3 × 5x ) x = ( lim 3 x ) × 5 = 1 × 5 = 5 (3分)
x→∞
x→∞
By Squeeze Theorem
1
lim (2x + 3x + 5x ) x = 5 (1 分)
x→∞
5. (9%) Find the linear approximation of the function
√ x−1
−1
g(x) = sin
− tan−1 x at the point x = 3
x+1
Solution:
Linear approximation at x = 3 is
g(x) ≈ g(3) + g 0 (3) · (x − 3)
(1%)
d
x−1
1
d x−1
(sin−1 (
)) = q
(
)
·
x−1 2 dx x + 1
dx
x+1
1 − ( x+1
)
1
x(x + 1)
√
d
1
d √
√ 2·
(tan−1 ( x)) =
( x)
dx
1 + ( x) dx
1
= √
2 x(x + 1)
=√
(1% for the chain rule term, 3% in total)
(1% for the chain rule term, 3% in total)
1
1
⇒ g 0 (x) = √
and g 0 (3) = √
2 x(x + 1)
8 3
(1%)
√
1
g(3) = sin−1 ( ) − tan−1 ( 3)
2
π π
π
= − =−
6
3
6
⇒ g(x) ≈ −
(1%)
π
1
+ √ (x − 3)
6 8 3
6. (10%) Let y = f (x) satisfy x3 + 2xy + y 3 = 13. Find y 0 and y 00 at the point x = 1, y = 2.
Solution:
x3 + 2xy + y 3 = 13
Implicit differentiation gives
3x2 + 2y + 2xy 0 + 3y 2 y 0 = 0 (∗)
Page 3 of 6
4分
y0 =
−3x2 − 2y
2x + 3y 2
−3 − 4
1
=−
2 + 12
2
Differentiate (*) once more we have
2分
At (x, y) = (1, 2), y 0 =
6x + 2y 0 + 2y 0 + 2xy 00 + 6y(y 0 )2 + 3y 2 y 00 = 0
2分
At (x, y) = (1, 2),
2
1
1
00
+ 2y + 6 · 2 · −
6+4 −
+ 3 · 4y 00 = 0
2
2
4 + 3 + 14y 00 = 0
ˆ
7. (10%) Evaluate lim+
x2
y 00 = −
1
2
2分
√
√
et t sin tdt + x cos x − x
0
x→0
.
x3
Solution:
√
´ x2 t √
e
t
sin
tdt + x cos x − x 0
0
lim
x→0+
x3
0
2
x
(e x sin x) · 2x + (cos x − x sin x − 1)
= lim+
x→0
3x2
2 x2
(2x e sin x) − x sin x + (cos x − 1)
= lim+
x→0 3x2
x sin x cos x − 1
2 x2
e sin x −
+
= lim+
x→0
3
3x2
3x2
1 cos x − 1
2 2
1 sin x
+ lim+
= lim+ ex sin x − lim+
x→0 3
x→0 3
x→0 3 x
x2
1 1
−1
=0− + ·
3 3
2
1
=− .
2
Grading Policy:
0
type: 3 points
Identify
0
ˆ x2 √
√
d
et t sin tdt: 2 points
dx 0
(x cos x − x)0 : 2 points
Three limits: each get 1 point
8. (15%) The minute hand (分針) on a clock is 8cm long and the hour hand (時針) is 4cm long. How
fast is the distance between the tips of the hands changing at two o’clock? Give your answer in the
unit cm/hour.
Page 4 of 6
Solution:
Let θ be the angle of minute hand and hour hand.
Let X be the distance of minute hand and hour hand.
By cosine theorem:[3 pts]
X 2 = 82 + 42 − 2 × 8 × 4 × cos θ
√
π
At 2 o’clock. θ = and X = 4 3. [2 pts]
3
π
The angle vector of minute hand is 2π and the angle vector of hour hand is [4 pts]
6
π
11π
Moreover, the angle vector of θ is − 2π = −
. [2 pts]
6
6
By diffrerntating: [2 pts]
dX
dθ
2X
= 64 sin θ
dt
dt
dX
22π
To end up,
=−
[2 pts].
dt
3
x2 2 ln x
−
, x > 0.
6
3
Answer the following questions and give your reasons (including computations). Put ”None” in
the blank if the item asked does not exist.
9. (20%) Let y = f (x) = x −
(a) (5%) Find the interval(s) on which f is increasing.
Answer:
(b) (4%) Find the local maximal point(s) and minimal point(s) of f , if any.
Answer: local maximal point(s) (x, y) =
local minimal point(s) (x, y) =
(c) (5%) Find the interval(s) on which f is concave up.
Answer:
(d) (2%) Find the inflection point(s) if any.
Answer:
(e) (4%) Sketch the graph of f . Indicate all information in (a)-(d).
Solution:
a. f 0 (x) > 0
x
2
⇒1− −
>0
3 3x
2
x
2
⇒x−
− > 0, (x > 0)
3
3
⇒1<x<2
(b) by (a)local maximal point (x, y) = (2,
5
local minimal point (x, y) = (1, )
6
4 − 2ln2
)
3
Page 5 of 6
(c) f 00 (x) > 0
−1
2
⇒
+ 2 > 0, (x > 0)
3
3x√
⇒0<x< 2
√ √
1 − ln2
(d) by (c) the inflection point (x, y) = ( 2, 2 −
)
3
√
√
(e) f is concave up on(0, 2), f is concave down on ( 2, ∞),and lim f (x) = ∞.
x→0
If you are totally right of (a),(b),(c),(d),I think you will get the full scores of (e).
Sketching the graph of f by yourself.
Page 6 of 6