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Entrance Exam 2006 -2007 Mathematics Duration : 3 hours

The distribution of grades is over 25

I-(2 pts) The adjacent table is that of a continuous function f
defined on IR. The représentative curve (C) of f x -∞ 1 +∞
admits at +∞ an asymptotic direction parallel to the straight f ̋ (x) + +
line of equation y = x. f ´ (x) + ½ +
1-Determine an equation of the tangent (T) to (C) at the point of f (x) 0 1 +∞
abscissa 1.
2- Draw (C) and (T).
3- Prove that, for all x Є ]1 ; +∞[ , (x +1)/2 < f (x) < x.

II-(4 pts) An urn contains 6 identical balls of which 4 are red and 2 are black.
1- We randomly draw two balls from the urn. Consider the three events:
A0 : the two drawn balls are red
A1 : the two drawn balls have different colors.
A2 : the two drawn balls are black.
Calculate the probability of each of A0, A1 and A2.

2- After the first drawing, the urn contains 4 balls. We randomly draw two new balls from the urn.
Consider the three events:
B0 : the two drawn balls are red
B1 : the two drawn balls have different colors.
B2 : the two drawn balls are black.
a) Calculate p(B0 / A0), p(B0 / A1) and p(B0 / A2). Deduce that p(B0) = 0.4
b) Calculate p(B1) and p(B2)
c) Knowing that only one black ball is obtained in the second drawing, calculate the probability that only
one black ball has been obtained in the first drawing.
3- Calculate the probability that, after the two drawing, the remaining two balls in the urn are red.

III- (6 pts) The space is referred to a direct orthonormal system ( O ; i , j, k )

Consider the point A (-1 ; 1 ; 0), the plane (P) of equation x- 2y +2z - 6 = 0 and the straight line (D) defined by
the system x = m+1 ; y = 2m+1 ; z = 3m+2.
1- Prove that A and (D) determine a plane (Q) and determine an equation of (Q).
2- a) Prove that (P) and (Q) intersect along the straight line (Δ) defined by x = 2 ; y = t - 2 ; z = t.
b) Determine the coordinates of A´, the orthogonal projection of A on (Δ).
3- M is a variable point of (Δ). Let α be a measure of the angle that (AM) makes with (P).
a) Prove that AM x sin α = 3
b) Determine the position of M so that α is maximum. Calculate sin α in this case.
c) What does this value of α represent for the two planes (P) and (Q)?

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4- Consider the circle (C) of center A tangent to (Δ) and lying in the plane (Q). The orthogonal projection
of (C) on the plane (P) is an ellipse (E).
a) Calculate the radius of (C).
b) Determine the coordinates of the center of (E).
c) Calculate the eccentricity of (E).
d) Determine a system of parametric equation of the focal axis of (E).
e) Determine the coordinates of each of the two foci of (E).
f) Calculate the area of the domain bounded by (E) and its auxiliary circle (γ) .

IV- (6 pts) The complex plane is referred to a direct orthonormal system ( O ; u , v )

Consider the points A0, A1 and A2 of respective affixes z0 = 5- 4i, z1 = -1-4i et z2 = -4-i. Let S be the
similitude that transforms A0 into A1 and A1 into A2
1-a) Determine the ratio of S.
b) Prove that the point I (2 ; 2) is the center of the circle (γ) circumscribed about the triangle A0 A1 A2.
c) Calculate the radius of the image of (γ) by (S).
1 i i 3
2-a) Prove that the complex expression of S is z  z .
2 2
b) Deduce the angle of S and the affix d of its center D.
3-Let M be any point of affix z, such that z ≠ d, and M ´, with affix z´ , its image by S.
a) Determine the nature of the triangle DMM´
b) Deduce that d - z´ = i (z - z´).
4-Consider the sequence of points (An) of first term A0 defined by An+1 = S(An)
Let (Un) be the sequence defined on IN by Un = An An+1
a) Plot the points A0, A1 , A2 and construct the points A3, A4 , A5
b) Prove that the sequence (Un) is geometric.
5- Consider the sequence of points (Pk) defined by Pk = Am+4k where m is a given natural number
a) Prove that all the points Pk are collinear.
b) Prove that, for all natural numbers k , Pk+1 = H (Pk) where H is a transformation to be determined.

V-(7 pts) A. Consider the differential equation (1) y  2 y 2e x  y  0 where y is a function defined on IR,
1
such that, for all x in IR, y(x) ≠ 0. Let z  where z is a differentiable function defined on IR.
y
1- Determine the differential equation (2) satisfied by z.
2- Solve the equation (2) and deduce the general solution of equation (1).
1
3- Determine the particular solution of equation (1) that satisfies the condition y(0) =
2

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1
B. Let f be the function defined on IR by f (x) = . Designate by (C) the representative curve of f in an
e  ex
x

orthonormal system ( O ; i , j )
1- Prove that f is an even function.
2- Set up the table of variations of f .
3-a) Set up the table of variations of the function g defined on [0 ; +∞[ by g(x) = f(x) –x.
b) Deduce that the equation f (x) = x admits in [0 ; +∞[ only one solution α. Verify that 0,4 < α < 0,5.
c) Draw (C) (Unit = 4 cm).
4-a) Prove that the restriction de f to the interval [0 ; +∞[ admits an inverse function f -1.
b) Determine the domain of definition of f -1 and calculate f -1 (x).
c) Draw the curve (γ) of f -1 in the same system as (C).
n

C. Let (Vn) be the sequence defined on IN by Vn=  f ( x)dx

0
x
1-a) Prove that, for all x ≥ 0, f (x) < e
b) Deduce that, for all n in IN, Vn ≤ 1- en
n1
2-a) Verify that Vn+1 –Vn =  f ( x)dx
n
b) Deduce that the sequence (Vn) is strictly increasing.
c) Prove that the sequence (Vn) is convergent to a limit  such that 0 ≤  ≤ 1.
ex
3-Verify that f (x) = then Calculate Vn in terms of n and determine 
1 e2 x
4- Calculate in cm2 the area of the domain bounded by (γ), y´y, x´x and the straight line of equation y = 2.

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Entrance Exam 2006 -2007 Solution of Mathematics Duration:3 hours

The distribution of grades is over 25

Exercice I
1
1- The equation of the tangent (T) to (C) at the point of abscissa 1is y = f(1) + f ´(1) (x-1) =1+ (x-1)
2
1 1 1 1
f ´(1) = and f(1) = 1 so y = x   ( x  1)
2 2 2 2

2-
(d) (C)

(T)
2

-1 1

x -∞ +∞

f ’(x) + +∞

f (x) 0

3) For x >1, the representative curve of f is above the tangent and below the straight line of equation y = x
( x  1)
hence  f ( x)  x
2

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Exercise II
C2 2
1-p(A0)= 42 
C6 5

C 41C 21 8
p(A1)= 
C62 15
C22 1
p(A2)= 2

C6 15
C22 1 C32 1
2-a) p(B0 / A0)=  , p(B0 / A 1 )=  , p(B0 / A2)= 1
C42 6 C42 2
1 2 1 8 1 2
p(B0) = p(B0 / A0).p(A0) + p(B0 / A1).p(A1)+ p(B0 / A2).p(A2)= .  .    0.4
6 5 2 15 15 5
2 2 1 8 8
b) p(B1) = p(B1 / A0).p(A0) + p(B1 / A1).p(A1)+ p(B1 / A2).p(A2)= .  .  0 
3 5 2 15 15
or p(B1 / A2) =0
1 2 1
p(B2)= . 
6 5 15
p( A1  B1 ) p( B1 / A1 )  p( A1 ) 1
c) p(A1/B1) =  
p( B1 ) p( B1 ) 2
2 1 1 8 1 2
3- p(2R)= p(A0) x p(B2 / A0)+p(A1) x p(B1 / A1)+ p(B0 / A2).p(A2)= .  .  1. 
5 6 2 15 15 5

Exercise III

1) The point A does not belong to (d) since if -1 = m+1, 1= 2m+1, 0= 3m+2
2
We get m = -2, m = 0, m = 
3
Hence A and (d) determine a plan (Q). B (1, 1, 2) is a point of (d) and M(x, y, z) is a variable point of
  
(Q). An equation of (Q) is AM .( AB  d )  0 which gives x+ y-z =0

 
2) a- n p (1;2;2) and nQ (1;1;1) are not collinear then (P) and (Q) intersect along the straight line (Δ)
Let M (2; t-2; t) be a variable point of (Δ)
M  (P) since x M 2 yM  2 z M  6  2  2t  4  2t  6  0
M  (Q) since x M  yM  z M  2  t  2  t  0
Hence x = 2, y = t-2, z = t is a system of parametric equations of (Δ).

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   
b- A is a point of (Δ) then A (2; t-2; t), AA (3; t-3; t) and   (0;1;1) are orthogonal, then AA .   0
3 1 3
which gives t-3+t = 0, then t = and consequently A(2; ; )
2 2 2

3) a- Let h be the orthogonal projection of A on (P), the angle that (AM) makes with (P) is AMH .

HA 1  2  6
sin  but HA  d ( A; P)  3
AM 1 4  4

3
then sin  and consequently AM .sin   3
AM
b-  is maximum when AM is minimum that is when M is confounded with A in this case
3 3 3
AA(3; ; ) therefore AA  6
2 2 2

HA 3 6
sin   
AA 3 3
6
2

c- (AH)  (P) then (AH)  (Δ) and since (AA’)  (Δ) then (Δ)  ( A’H) hence  is the plane angle of
the dihedral of (P) and (Q).

3
4) a- The radius of (C) is r = AA  6
2

b- The center of (E) is the point H. The vector r n p (1;2;2) is a direction vector of (AH). A system of
parametric equations of (AH) is x   1, y  2  1, z  2 . H is the point of intersection of (AH)
and (P), then  1  4  2  4  6  0 which gives  =1 so H(0; -1; 2)

c- We know that a = r, b = rcos  then, c 2  a 2  b2  r 2  r 2 cos2   r 2 sin2  so c = rsin  = 3

c r sin 6
e  sin 
a 5 3

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d- The focal axis of (E) is the straight line passing through the center of (E) and parallel to the line (Δ)

then   (0;1;1) is the direction vector of the focal axis, a system of parametric equation of the focal
axis is : x = 0, y = k-1, z = k+2

e- Let F be one focus of (E) , F belongs to the focal axis then: F(0; k-1; k+2), HF = c =3 so HF2 = 9.

3 2 3 2 3 2 3 2
But HF (0; k ; k ) so k2 + k2 = 9 which gives k  or k   therefore F(0;  1;  2)
2 2 2 2
3 2 3 2
And F(0;   1;  2)
2 2

f- The area of the auxiliary circle is S1    a 2    r 2 and the area of the ellipse is
S 2   a b    r  r cos , then the area of the domain bounded by (E) and its auxiliary circle is:
3 3
S = S1 – S2 =   r 2    r 2 cos    r 2 (1  cos ) but cos  1  sin 2   and r  6
3 2
27 3
then, S   (1  ) square units.
2 3

Exercise IV

A1 A2 z 2  z1  3  3i 2
1) a- k   
A0 A1 z1  z0 6 2

b- IA0= z0  z I  5  4i  2  2i  3  6i  3 5 , IA1= z1  z I   1  4i  2  2i   3  6i  3 5 ,
IA2= z2  z I   4  i  2  2i   6  3i  3 5
Then IA0 = IA1= IA2 consequently I (2; 2) is the center of circle ( ) circumscribed about triangle A0A1A2

2 3
c- The image of ( ) by S is a circle of radius, R’ = K.R = . 3 5 = 10
2 2
2) a- The complex expression of a similitude is z´= az + b ; S(A0) = A1 gives : zA1= azA0+b and S(A1) = A2
gives : zA2= azA1+b we get the system :
(5-4i) a+b= -1- 4i
1 1 3 1 1 i i 3
(-1-4i) a+b= -4-i that has no solution a   i and b    i then z   z
2 2 2 2 2 2

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
1  i  2 2 2 2 i 4
b- a  ( i ) e
 2  2 2 2 2

 b
then an angle of S is  the affix d of the center D of S is d   1  2i
4 1 a

2    2
3) a- DM ' DM and ( DM ; DM '   (2 ) . Let DM   so DM'   then
2 4 2

MM '2  DM 2  DM '2 2 DM  DM ' cos( )
4
2
 2 2 2 2
MM '2   2   2     so MM '  then the triangle DMM’ is isosceles at M’ and
2 2 2 2 2
 
since M D M '  then DMM’ is right isosceles at M’.
4

z   d DM ' i ( MM
  
b-  e '; DM ')
 1e i 2  i
z   z MM '

z'd  i( z' z) and d - z’ = i(z - z’)

4) a-

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U n1 A( n1) A( n2) A( n1) A( n2) 2

b-  but An1  S ( An )and An2  S ( An1 ) then  and consequently
Un A( n ) A( n1) A( n ) A( n1) 2
U n 1 2 2
=  ( U n ) is a geometric sequence of common ratio and of first term U0 = A0A1 = 6
Un 2 2

    ( DAm4 k  DAm4 k 1 )  DAm 4 k 1 ; DAm4 k 2 )  ( DAm4 k 2 ; DAm4 k 3 )

5) a- ( DPk ; DPk 1 )  ( DAm 4 k  DAm 4 k  4 )
   
=
 ( DAm4 k 3 ; DAm4 k 4 )        (2 )
4 4 4 4

Then the points D, Pk and Pk+1 are collinear; consequently all the points Pk belong to the same straight
line passing through D.

DP( k 1) DA( m 4 k  4 ) DA( m 4 k  4 ) DA( m 4 k 3) DA( m 4 k  2 ) DA( m 4 k 1)

b-   . . .
DP( k ) DA( m 4 k ) DA( m 4 k 3) DA( m 4 k  2) DA( m 4 k 1) DA( m 4 k )

2 2 2 2 1 1
      ( DPk ; DPk 1 )   (2 ) then DPk 1   DPk
2 2 2 2 4 4
Consequently Pk+1 = H(Pk) where h is the dilatation of center D and ratio -1/4

Exercice V
1
A-1) z =  y = 1/z so
y
z'
y  2 then y   2 y 2 e x  y  0
z
z' 1 1
Gives  2  2 2 e x   0 that is  z'2e x  z  0 and (  ); z ' z  2e x
z z z

2) The general solution of z´+z = 0 is z1 = C e-x

z 2  e x is a particular solution of the equation (β) then

z = z1 = z2 = Ce-x + ex is a general solution of (β)
1 1
y ( x)   x is a general solution of (α)
z ( x) Ce  e x

1 1
3) y(0)   c  1  y( x)  x  x is a particular solution of (α)
2 e e
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1
B- 1) The domain of f is centered at O , and f (-x) =  f ( x)  f is an even function
e  ex
x

e x  ex 1 e2x
2) f´(x)=  , f ' ( x)  0 for x  0 than the table of variations of f is :
(e x  e  x ) 1  e 2 x
lim f ( x)  lim f ( x)  0 x -∞ 0 +∞
x x
f ´ (x) + 0 -
f (x) 1/2

3) a- g´(x) = f ´(x) -, then for x > 0, f ´(x) < 0 so

x 0 +∞
g ´ (x) -
x > 0 g´(x) < 0 than the table of variations of g is g (x) 1/2 -∞
lim g ( x)  lim [ f ( x)  x]  0    
x x

1
b- g is continuous and strictly decreasing over [0 ; +∞[, it decrease from to -∞ then its representative
2
curve cuts the axis x’x at a unique point, consequently g(x) = 0 has one root  , so the equation

f(x) = x has one solution  over [0 ; +∞[

g (0,4)  f (0,4)  0,4  0,0625  0

g (0,6)  f (0,6)  0,6  0,056  0 therefore0,4    0,5

c)

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4) a- f is continuous and strictly decreasing over [0; +∞ [, then it admits an inverse function f -1

 1
b- The domain of definition of f 1 is 0; 
 2
x
1 e
y= x x
y then ye2 x  y  e x  0
e e 1  e2 x
1  1  4y2
quadratic equation in e , Δ=1-4y ; e 
x
x 2
; which gives
2y
 1  1
1 1   1 1 
1 1 4 y 2
1 4   ln(2  3)  0 or x  ln  4   ln(2  3)  0
x  ln( ) for y  ; x  ln 
2y 4  1   1 
   
 2   2 
1  1  4 y 2  1  1  4 x 2 
so the accepted solution is x  ln    f 1 ( x)  ln  
 2y   2x 
c- Drawing of the graph (  ) of f -1 in the same system as that of (C).

n
 n  is the sequence defined on IN by  n =  f ( x)dx
0

1 1  1  e 2 x  e 2 x
C- 1) a- f ( x)  e  x   e x
   0 then f ( x)  e  x
e x  ex e x  ex e x  ex

for all x and in particular for x  0

 
n b n

 f ( x)dx   e dx so  f ( x)dx   e  x
x x n
b- f ( x)  e then 0
0 0 0
n

 f ( x)dx  1  e
n
 Vn  1  e n
0

n 1 n 0 n 1 n 1
2) a- Vn+1 –Vn =  f ( x)dx   f ( x)dx   f ( x)dx  
0 0 n 0
f ( x)dx   f ( x)dx
n

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n 1
b  Since f ( x)  0 then  f ( x)dx  0 so
n
n +1 - n  0 then  n +1   n consequently the sequence

( n ) is strictly increasing.

c  The sequence ( n )is increasing and bounded above by 1then Vn  1  e  n  1 so it is convergent to a limit .
Since 0   n  1 then 0    1

1 ex
3) f ( x)  
e x  ex e2 x  1


 
n
ex
vn  
n
dx  arctan e x  arctane n  arctan1  arc tan e n 
e 1
2x 0
0
4

    
lim vn  lim arctan e n     since lim arctan e n  arctan() 
n n 4 2 4 4 n 2
2
 
4) The required area is  f ( x)dx   2  arctan e 2  square units  16  (arctan e 2  )cm 2
0
4 4

Faculty of Engineering – Lebanese University

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