امتحانات الشھادة الثانویة العامة الفرع : علوم عامة مسابقة في مادة الریاضیات المدة أربع ساعات

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1 سنة ۲۰۰۷ الا كمالیة الا ستثناي یة I ( points) وزارة التربیة والتعلیم العالي المدیریة العامة للتربیة داي رة الامتحانات امتحانات الشھادة الثانویة العامة الفرع : علوم عامة دورة الاسم: الرقم: مسابقة في مادة الریاضیات المدة أربع ساعات عدد المساي ل : ست ملاحظة : یسمح با ستعمال آلة حاسبة غیر قابلة للبرمجة او اختزان المعلومات او رسم البیانات یستطیع المرشح الا جابة بالترتیب الذي یناسبھ ) دون الالتزام بترتیب المساي ل الوارد في المسابقة) The comple plane is referred to a direct orthonormal system ( O ; u, v ) Let be the nonero comple number defined by its eponential form is denoted by Consider the points, and C of respective affies =, = r e = and i, whose conjugate = C - etermine the eponential form of each of the numbers and C in terms of r and - etermine a measure of the angle ( O ; OC ) in terms of educe the values of such that O, and C are collinear and O belongs to [C] - Suppose in this part that = a) Verify that = C b) Let be the point of affi such that = Calculate each of the numbers and C in terms of r and prove that the straight lines ( ) and (C ) are parallel c) Prove that C is an isosceles trapeoid II ( points) The space is referred to a direct orthonormal system ( O ; i, j, k ) Consider the points (- ; ; 0), ( ; ; 0) and C(0 ; 0 ; ) - Calculate the area of triangle C - Calculate the volume of the tetrahedron OC educe the distance from O to plane (C) - a) Write an equation of the plane (C ) b) Show that the point O ' ; ; is the symmetric of O with respect to plane (C ) c) Calculate cos (OO ) as well as the cosine of the angle between the line (O ) and the plane (C ) - Let J be the midpoint of [] a) Verify that the plane (COJ ) is the mediator plane of [] b) Calculate the cosine of the acute angle between the two planes (COJ) and (O)

2 III ( points) C is a trapeoid of bases [] and [C] such that: [] is fied and = ; [C] is variable and C = 6 Let F be the mid point of [] - a) Prove that if the perimeter of C remains equal to 8, then moves on an ellipse (E) of foci and F b) raw (E) In all what follows, the plane is referred to the orthonormal system ( ; i, j ) such that ( ; 0) ( ) y - a) Prove that + = is an equation of the ellipse (E ) 5 6 b) Calculate the eccentricity of (E) and determine an equation of the directri (d ) associated to - Let L be one of the points of intersection of (E ) with the ais of ordinates a) etermine an equation of the tangent (T ) to (E ) at L b) Show that (T) cuts the focal ais of (E ) at a point belonging to the directri (d ) F C IV ( points) In order to ensure that the cars in a given city are functioning well, a certain company is inspecting all the cars in this city It is known that 0 % of these cars are under guarantee mong the cars under guarantee, the probability that a car has a defect is 00 mong the cars not under guarantee, the probability that a car has a defect is 0 - Calculate the probability of each of the following events: : «The inspected car is under guarantee and has a defect» : «The inspected car has a defect» - Prove that the probability that an inspected car is under guarantee knowing that it has a defect is equal to - The car inspection is for free if the car is under guarantee; it costs LL if the car is not under guarantee and does not have a defect; it costs LL if the car is not under guarantee and has a defect enote by X the random variable that is equal to the cost of inspection of a car a) What are the possible values of X? b) etermine the probability distribution of X and calculate the epected value of X - The company inspects an average of 50 cars per day Estimate the daily inspection cost collected by this company

3 V ( points) Given a triangle C such that = 6, C = and ( ; C) = () Let I be the orthogonal projection of on (C ) - Let h be the dilation of center I that transforms C onto Construct the image (d ) of the line (C ) under h educe the image of under h - Let S be the similitude that transforms onto, and C onto a) etermine the ratio and an angle of S b) etermine the image by S of each of the two straight lines (I ) and (C ) educe that I is the center of S c) etermine the image of ( ) by S educe that S ( ) = - a) etermine the nature and the characteristic elements of S S b) Prove that S S( ) = h( ) c) Prove that S S = h - Let E be the mid point of [C ] a) etermine the points F and G such that F = S(E ) and G = S(F ) b) Show that the points E, I and G are collinear C I VI (7 points) - Consider the differential equation (I ) : y y = ln - Verify that y = + ln is a particular solution of the equation (I ) - etermine the general solution Y of the differential equation y ' y = 0 - a) Verify that Y + y is the general solution of the differential equation (I ) b) etermine the particular solution y of the equation (I ) such that y ( ) = The figure below shows, in an orthonormal system, the representative curve (γ ) of the function h defined on the interval ] 0; + [ by h( ) = + ln y O (γ)

4 - a) Prove that 5 < < 5 b) etermine the maimum of h () - a) Using integration by parts, calculate ln d in terms of b) educe the area S () of the shaded region bounded by (γ ) and the ais of abscissas + ln C- Let f be the function defined on ] 0 ; + [ by f ( ) = esignate by (C ) the representative curve of f in an orthonormal system ( O ; i, j ) - a) etermine the point of intersection of (C ) with the ais of abscissas b) Prove that the aes of the system are the asymptotes of (C ) - a) Set up the table of variations of f and prove that f ( ) = b) raw (C ) - a) Prove that the restriction of f on the interval [ ; + [ has an inverse function f b) etermine the domain of definition and the domain of differentiability of f c) Solve the inequality f ( ) > n+ - Let ( I n ) be the sequence defined, for n, by I n = f ( ) d n -Prove that, for all in the interval [ ; + [, 0 f ( ) - educe that, for all natural integers n, n + 0 I n ln n - etermine the limit of the sequence ( I n )

5 وزارة التربیة والتعلیم العالي المدیریة العامة للتربیة داي رة الامتحانات معیار التصحیح امتحانات الشھادة الثانویة العامة الفرع : علوم عامة مسابقة في مادة الریاضیات الدورة الا ستثناي یة للعام ۲۰۰۷ I- ( points) a b = = e r i ; nswer r e = r e i i C = = r e i ( O ; OC ) = ( u ; OC ) ( u ; O ) = ( ) = O, and C are collinear and O [C ] is equivalent to ( O ; OC ) = + k Therefore = + k where k IZ i i i i C = e r e = e = e = r r cos + = + = = = ; r r C = = ( ) = i ( i r sin ) = r = ( real number ) C r being a real number, then ( ) and (C ) are parallel C or Each of the numbers and C is a real number, then each of the straight lines ( ) and (C ) is parallel to the ais of abscissas ; consequently they are parallel 5 y c C O = OC = r and O = O = r = = r Hence, O, and are collinear Therefore C is an isosceles trapeoid since its diagonals intersect and determine isosceles triangles O

6 II - ( points) c nswer ( ; - ; 0 ) ; C ( ; - ; ) ; C = - i - 9 j 5k The area of triangle C is S = = 5 units of area C ( ; 9 ; 5) and O ( ; ; 0) ; then O ( C ) = 5 The volume of tetrahedron OC is V = O ( C ) = 5 = units of volume d 5 If d is the distance from O to plane (C ), then V= d S= Therefore d = 5 = a ( C ) : + 9y = 0 b u ( ; 9 ; 5) is a direction vector of the straight line (OO ') ; ( OO ') : = t ; y = 9t ; = 5t ( OO ') ( C ) : t = ; hence ( OO ') ( C ) = H ; ; H being the mid point of [OO '] ; Therefore O ' ; ; O O ' 5 cos (OO') = = O O ' Let be the angle of (O ) and (C ) O O ' = 5 a + cos cos = = Since is acute, cos = J ; ; 0 ; OJ = 0 hence ( ) ( OJ ) ( or notice that C is isosceles ) lso ( ) ( OC ) since OC = 0 ( or notice that ( ) ( O y ) and C ' ) Therefore the plane (COJ ) is the mediator plane of [] b j ( O ) and (COJ ) ; therefore j cos β = = 0

7 III- ( points) nswer a If + C + C + = 8 then F + = 8 Hence F + = 0 > F The point varies on the ellipse (E ) of foci and F and of length of focal ais a = 0 y (T) (d) L K b I F a b a b I( ; 0) is the center of (E ) ; a = 5 and c = F = Then b = ( ) y The focal ais of (E ) being ', then ( E ) : + = 5 6 c a 5 6 e = = and ( d ) : = I = ; ( d ) : = a 5 c 6 L (0 ; ) ( T ) : 6( L )( ) + 5( yl ) y = 00 5 ( T ) : ( ) + 5y = 5 ; -+5y = 6 6 (T ) cuts ' at K ( ; 0) that belongs to the directri (d ) of (E ) IV- ( points) nswer Let G be the event : «the car is guaranteed» We can construct the following tree : 5

8 0 G G 0 09 = G, then p() = p (G) p( / G ) = 0 00 = 000 ( ) P( G ) + P( G ) P = = = = 0 08 G p G 000 p = = = = p() The possible values of X are : 0 ; and ( ) ( ) a b ( X = 0 ) = p( G) = 0 ( X = 50000) = p(g ) = P(G) P( ) = p p = G p (X = ) = p(g ) = 0 08 = 008 i P( X= i ) E (X ) = pii = = = 8000LL 5 The average control cost of a car is equal to LL Therefore, if the company has controlled 50 cars, we can estimate a cost of = LL V- ( points) (d) G nswer F I C E h ( C ) = The image (d ) of (C ) by h is the parallel to (C ) that passes through The image of by h is the point of intersection of (I ) and (d )

9 a b = ; its ratio is The angle of S is ( C ; ) = [ ] S ( ) = and =, hence S (I ) is the perpendicular to (I ) passing through ; S ( I ) = ( C ) k = C Similarly, S ( C ) = ( I ) S ( I ) = ( C ) and I (I ) hence S( I ) ( C ) 5 S ( C ) = ( I ) and I (C ) hence S( I ) ( I ) Therefore S ( I ) = I and I is the center of S c S ( ) = and =, hence S ( ) is the perpendicular to ( ) passing through ; S () = (d ) S ( ) = ( d ) and ( ) hence S( ) ( d ) ; (C), then S() (I),therefore S() = (I) (d) and S() = a 9 S S = S( I,, ) S( I,, ) = S( I,, ) Hence S S is the 9 dilation h ( I, ) b S S( ) = S( S( )) = S( ) = c S S and h are two dilations of same center I and S S( ) = h( ) = hence S S = h a S ([ C ]) = [ ] and E is the mid point of [C ]; hence S (E ) is the mid point F of [ ] S ([ ]) = [ ] and F is the mid point of [ ] ; hence S(F ) is the mid point G of [ ] b G = S( F ) = S S( E ) = h( E ) Then E, I and G are collinear = VI- (7 points) nswer y' y = n = n y = 0 is a particular solution of the equation ( ) : y ' y = 0 ; y If y 0, = ;ln y = ln + k; y = a y The general solution of ( ) is Y = a where a IR a Y + y = + a + n depends on an arbitrary constant and satisfies the equation ( ) Therefore y = + a + n is the general solution of ( ) b y ( ) = 0 iff a = ; y = + n h ( 5) h(5) (000)( 000) < 0 Hence 5 < < 5 a b The maimum of h() is h( ) = + n a n d = [ n ] d = [ n ] = n + 5

10 b Ca Cb h( ) d = + ( + ) = + n units of S = n area ( ; 0) e f () = lim and + lim f ( ) = ; ' and y ' y are asymptotes to (C ) Ca,5 f n '( ) = f + n h( ) + ) = = ( = y Ο,5 Cb Ca f is continuous and strictly decreasing on [ ; + [ ; f admits an inverse function f Cb f is defined on f ([ ; + [) = ]0 ; ] f is differentiable on [ ; + [ and the equation f '( ) = 0 admits a unique solution = Therefore f is differentiable on f (] ; + [) = ]0 ; [ Cc f ( ) > is equivalent to f ( f ( )) < f ( ) Hence < ; ie ]0 ; [ The figure drawn in Cb shows that, for all [ ; + [, f ( ) > 0 + n h( ) In addition, f ( ) = = ; for [ ; + [, > and h() 0 Then, for all [ ; + [, 0 f () 6

11 n+ n+ d n + For all n, 0 f ()d ; then 0 I n n n n n n + lim n = n() = 0 ; then lim I n = 0 n + n n + 7