1. : = lim 2 sin x(cos x − 1) arctan(tan(arctan(2x) − 2 arctan x)) = lim

2006 d d d d d d
dd1. d :
2 sin x(cos x − 1)
arctan(tan(arctan(2x) − 2 arctan x))
dd = lim
x→0
2
2x · (− x2 )
= lim
2x
2x− 1−x
2
x→0
¡
arctan( 1+2x·
√
2x
1−x2
−x3
1
=
3
−2x
x→0 arctan
2
1+3x2
= lim
)
¢
¡√ √1 ¢
ln
x
√x
lim
= e x→+∞ x = e2 ln x
= e2 ln 1 = 1
1
−
x
1
1
x2
4. d : y 0 = arctan + x ·
= arctan −
x
x 1 + x2
1 + x12
− x12
−2x
−2
1
y 00 =
1 − 1 + x2 − x · (1 + x2 )2 = (1 + x2 )2
1 + x2
y 0 (t)
2t + 2t2
d2 y
4(1 + t)2
dy
4(1 + t)
2
5. d :
= 0
=
=
2(1
+
t)
,
=
=
.
1
1
dx
x (t)
dx2
t
1 − 1+t
1 − 1+t
Z
1
x2
+ C,
6. d : dddd(f (ex ))0 = f 0 (ex ) · ex = e2x , df (ex ) = e2x dx = e2x + C, f (x) =
2
2
2
x
df (0) = 1, dC = 1, f (x) =
+ 1.
2
¯ 2
¯ ¯
¯
¯ 2x + x
¯ ¯
¯
3
3
x+2
¯<
ddd
dd . ∀ε > 0, dN =
+ 1, dx > N d, ¯¯ 2
− 2¯¯ = ¯¯
< ε.
2ε
x −1
(x + 1)(x − 1) ¯ 2(x − 1)
1
1 −1
1
1
ddd
d: dx > 0ddf 0 (x) = 2x sin + x2 cos · 2 = 2x sin − cos ;
x
x x
x
x
dx < 0ddf 0 (x) = 2x − sin x − x cos x;
x2 sin x1
x(x − sin x)
dx = 0ddf 0 (0 + 0) = lim
= 0, f 0 (0 − 0) = lim
= 0, df 0 (0) = 0.
x→0+0
x→0−0
x
x
ddddf 0 (x)dx = 0dddd:
µ
¶
1
1
1
lim f 0 (x) = lim
2x sin − cos
= lim − cos dddddf 0 (x)dx = 0dddd
x→0+0
x→0+0
x→0+0
x
x
x
dd
lim f 0 (x) = 0 − 0 − 0 = 0 = f 0 (0), df 0 (x)dx = 0ddddd
x→0−0
Z
Z
1
dd1. d :
cos3 x dx = (1 − sin2 x) d sin x = sin x − sin3 x + C.
3
Z
Z
Z
2. d :
ex
x
x
x
x
x
x
e arctan e dx = arctan e de = e arctan e − ex
dx
1 + e2x
Z
1
1
1
= ex arctan ex −
de2x = ex arctan ex − ln(1 + e2x ) + C.
2
1 + e2x
2
3. d : dd = e
limx→+∞
ln x+ln
1+
√
x
1+x2
x
2007 d d d d d d
¯
¯
¯ 2x − 5
¯
7
7
7
dd1. d d . ∀ε > 0, dN = , dx > N d, ¯¯
− 2¯¯ =
< < ε.
ε
x
+
1
x
+
1
x
³ ε´
2. d d . ∀ε > 0, dδ = min 1,
, d0 < |x − 1| < δd, 0 < x < 2
6
¯ 2
¯
¯x + 3x − 4¯ = |x − 1| |x + 4| < 6 |x − 1| < ε.
ddd
d:
limπ (sin x)tan
2
x→ 2
¡
¢
π
y = −x
2
x
¡
¢ 1 · sin x−1
= limπ 1 + sin x − 1 sin x−1 cot2 x
x→ 2
=e
limy→0
cos y−1
tan2 y
=e
limy→0
y
−2 sin2
2
y2
1
= e− 2 .
2
2 ln(1 + t)
ddd
dd . ddddddd
+
ddddd
1
+
t
¶
Z xµ
Z x 1+t
2
2 ln(1 + t)
2
2 ln(1 + t)
+
dt <
2 dt, ddt > 0d
+
< 2, dddln(1+t) <
1
+
t
1
+
t
1
+
t
1+t
0
0
Z t
Z t
t. dt > 0dt =
1 ds <
es ds = et − 1ddd
0
0
ddd
d: 1) dx > 0ddf 0 (x) = 2x +
2x
.
1 + x2
dx < 0ddf 0 (x) = 2 sin x + 2x cos x.
x2 + ln(1 + x2 ) − 0
2x sin x
dx = 0ddf 0 (0 + 0) = lim
= 0, f 0 (0 − 0) = lim
= 0, df 0 (0) =
x→0+0
x→0−0
x
x
0.
µ
¶
2x
0
d lim f (x) = lim
2x +
= 0, lim f 0 (x) = lim (2 sin x + 2x cos x) = 0, df 0 (x)d0d
x→0+0
x→0+0
x→0−0
x→0−0
1 + x2
ddd
2x
2x + 1+x
f 0 (x) − f 0 (0)
f 0 (x) − f 0 (0)
2 sin x + 2x cos x
2
2) lim
= lim
= 4, lim
= lim
= 4.
x→0+0
x→0+0
x→0−0
x→0−0
x
x
x
x
00
df (0) = 4.
2006 d d d d d d
¡ k ¢2
Z 1
n
X
1
x2
ln(1 + x3 ) ¯¯1
ln 2
n
dd2. d : dd =
·
dx
=
.
¯ =
¡ k ¢3 =
3
n 1+
3
3
0
0 1+x
k=1
n
Z
Z
Z
dd1. d :
d sin x
1
−d cos x
I1 =
+
= (1) + (2) + (3).
dx +
2 − cos2 x
1 + 12 (1 − cos 2x)
1 + sin2 x
Z
Z
√
√
2
1
dt
dt
1
√
√
(tan x = t) (1) =
arctan
2t
+
C
=
arctan(
2 tan x) + C
=
=
2
1 + t2
1 + 2t2
2
2
3 − 1−t
1+t2
√
1
2 − cos x
(2) = √ ln √
+ C, (3) = arctan sin x + C.
2 2
2 + cos x
√
√
1
1
2 − cos x
dI1 = √ arctan( 2 tan x) + √ ln √
+ arctan sin x + C.
2
2 + cos x
Z2 π2 2
− sin t
√
2. d : dx = π + t, I2 =
dt = 0, dddddddddd.
1
− cos 2t
−π
2
¡
¢0
1
ln x
ddd
d: f (ln x) = f 0 (ln x) · =
,d
x
(1 + ln x)2
Z
Z
Z
ln x
−1
1
f (ln x) =
dx
=
dx
+
dx
(1 + ln x)2
(1 + ln x)2
1 + ln x
Z
Z
x · x1
x
−1
=
dx
+
+
dx
2
(1 + ln x)
1 + ln x
(1 + ln x)2
x
=
+ C,
1 + ln x
df (0) = 1, df (ln 1) = 1 + C = 1, dC = 0, f (1) = f (ln e) =
e
e
= .
1 + ln e
2
2007 d d d d d d
¯
¯
Z
¯x − 1¯
1
ln x
ln x
1 1
¯
¯ + C.
ddd
d: 1. dd = ln x d
=
−
dx =
+ ln ¯
1−x
1−x
1−xx
1−x
x ¯
1
tan x − sin x
tan x(1 − cos x)
√
= .
2. dd = lim 3 √
=
3
x→0 x ( 1 + tan x +
2x
4
1 + sin x)
¶
Z 2
Z 2µ
1
1
t ¯¯2
2t
dt
4
3. dx = t2 , dd =
=
2
−
dt
=
2
ln
¯ = 2 ln .
2 (1 + t)
t
t
1
+
t
1
+
t
3
1
1
1
µ
¶(10)
µ
¶
µ
¶(8)
µ
¶(n)
(9)
1
1
1
1
n!
4. y (10) =
x2 + 10
2x + 10 × 9
. d
=
dd
1−x
1−x
1−x
1−x
(1 − x)n+1
¢
10! ¡ 2
10!
y (10) =
x + 2x(1 − x) + (1 − x)2 =
(1 − x)11
(1
−
x)11
Z x
Z x
3
x
−
1
2
, d1 < x ≤ 2d, F (x) =
1 dt = x − 1.
8. d0 ≤ x ≤ 1ddF (x) =
t dt =
3
1
1
Z
p
√
√
ddd
d: xn+1 = 4 − (xn − 2)2 , d0 ≤ xn+1 ≤ 2 (n ≥ 1d).pdn > 1ddxn+1 > xn · 2 > xn · xn = xn ,
dn > 1dxn ddddddddddddddddddA, dA = A(4 − A)dA = 2.
ddd
d: dddddg(f (x))f 0 (x) = ex (x2 + 2x) = xf 0 (x)d
Z
f 0 (x) = ex (x + 2), f (x) = (x + 2)ex dx = (x + 2)ex − ex + C,
df (0) = 0, df (x) = (x + 1)ex − 1.
Z 1
n
X
1
1
dx
π
ddd
d: dd = lim
=
= .
¡
¢
2
2
k
n→∞
n1+
4
0 1+x
k=1
n
¶
Z 2µ
y2
ddd
d: ddddddddd(2, 2), (8, −4)ddS =
dy = 18.
4−y−
2
Z x
Z −4
x
ddd
dd . dx ∈ [0, 1]. dxf (x) =
(tf (t))0 dt =
(tf 0 (t) + f (t)) dt d
0
0
Z
x
x |f (x)| ≤ x
Z
0
Z
1
d(1 − x)f (x) = −
x
|f (t)| dt.
0
Z
((1 − t)f (t))0 dt = −
x
|f 0 (t)| dt +
1
((1 − t)f 0 (t) − f (t)) dt d
x
Z
1
(1 − x) |f (x)| ≤ (1 − x)
Z
|f 0 (t)| dt +
x
Z
dddddddd|f (x)| ≤ x
1
|f (t)| dt.
x
Z
1
|f 0 (t)| dt + (1 − x)
0
1
Z
1
|f 0 (t)| dt +
0
|f (t)| dt = dddddd
0
2008 d d d d d d
Z
Z
1
2 ln x
1
1 + ln x
=−
+2
dx = −2
+ C.
2
x
x
x
x
ex ln(1+x) − 1
x ln(1 + x)
1
2. dd = lim x
= lim
= .
x→0 (e − 1) ln(1 + 2x)
x→0
x · 2x
2
µ
¶ Z 1
n
X
1
k−1
3. dd = lim
sin a +
b =
sin(a + bx) dx. db = 0ddd = sin a; db 6= 0,
n→∞
n
n
0
k=1
¯1
¢
1
1¡
cos a − cos(a + b) .
ddd = − cos(a + bx)¯0 =
b
b
¯
³
7π ´¯¯
¯
3
4. f (10) (0) = C10
(sin x)(7) · (x3 )(3) ¯
= 10 × 9 × 8 sin x +
= −720.
¯
2 x=0
x=0
Z 1
¯1
x
8. ddddddd = 2
dx = ln(1 + x2 )¯0 = ln 2.
2
0 1 +Zx
1
¯1
√ x
π
2t2
dt = 2 − 2 arctan t¯0 = 2 − .
10. d e − 1 = tdddd =
2
1
+
t
2
0
1
1
ddd
d: 1. dddd(x0 , y0 ). dddddddddd2ax0 =
ddx0 = √ . dy0 = ax20 = ln x0 ,
x0
2a
√
1
1
dda = , x0 = e, y0 = .
2e Z 1
2
√
√
³
p ¢
2 ¡
2 e
2 2e 3 ´¯¯ 12
y
y
2. S =
e − 2ey dy = e −
y2 ¯ =
− 1.
3
3
0
0 Z
Z
Z
0
a
0
f (−t)
f (t)et
f (x)
dx
=
−
dt
=
dt.
ddd
dd .
x
−t
t
a 1+e
0 1+e
−a 1 + e
Z a
Z 0
Z a
Z a
f (x)
f (x)
f (x)
dx =
dx +
dx =
f (x) dx. dd.
d
x
x
x
0 1+e
−a 1 + e
0
−a 1 + e
ddd
d: 1. dd = −2
ln x d
3