But fx( ) fx( ) xt
x

xt

(^21) x
2
2
2
1
1
2
2
1
2
1
− = −

• − −


• 
  

  =
  − + − +
  ++
  ()[() ]
  ()()
  ,
  xxtxx xx
  xx
  21 12 12
  2
  2
  1
  2
  22
  11
  and t(x 1 + x 2 ) – 2x 1 x 2 + 2

  
  

  
  

  t(x 1 + x 2 )− 2 +>^1
  2
  xx 12 0 , thus f(x 2 – f(x 1 ) > 0.
  Consequently, f(x) is an increasing function on
  the interval [α, β].
  Since α + β = t and αβ=−^14 ,
  g(t) = max f(x) –min f(x) = f(β) – f(α)

  
  

  ++⎛⎝⎜ ⎞⎠⎟

  
  

  
  


• 
  

  =
  ++

  
  


• 
  

  tt
  t
  tt
  t
  22
  2
  22
  2
  (^152)
  25
  16
  8125
  16 25
  ()
  () (tan )
  cos cos
  cos
  2
  (^823)
  (^169)
  2
  2
  gu
  u u
  u
  i
  i i
  i

  
  

  ⎛ +
  ⎝⎜
  ⎞
  ⎠⎟

  
  


• 
  

  =

  
  


• 
  


• 
  

  (^1624)
  16 9 2
  cos
  cos
  cos
  u
  u
  u
  i
  i
  i
  ≥ ×

  
  


• 
  

  =

  
  


• 
  

  216 24 =
  16 9
  16 6
  16 9
  22123
  cos cos
  (,,)
  uu
  i
  ii
  So,
  (tan )
  (^1) (cos)
  16 6
  16 9
  1
  (^32)
  1
  3
  gu
  u
  i i
  i
  ==i
  ∑∑≤
  (^1) +
  =×+×−
  ⎛
  ⎝
  ⎜
  ⎞
  ⎠
  ⎟

  
  

  ∑
  1
  16 6
  16 3 9 3 9 2
  1
  3
  sin ui
  i
  Since sinuui and , ,
  i
  i

  
  

  ∑ = ∈
  ⎛
  ⎝⎜
  ⎞
  1 ⎠⎟
  3
  10
  2
  π
  i = 1, 2, 3, we obtain
  312
  1
  3
  1
  3 2
  sin uui sin.
  i
  i
  ==i
  ∑∑>
  ⎛
  ⎝
  ⎜
  ⎞
  ⎠
  ⎟ =
  Thus,
  (tan ) (tan ) (tan )
  111
  gu gu gu 123
  ++
  < − ×
  ⎛
  ⎝⎜
  ⎞
  ⎠⎟=
  1
  16 6
  75 9 1
  3
  3
  4

  
  

6. 
   

   Remark : Part (1) of this problem is well-known,
   we put in an inequality to increase the level of
   difficulty.

   
   


7. 
   

   1st solution :
   Let z = x + yi (x, y ∈ R), then
   xyia ti+=cos^422 ⋅ ++ 2 ⎛⎝⎜^1 bi⎠⎟⎞cos t⋅sin t
   2
   +(1 + ci) sin^4 t.
   Separating real and imaginary parts, we get
   x = cos^2 t · sin^2 t + sin^4 t = sin^2 t,
   y = a(1 – x)^2 + 2b(1 – x)x + cx^2 (0 ≤ x ≤ 1)
        

   
   

 





That is,

y = (a + c – 2b)x^2 + 2(b – a)x + a (0 ≤ x ≤ 1) ...(i)

Since A, B, C are non-collinear, a + c –2b ≠ 0.

So equation (i) is the segment of a parabola

(see the diagram). Furthermore, the mid points

of AB and BC are D^1 ab E bc

42

3

42

⎝⎛⎜ ,and,+ ⎠⎟⎞ ⎛⎝⎜ + ⎞⎠⎟,

respectively. So the equation of line DE is

ycax=() (− ++^1 abc− )

4

32 ..(ii)

Solving equations (i) and (ii) simultaneously, we

get ()ac bx+ −− (^2) ⎝⎜⎛^1 ⎠⎟⎞ =
2
0
2

. Then x=^1
2

,

since a + c –2b ≠ 0. So the parabola and line DE have

one and only one common point P^1 ac b

2

2

4

⎛⎝⎜ ,.++⎞⎠⎟

Notice that^1

4

1

2

3

4

<<, so point P is on the segment

DE and satisfies equation (i), as required.

2 nd solution :

We can solve the problem using the method of

complex numbers directly. Let D, E be the mid

points of AB, CB, respectively.

Then the complex numbers corresponding to

D, E are^1

2

1

(^0142)
()zz+=+ab+ i,
1
2
3
(^1242)
()zz+=+bc+ i, respectively. So, complex