Một số bài toán tổng hợp về hàm số - Pdf 27

class="bi x0 y0 w1 h1"
y =
ax
2
+ bx + c
dx + e
y =
x
2
− mx + 1
x − 1
y =
x
2
− mx + 1
x − 1
(∗)
y = ax
3
+ bx
2
+ cx + d
class="bi x1 y1 w3 h6"
y = f(x) x → x
0
y = f(x) x
0
x
0
∀ε > 0, ∃δ = δ(ε) : 0 < |x − x
0

); f(x
2
); f(x
3
) ; f(x
n
)
lim
x→0
x.sin
1
x
= 0
f(x) = x.sin
1
x
x
0
= 0
x
0
= 0 x
n
(
−π
4
;
π
4
)

x→0
x.sin
1
x
= 0
lim
x→1
sin
1
x − 1
x
n
= 1 +
1

n
= 1 +
2
(4n + 1)π
lim
n→∞
x
n
= 1; lim
n→∞
n
= 1
f(x
n
) = sin

x − 1
y = f(x
n
) x
0
x → x
0
A =
lim
x→(x
0
+0)
f(x) = f(x
0
+ 0) ∀ε > 0, ∃δ = δ(ε) > 0 ∀x : 0 < x − x
0
< δ ⇒
|f(x) − A| < ε
y = f(x) x
0
x
0
f(x) x → x
0
A = lim
x→(x
0
−0)
f(x) = f(x
0

) ∃ M > 0
|f(x)| ≤ M, ∀x ∈ V (x
0
), x = x
0
V (x
0
)
1 > |f(x) − A| ≥ |f(x)| − |A|
⇒ |f(x)| < 1 + |A| 1 +|A|
lim
x→x
0
f(x) = A, A = 0
V (x
0
) |f(x)| >
|A|
2
, ∀x ∈ V (x
0
), x = x
0
lim
x→x
0
f
1
(x) = A
1

2
(x) = A f
1
(x) ≤ ϕ(x) ≤
f
2
(x), ∀x ∈ V (x
0
), x = x
0
lim
x→x
0
ϕ(x) = A
∃ lim
x→x
0
f(x)
y = f(x) x
0
x
0
∀ε > 0
∃ V (x
0
) |f(x

) − f(x”)| < ε ∀x

, x” ∈ V (x

x
= 1
f(x) =
sinx
x
x
0
= 0
V (x
0
) = x : 0 < |x| <
π
2
0 < x <
π
2
S
AOM
< S
quatAOM
< S
AOT

1
2
OA.MH <
1
2
.OA


2
< x < 0
lim
x→0
1
cosx
= 1 ⇒ lim
x→0
sinx
x
= 1
x
0
x
0
lim
x→x
0
= f(x
0
) x
0
x
0
x
0
x
0
f(x
0

) = 0 c = u
0
f(u
0
< 0)
α
1
= u
0
, β
1
= β
0
f(u
0
> 0) α
1
= α
0
, β
1
= u
0

1
, β
1
] f(α
1
).f(β

, β
n+1
= β
n
f(u
n
> 0)
α
n+1
= α
n
, β
n+1
= u
n
α
n
, β
n
f(x)
µ ∈ [m, M]
m = min f(x), M = maxf (x) ∃ξ ∈ (a, b) : f (ξ) = µ
y = f(x) f(x)
c ∈ (a, b)
f

(c) f

(c) = 0
y = f(x) [a; b]

∈ [2; 3]; ; c
2013
∈ [2013; 2014]
f

(c
1
) = 0, f

(c
2
) = 0, , f

(c
2013
) = 0
⇒ f

(x) ⇒
f

(x) = 0 C
1
; C
2
; C
3
; ; C
2013
f(x) = x

∃c ∈ (a, b) h

(c) = 0
h

(x) = f

(x) −
f(b) − f(a)
b − a
⇒ h

(c) = f

(c) −
f(b) − f(a)
b − a
= 0 →
f

(c) =
f(b) − f(a)
b − a
, c ∈ (a, b)
0 < b < a
a − b
a
< ln
a
b

<
ln
a
b
a − b
<
1
b

a − b
a
< ln
a
b
<
a − b
b
y = f(x)
[a, f(a)] , [b, f(b)]
y = f(x), y = g(x)
g

(x) = 0, ∀x ∈ (a, b)
∃c ∈ (a, b) :
f(b) − f(a)
b − a
=
f

(c)


f

(c)
g

(c)
=
f(b) − f(a)
g(b) −g(a)
y =
ax
2
+ bx + c
dx + e
y =
x
2
− mx + 1
x − 1
m = 1
y =
x
2
− x + 1
x − 1
= x +
1
x − 1
. (∗∗)

y = 0 → x
2
−x + 1 = 0 →
x = 0 → y = −1 →


x = X + 1
y = Y + 1
Y + 1 = X + 1 +
1
X + 1 −1
⇔ Y = X +
1
X
1
x − 1
⇒ x − 1 = ±1 ⇒

x
1
= 1, y
1
= −1
x
2
= 2, y
2
= 3.
M
1

− x + 1
x − 1
= −2x ⇔ 3x
2
−3x + 1 = 0

x
2
− x + 1
x − 1
= 2x ⇔ x
2
− x − 1 = 0
⇒ x =
1 ±

5
2
x =
1 ±

5
2

d
1
= OD =

(x
2

0
+ (−1)y
0
|

1
2
+ (−1)
2
=
1

2
.|x
0
− (x
0
+
1
x
0
− 1
)| =
1

2
.
1
|x
0

0
− 1|
=
2
4

2
|x
0
− 1| =
1

2.|x
0
− 1|
⇔ (x
0
− 1)
2
=
1

2
⇒ x
0
= 1 ±
1
4

2

1
β

MN =

(x
M
− x
N
)
2
+ (y
M
− y
N
)
2

(α + β)
2
+ (α + β +
1
α
+
1
β
)
2
(α + β).


+ 2 2

(2.

2 + 2)
⇔ α = β =
1
4

2
x
M
= 1 +
1
4

2
x
N
= 1 −
1
4

2
M(x
0
, y
0
)
d


2
.
1
|x
0
− 1|
d
1
.d
2
=
1

2
M(x
0
, y
0
)


(x
0
− 1)
2
+ (y
0
− 1)
2

2

(2(x
0
− 1)
2
.
1
(x
0
− 1)
2
+ 2
⇔ 2(x
0
− 1)
2
=
1
(x
0
− 1)
2
⇔ (x
0
− 1)
4
=
1
2

⇒ Y = X +
1
X
. (i)

⇒ M
,
(−X, −Y )
M
,
M
,

M
,
⇔ Y (−X) = Y
,
(X)
Y
,
(X) = 1 −
1
X
2
= 1 −
1
(−X)
2
= Y
,

0
)
⇔ y =

1 −
1
(x
0
− 1)
2

.x +
x
0
(x
0
− 1)
2
+
1
x
0
− 1
⇔ y =

1 −
1
(x
0
− 1)

E

1, 1 +
2
X
0
− 1

⇒ IE =
2
|x
0
− 1|
F (2x
0
− 1, 2x
0
− 1)
⇒ IF =

(2x
0
− 2)
2
+ (2x
0
− 2)
2
=


0
⇒ y
,
(x
0
) = −2 ⇔ 1 −
1
(x
0
− 1)
2
= −2 ⇔ (x
0
− 1)
2
=
1
3
⇒ x
0
= 1 ±
1

3
x
0
= 1 ±
1

3

0
, 0) ∈
(x
0
, 0) y = k(x − x
0
)
x
2
− x − 1
x − 1
= kx − kx
0
⇔ f(x) = (k − 1)x
2
− (k − kx
0
+ 1)x + kx
0
− 1 = 0
x = 1 ⇒  = 0
⇔ (k + kx
0
− 1)
2
− 4(k −1)(kx
0
− 1) = 0
⇔ (x
0

= −1 ⇒
−3
(x
0
− 1)
2
= −1 ⇔ (x
0
− 1)
2
= 3 ⇒ x
0
= 1 ±

3
x
0
= 1 ±

3
(x
0
, 0) ∈
(x
0
, 0) y = k(x − x
0
)
x
2

2
+ 4k + 4kx
0
− 4 = 0
⇔ (x
0
− 1)
2
k
2
+ 2(x
0
+ 1)k −3 = 0 (i)
x
0
= 1 ⇒ ∃ k =
3
4
x
0
= 1 
,
≥ 0
⇔ (x
0
+ 1)
2
+ 3(x
0
− 1)

0

1
x
0
− 1
= 1 − x
0
+
x
0
− 1
(x
0
− 1)
2
⇔ −
1
x
0
− 1
=
1
x
0
− 1

2
x
0











k = 1
(k −1)f (−1) > 0
1 <
S
2










k = 1
(k −1)(−1) > 0
1 <
−2
2(k −1)









k = 1
(k −1)f (1) > 0
S
2
< 1










k = 1
(k −1)(−1) > 0
−2
2(k −1)
< 1









k = 1
∀k
1 + (k −1)(k + 2) = 0


k = 1
k
2
+ k −1 = 0
⇔ k =
−1 ±

5
2
f(x, y) = kx −y + k + 1 = 0
⇔ −2 < k <
2
3
⇔ k < −2
k >
2
3
⇔ k = −2
k =

x − 1
− log
2
a = 0
x
2
− x + 1
x − 1
= log
2
(a)
y = log
2
a
≤ 0 ⇒ ∃log
2
a ⇒
log
2
(a) < −1 = log
2
(2)
−1
= log
2
(
1
2
⇔ 0 < a <
1

log
2
a = 3 ⇔ a = 8 ⇒ y = log
2
a

log
2
a > 3 ⇔ a > 8 ⇒

x
2
− x + 1
x − 1
=
a
2
− a + 1
a − 1

−∞ < a < 1
a = 0

a
2
− a + 1
a − 1
< −1
⇒ y =
a

2
− a + 1
a − 1

45
0
⇒ 45
0
y
,
(x) = ±1 ⇒ 1 −
1
(x − 1)
2
= ±1 ⇒ 1 −
1
(x − 1)
2
= −1
⇔ 2 =
1
(x − 1)
2
⇔ (x − 1)
2
=
1
2
⇒ x = 1 ±
1


2
)
x
2
− x + 1
x − 1
= 6 x
A
, x
B
: |x
A
− x
B
| = 6
⇔ x
2
−(b+1)x+1+b = 0 x
A
, x
B
: |x
A
−x
B
| = 6
⇔ x
A
, x

y =
x
2
− 2x + 2
x − 1
. (33)
⇔ y =
x
2
− x + 1
x − 1
− 1.
y =
x
2
x − 1
. (34)
⇔ y =
x
2
− x + 1
x − 1
+ 1
y =
x
2
− 3x + 3
x − 2
. (35)
⇔ y =








x
2
− x + 1
x − 1
,
x
2
− x + 1
x − 1
> 0

x
2
− x + 1
x − 1
,
x
2
− x + 1
x − 1
< 0
⇔ y =



x
2
− x + 1
x − 1
, x ≥ 0
x
2
+ x + 1
−x − 1
, x < 0
|y| =
x
2
− x + 1
x − 1
. (39)
x
2
− x + 1 > 0, ∀x ⇒ (39) ⇔ x − 1 > 0 ⇔ x > 1
⇔ y = ±
x
2
− x + 1
x − 1
,
x
2
− x + 1
x − 1


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