Một số bài toán tổng hợp về hàm số

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
y = f(x) x → x
0
y = f(x) x
0
x
0
∀ε > 0, ∃δ = δ(ε) : 0 < |x − x
0

| < δ ⇒ |f(x) − A| < ε
lim
x→1
(2x + 3) = 5
|(2x + 3) − 5| < ε ⇔ 2|x − 1| < ε ⇔ |x − 1| <
ε
2
∀ε > 0, ∃δ(ε) =
ε
2
⇒ ∀x : |x − 1| < δ ⇒ |(2x + 3) − 5| < ε
lim
x→1
(2x + 3) = 5
y = f(x) x
0
x
0
x → x
0
x
n
, x
n
= x
0
x
0
f(x
1

); 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
)

lim
n→∞
x
n
= 0
0 ≤ |f(x
n
)| = |x
n
.sin
1
x
n
| ≤ |x
n
|
lim
n→∞
x
n
= 0 → lim
n→∞
|x
n
| = 0 ⇒ lim
n→∞
f(x
n
) = 0
lim

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

1
1 +
1
nπ
− 1
= sinnπ = 0
f(
n
) = sin
1
1 +
2
(4n + 1)π
= sin(
π
2
+ 2nπ) = 1
⇒ lim
n→∞
f(x
n
) = 0; lim
n→∞
f(
n
) = 1
lim
x→1
sin
1

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

− 0) ∀ε > 0 ∃δ = δ(ε) > 0 : ∀x ∈
0 < x
0
− x < δ ⇒ |f(x) − A| < ε
f(x) = 2014 +
1
1 + 5
1
x − 1
, x → 1
1
1 − x
→ +∞ x → 1 − 0
1
1 + 5
1
x − 1
→ 0 f(1 − 0) = 2014 x → 1 + 0
1
1 − x
→ −∞ 5
1
x − 1
→ 0 f(1 + 0) = 2015
lim
x→x
0
f(x) = A f(x)
V (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

, lim
x→x
0
f
2
(x) = A
2
V (x
0
) : f
1
(x) ≤ f
2
(x), ∀x ∈ V (x
0
), x = x
0
A
1
≤ A
2
lim
x→x
0
f
1
(x) = A, lim
x→x
0
f

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

0
); x

, x” = x
0
lim
x→x
0
f(x) = A, lim
x→x
0
g(x) = B; A, B
lim
x→x
0
[f(x) ± g(x)] = A ± B; lim
x→x
0
[f(x).g(x)] = A.B B = 0
lim
x→x
0
f(x)
g(x)
=
A
B
lim
x→0
sinx

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


AM <
1
2
.OA.AT
⇔ MH <

AM < AT ⇔ sinx < x < tanx ⇔ 1 <
sinx
x
<
1
cosx
−π
2
< x < 0 x = −t ⇒ 0 < t <
π
2
cosx = cos(−t) = cost;
sinx
x
=
sin(−t)
−t
=
sint
t
0 < t <
π
2
−π

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) = f (x
0
) f(x
0
+ 0) = f (x
0
)
[a, b]
∀x ∈ (a, b)
f(a).f(b) < 0
∃c ∈ (a, b) : f (c) = 0
α
0
= a, β
0
= b ⇔ f(α
0
) < 0; f(β
0
) > 0
u
0
=
α
0
+ β
0
2
f(u
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
[α
1
, β
1
] f(α
1
).f(β

1
) < 0
u
1
=
α
1
+ β
1
2
[α
n
, β(n)], u
n
=
α
n
+ β
n
2
f(u
n
) = 0 c = u
n
f(u
n
) < 0 α
n+1
= u
n

, 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]

f(a) = f(b) c ∈ (a; b)
f

(c) = 0
⇒ ∃M = max f (x)
m = minf(x)
f(a) = f(b) = m = M ⇒ f(x) = C(const) ∀x ∈ (a; b).
⇒ f

(n) = 0 ∀x ∈ (a; b) ⇒ f

(x) = 0, ∀x ∈ (a; b)
c ∈ (a; b) f

(c) = 0
f(x) = (x − 1)(x − 2) (x − 2014)
f

(x) = 0
f
1
= f
2
= = f
2014
[1; 2]; [2; 3]; ; [2013; 2014]
⇒ ∃c
1
∈ [1; 2]; c
2

∈ [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

2
−xsinx −cosx = 0
f

−Π
2

> 0, f (0) < 0, f

Π
2

> 0
f
,
(x) = 0 f
,
(x) = 2x −sinx−xcosx + sinx =
x(2 − cosx) = 0
y = f(x)
c ∈ (a, b)
f

(c) =
f(b) − f(a)
b − a
h(x) = f (x) − f(a) − (x − a)
f(b) − f(a)
b − a
∀x ∈ [a, b]

∃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

<
a − b
b
f(x) = lnx [a, b]
[b, a], f

(x) = 1, ∀x ∈ (b, a) ∃c ∈ (b, a)
f

(c) =
f(a) − f(b)
a − b
⇔
1
c
=
lna − lnb
a − b
0 < b < c < a →
1
a
<
1
c
<
1
b
⇒
1
a

<
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)

g

(c)
∃c
0
∈ (a, b)
g(b) −g(a) = g

(c
0
)(b − a) g

(c
0
) = 0 g(a) = g(b)
h(x) = f (x) − f(a) −
f(b) − f(a)
g(b) −g(a)
[g(x) −g(a)]
[a; b] ⇒ ∃c ∈ (a, b) : h

(c) = 0
⇔ f

(c) −
f(b) − f(a)
g(b) −g(a)
.g

(c) = 0

⇒
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
. (∗∗)

∀x = 1
x = 1 lim
x→1
x
2
− x + 1
x − 1
= ∞
y = x lim
x→1

x
2
− x + 1
x − 1

− x = lim
x→∞
1
x − 1
= 0
y
,
=
x
2
− 2x
(x − 1)
2
= 0 ⇔ x = 0, x = 2

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

(0, −1); M
2
(2, 3)
x
2
+ y
2
= 1
M
1
(0, −1)
x
2
+ y
2
= 1 M
1
(0, −1)
x
2
− x + 1
x − 1
= −x ⇔ 2x
2
−2x + 1 = 0 ⇒
⇒
M(x, y) ∈ Oxy
⇒ y = ± 2x
x
2

− 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

D
+ y
2
D
) =
√
13
⇒ d
2
=
OT = 1
⇒ d > d
1
; d > d
2
⇒
M(x
0
, y
0
)
d
1
= |x
0
− 1|
d
2
=
|1.x

0
+ (−1)y
0
|

1
2
+ (−1)
2
=
1
√
2
.|x
0
− (x
0
+
1
x
0
− 1
)| =
1
√
2
.
1
|x
0

− 1|
d
1
+ d
2
= |x
0
− 1| +
1
√
2.
1
|x
0
− 1|
d
1
+ d
2
≥ 2

|x
0
− 1|.
1
√
2
.
1
|x

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

d
1
+ d
2
2
4
√
2
M(x
0
, y
0
)
x
0
= 1 +
1
4
√
2
M

1 + α, 1 + α +
1
α

(α > 0)
N

1 − β, 1 − β −

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).


(1 + (1 +
1
αβ
)
2
≥ 2
√
αβ.

2 +
2
αβ
+
1
α
2
β
2
2

2αβ +
1
αβ
+ 2 ≥ 2

2

2αβ.
1
αβ

+ 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

1
= |x
0
− 1|
x − y = 0
d
2
=
|x
0
− y
0
|
√
2
=
|x
0
− (x
0
+
1
x
0
− 1
)|
√
2
=
1

√
2
.
1
|x
0
− 1|
d
1
.d
2
=
1
√
2
M(x
0
, y
0
)
⇒

(x
0
− 1)
2
+ (y
0
− 1)
2


(x
0
− 1)
2
+ (x
0
− 1 +
1
x
0
− 1
)
2

2(x
0
− 1)
2
+
1
(x
0
− 1)
2
+ 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

⇒ x
0
= 1 ±
1
4
√
2
x
0
= 1 −
1
4
√
2
x
0
= 1 +
1
4
√
2
−→
OI
⇒

x = X + 1
y = Y + 1
⇒ Y + 1 = X + 1 +
1
X + 1 −1

⇒ 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
,

(−X)
k = y
,
(x) = 1 −
1
(x − 1)
2
< 1
x
0
y −

x
0
+
1
x
0
− 1

=

1 −
1
(x
0
− 1)
2

(x − x

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)

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

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
=


(8(x
0
− 1)
2
= 8|x
0
− 1|
S =
1
2
.IE.IF. sin 45
0
=
1
2
.
2
|x
0
− 1|
.8|x
0
− 1|.
√
2
2
= 4
√
2
x

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

y =
1
2
x + 2014
x
0
⇒ y
,
(x
0
).
1
2
= −1 ⇒ y
,
(x
0
) = −2 ⇔ x
0
= 1 ±
1
√
3
x
0
= 1 ±
1
√
3
(x

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)
2
k
2
− 2(x
0
+ 1)k + 1 − 4x
0
k
2
+ 4k + 4kx
0
− 4 = 0
⇔ (x
0
− 1)
2
k
2
+ 2(x
0
+ 1)k −3 = 0 (i)
k
1
, k
2
k
1
.k
2

= −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

− 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)
2
k
2
− 2(x
0
+ 1)k + 1 − 4x
0
k

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)

2
≥ 0 ∀x
0
= 1
x
0
= 1
y −

x
0
+
1
x
0
− 1

=

1 −
1
(x
0
− 1)
2

[x − x
0
]
1 − x

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

− 1
= 0
∀x
0
= 1
y = kx + k + 1
x
2
− x + 1
x − 1
= kx + k + 1 ⇔ f (x) = (k − 1)x
2
+ 2x − (k + 2) = 0
x
1
< 1 < x
2
⇒

k = 1
(k −1)f (1) < 0
⇔

k = 1
(k −1)(−1) < 0
⇔ k > 1
1 < x
1
< x
2

⇒









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
1
k + 1
+ 1 < 0
⇔



k < 1
k
k −1
< 0
⇔ 0 < k < 1
x
1
< x
2
< 1 ⇒










k = 1
(k −1)f (1) > 0
S
2
< 1
⇔









k = 1
(k −1)(−1) > 0
−2
2(k −1)
< 1
⇔




k < 1
1
k + 1
+ 1 > 0
⇔



k < 1
k
k −1
> 0
⇔ k < 0
x = 1 ⇒







k = 1
f(1) = 0

,
= 0
⇔








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 =

2
3
x
2
− x + 1
x − 1
= a(x − 1) + 1

 ⇔ 
⇔ a > 1
⇔ a > 1
x
2
− (a + 1)x + 1 + a = 0.
x
2
− x + 1 = a(x − 1)
⇔
x
2
− x + 1
x − 1
= a
⇒
⇒
⇒
⇒
x
2
− x + 1

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

2
y = log
2
a ⇒
log
2
(a) = −1 ⇔ a =
1
2
⇒ y = log
2
a
⇒
−1 < log
2
a < 3 ⇔ log
2
1
2
< log
2
(a) < log
2
8 ⇔
1
2
< a < 8
⇒ y = log
2
a ⇒

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
⇒
a
2
− a + 1
a − 1
= −1 ⇔ a = 0 ⇒ y =
a
2
− a + 1
a − 1
⇒
a
2
− a + 1
a − 1
= 3 ⇔ a = 2 y =
a
2
− a + 1
a − 1
⇒
a
2
− a + 1
a − 1
> 3 ⇔ a > 1, a = 2 ⇒ 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 = 1 ±
1
√
2
⇒ y −y

1 +
1
√
2

= −

x − 1 −
1
√
2

x = 1 −
1
√
2
⇒ y −y(1 −
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

B
(x
A
+ x
B
)
2
− 4x
A
x
B
= 36
⇒

 > 0
(b + 1)
2
− 4(1 + b) = 36
⇔

(b + 1)
2
− 4(b + 1) > 0
b
2
− 2b − 3 = 0
⇔

(b + 1)(b − 3) > 0
b = −1, b = 3

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 − 1)
2
− (x − 1) + 1
(x − 1) − 1
y =
x
2
+ x + 1
x
. (36)
⇔ y =
(x + 1)
2
− (x + 1) + 1
(x + 1) − 1
y =
x
2
− x + 1
|x − 1|
. (37)
x
2
− x + 1 > 0, ∀x ⇒ ⇔ y = |
x
2
− x + 1
x − 1
|
⇔ 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 > 1
−
x
2
− x + 1
x − 1
, x < 1
y =
x
2
− |x| + 1
|x| − 1
. (38)
⇔ 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

> 0
y = |
x
2
− |x| + 1
|x| − 1
|. (40)