Moreover, when t = –4, for any x ∈ [1, 9], we

have always

(x – 1) (x – 9) ≤ 0

⇔−^1 + ≤

4

(),xx 412 that is f(x – 4) ≤ x.

Therefore, the maximum value of m is 9.

2. 1st solution :
   The slope of the tangent line passing through A
   is y′ = 2x|x = 1 = 2. So the equation of the tangent
   line AB is y = 2x –1.
   Hence the coordinates of B and D are B(0, –1),
   D^1
   2

⎛⎜⎝ , 0 ⎞⎠⎟. Thus D is the midpoint of line segment

AB.

Consider P(x, y), C(x 0 , x 02 ), E(x 1 , y 1 ), F(x 2 , y 2 ).

Then by

AE

EC

=λ 1 , we know x 1 10 x

1

1

1

= +

+

λ

λ

,

y 1 10 x

2

1

1

1

= +

+

λ

λ

. From BF
FC

=λ 2 , we get

x 2 20 x y x

2

2 20

2

(^12)
1
1

• =− +


• 
  

  λ
  λ
  λ
  λ
  ,.
  Therefore the equation of line EF is
  y
  x
  xx
  x
  x
  x
  −

  
  


• 
  


• 
  

  − +

  
  


• 
  

  − +

  
  


• 
  

  =
  −

  
  


• 
  


• 
  


• 
  

  1
  1
  1
  1
  1
  1
  1
  1
  1
  102
  1
  20
  2
  2
  10
  2
  1
  10
  1
  20
  2
  λ
  λ
  λ
  λ
  λ
  λ
  λ
  λ
  λ
  λ
  −− +

  
  


• 
  

  1
  1
  10
  1
  λ
  λ
  x.
  Simplifying it, we get
  [(λ 2 – λ 1 )x 0 – (1 + λ 2 )]y
  = [(λ 2 –λ 1 ) x 02 – 3] x + 1 + x 0 – λ 2 x 02. ...(i)
  When x 0 1
  2
  ≠ , the equation of line CD is
  y
  xx x
  x

  
  

  −
  −
  2
  21
  0
  2
  0
  2
  0
  ...(ii)
  From (i) and (ii), we get
  x x
  y x
  = +

  
  

  ⎧
  ⎨
  ⎪⎪
  ⎩
  ⎪
  ⎪
  0
  0
  1
  3
  3
  ,
  .
  Eliminating x 0 , we get the equation of the trail of
  point P as yx= 31 ()^31 −^2.
  When x 0
  1
  2
  = , the equation of EF is
  −^3 =⎛⎝⎜ −−⎠⎟⎞ + −
  2
  1
  4
  1
  4
  3 3
  2
  1
  21 4 2
  yxλλ λ, the equation
  of CD is x=^1
  2

  
  

. Combining them, we conclude that

(, )xy=⎝⎛⎜^1 , ⎞⎠⎟

2

1

12

is on the trail of P. Since C and

A cannot be congruent, xx 0 1 2

3

≠≠,.

Therefore the equation of the trail is

yx x=^1 −≠

3

31 2

3

(),.^2

2 nd solution :

From 1st solution, the equation of AB is

y = 2x –1, B(0, –1), D^1

2

⎛⎝⎜ , 0 ⎞⎠⎟.

Thus D is the midpoint of AB.

Set γλλ===+==+CD, ,.

CP

t CA

CE

t CB

(^1122) CF
11
Then t 1 + t 2 = 3.
Since AD is a median of ΔABC, SΔCAB = 2SΔCAD
= 2SΔCBD where SΔ denotes the area of Δ.
But^1
tt 12 22
CE CF
CA CB
S
S
S
S
S
S
CEF
CAB
CEP
CAD
CFP
CED
= ⋅
⋅
== +Δ
Δ
Δ
Δ
Δ
Δ
=+⎛
⎝⎜
⎞
⎠⎟
(^1) = + =
2
11
2
3
12 2
12
tt 12 12
tt
γγ γ γtt tt
,
So γ=^3
2
and P is the center of gravity for
ΔABC.
Consider P(x, y) and C(x 0 , x 02 ). Since C is different
from A, x 0 ≠ 1. Thus the coordinates of the center
of gravity P are x=^01 ++xx= + x≠
3
1
3
2
3
(^00) ,,
y=−^11 ++xx=
33
0
2
0
2
.
Eliminating x 0 , we get yx=^1 −
3
() 312.
Thus the equation of the trail is
yx x=^1 −≠
3
31 2
3
(),.^2

3. (1) Let α ≤ x 1 < x 2 ≤ β, then
   4 x 12 – 4tx 1 – 1 ≤ 0, 4x 22 –4tx 2 –1 ≤ 0.
   Therefore, 4(x 12 + x 22 ) –4t(x 2 + x 2 ) – 2 ≤ 0,
   2 1
   2

xx 12 −tx(). 1 +x 2 − < 0