1011微微微甲甲甲07-11班班班期期期中中中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. (15%) Find lim
x→0−
and lim
x→0+
of the following functions:
(a) sin(|x|)
x , (b) cos x − 1
sin(x sin x), (c) cos(sin x) − 1 tan2x
Solution:
(a) 5 points
lim
x→0−
sin |x|
x = lim
x→0−
sin (−x)
x = lim
x→0−−sin x x = −1 lim
x→0+
sin |x|
x = lim
x→0+
sin x x = lim
x→0+
sin x x = 1 (b) 5 points
x→0lim
cos x − 1
sin (x sin x) = lim
x→0
cos x − 1
sin (x sin x)(cos x + 1 cos x + 1) =
x→0lim
cos2x − 1
[sin (x sin x)](cos x + 1) = lim
x→0
− sin2x [sin (x sin x)](cos x + 1)
= lim
x→0
− sin2x [sin (x sin x)](cos x + 1)
x sin x x sin x = lim
x→0[− sin (x)/x] lim
x→0[(x sin (x)/ sin(x sin (x))] lim
x→0( 1
cos x + 1) =−1 2 (c) 5 points
x→0lim
cos (sin x) − 1 tan2x = lim
x→0
cos2(sin x) − 1
tan2x(cos (sin x) + 1) = lim
x→0
− sin2(sin x) cos2x sin2x(cos (sin x) + 1)
= lim
x→0[− sin2(sin x)/ sin2x] lim
x→0
cos2x
cos (sin x) + 1 = −1 2
2. (10%) Show that | tanx
2 − tany
2| ≥|x − y|
2 for x, y ∈ (−π, π).
Solution:
Let x, y ∈ (−π, π), W.L.O.G, set x < y.
Let f (t) = tant
2, then f is continuous on [x, y] and differentiable on (x, y). (2 point) By Mean Value Theorem, there is a number c between x and y such that
f (x) − f (y)
x − y = f0(c) (2 point)
Since f0(c) = 1 2sec2 c
2 (2 point)
⇒ | tanx2 − tany2 |
| x − y | =| 1 2sec2c
2 |
⇒ | tanx
2 − tany 2 |=| 1
2sec2 c
2 || x − y | (2 point) We know that | 1
2sec2c 2 |≥ 1
2 , (1 point)
⇒| tanx
2 − tany
2 | ≥ | x − y |
2 (1 point)
3. (10%) A rhombus (菱形) has sides 10in. long. Two of its opposite vertices are pulled apart at a rate of 2 in. per second. How fast is the area changing when the vertices being pulled are 16 in apart?
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Solution:
Let the distance between the two pulled vertices is x in., and the length of another diagonal is y in.. The change rate of x is dx
dt = 2.
By Pythagorean theorem, (x 2)2+ (y
2)2= 102= 100 =⇒ x2+ y2= 400 =⇒ y =p
400 − x2 So the area A of the rhombus is xy
2 =1 2xp
400 − x2 dA
dt = dA dx ·dx
dt = 1 2
p400 − x2−1
2 · x2
√
400 − x2
× 2 =p
400 − x2− · x2
√
400 − x2 When x = 16, dA
dt x=16
= 12 −256 12 = −28
3 (in2/sec)
評分標準如下:
寫出長度或者角度之間的關係 (2分) 寫出面積與所設變數之間的關係式 (2分)
將面積對變數作微分 (4分)
代入欲求取之值 (2分) 其餘錯誤酌量扣分。
4. (10%) Let f (x) = 1 + cos x
1 + sin x. Use a differential to estimate f (44◦).
Solution:
f (x + h) ' f (x) + f0(x) · h
f0(x) = − sin x(1 + sin x) − cos x(1 + cos x)
(1 + sin x)2 = − sin x − cos x − 1
(1 + sin x)2 (4 points) f (44◦) = f (45◦+ (−1◦)) ' f (45◦) + f0(45◦) · (−1◦) (2 points)
= f (π
4) + f0(π 4) · −π
180 (2 points)
= 1 + (2 − 2√ 2) ·−π
180
= 1 +(√ 2 − 1)π
90 . (1 point) 其中:
f (π
4) = 1 + cosπ4 1 + sinπ4 =1 +
√2 2
1 +
√ 2 2
= 1
f0(π 4) = −
√2 2 −
√2 2 − 1 (1 +
√2 2 )2
= −√ 2 − 1
3 2+√
2 =−2√ 2 − 2 3 + 2√
2 · 3 − 2√ 2 3 − 2√
2
= (−2√
2 − 2)(3 − 2√ 2)
= 2 − 2√
2. (1 points)
5. (25%) Let f (x) = (x + 1)2 x2+ 1 . (a) (5%) Find f0 and f00.
(b) (10%) Find the intervals on which f increases and the intervals on which f decreases. Indicate local extreme values and absolute extreme values.
(c) (5%) Find the intervals on which the graph of f is concave up and the intervals on which the graph of f is concave down. Indicate points of inflection.
(d) (5%) Find vertical and horizontal asymptotes if any. Sketch the graph of f .
Solution:
(a)
f (x) = (x2+ 1)−1(x + 1)2
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f0(x) = −(x2+ 1)−2(2x)(x + 1)2+ 2(x2+ 1)−1(x + 1) =−2(x + 1)(x − 1) (x2+ 1)2 (3%) f00(x) = −2[−2(x2+ 1)−3(2x)(x + 1)(x − 1) + (x2+ 1)−2(2x)] =4x(x2− 3)
(x2+ 1)3 (2%) (b)
From the above chart we know that (1) f is increasing on [−1, 1] (2%)
(2) f is decreasing on (−∞, −1] and [1, ∞] (2%)
(3) Local minimum: f (−1) = 0 (2%); local maximum: f(1) = 2 (2%)
(4) Absolute minimum: f (−1) = 0 ; absolute maximum: f(1) = 2 (1%) since lim
x→±∞f (x) = 1 and f (x) is finite for any real x (reasoning 1%).
(c) f00(x) is 0 and changes its sign at x = 0, ±√ 3
⇒ inflection points: x = 0, ±√ 3 (3%).
From the chart in (b), we know that f is concave up on (−√
3, 0) and (√
3, ∞) (1%), concave down on (−∞, −√
3) and (0,√
3) (1%).
(d) Since f (x) and f0(x) are finite for any real x, the graph y = f (x) does not have any vertical asymp- totes.
Since lim
x→±∞f (x) = 1 (1%), y = f (x) has a horizontal asymptote y = 1 (1%).
Sketch of f (x):
6. (10%) Consider all the rectangles with base on the line y = −2 and with two upper vertices on the ellipse x2+y2/4 = 1 and symmetric with respect to the y-axis. Find the maximal possible area for such a rectangle.
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Solution:
f (θ) = 2 cos θ(2 sin θ + 2) (3pt.) 0 ≤ θ ≤ π 2 (1 pt.) f0(θ) = −4(2 sin θ − 1)(sin θ + 1) (2 pt.)
sin θ = 1 2 C.P. at θ = π
6 f (0) = 4 f (π
6) = 3√ 3 f (π
2) = 0 (2 pt. ) f (π
6) = 3√
3 is the maximum (2 pt.)
7. (10%) Find f0(2) given that f (x) = Z x3−4
2x
x 1 +√
tdt.
Solution:
f (x) = Z x3−4
2x
x 1 +√
tdt = x Z x3−4
2x
1 1 +√
tdt By the fundamental theorem of calculus,
d
dxf (x) = Z x3−4
2x
1 1 +√
tdt + x
1
1 +√
x3− 4· (3x2) − 1 1 +√
2x· (2)
(3pts) (2pts) (2pts)
f0(2) = Z 4
4
1 1 +√
tdt+ 2
3 ˙22 1 +√
4− 2 1 +√
4
= 20 3
(2pts) (1pts)
8. (10%) Calculate
Z csc22x
√2 + cot 2xdx.
Solution:
Let u = 2 + cot 2x, then du = −2 csc22x dx (2 point) Z csc2x
√2 + cot 2x dx
= Z −1
√2
u du (2 point)
=−1 2
Z 1
√u du
= - u12+ C (5points)
= -√
cot 2x + 2 + C, where C is a constant. (1 point)
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