Deduce, from this equality, the area of the region bounded by (E).

Transcription

1 وزارة التربية والتعليم العالي المديرية العامة للتربية دائرة االمتحانات امتحانات شهادة الثانوية العامة فرع العلوم العامة دورة سنة العادية مسابقة في الرياضيات عدد المسائل : ستة المدة : أربع ساعات مالحظة : ي سمح بإستعمال آلة حاسبة غير قابلة للبرمجة أو إختزان المعلومات أو رسم البيانات. يستطيع المرشح اإلجابة بالترتيب الذي يناسبه )دون االلتزام بترتيب المسائل الوارد في المسابقة( االسم : الرقم : I - ( points) Let f be the function defined, on [ ; ], by f() 9 and (C) be its representative curve in an orthonormal system (O ; i, j ). ) Calculate f '() and set up the table of variations of f. ) Draw the curve (C). y ) Consider the ellipse (E) of equation. 9 Show that (C) is a part of (E) and draw (E). ) Using the change of variable = cos, we get : π f() d 6 sin θ dθ (It is not required to prove this equality). Deduce, from this equality, the area of the region bounded by (E). II - ( points) The space is referred to a direct orthonormal system (O ; i, j, k ). Consider the lines (d) and (d') defined by : t m d : y t and d' : y m ( t and m are two real parameters). z t z m ) Show that the lines (d) and (d') intersect at the point A ( ; ; ) and that they are perpendicular. ) Write an equation of the plane (P) determined by (d) and (d'). ) Consider, in plane (P), the line (D) defined by: λ D: y ( λ is a real parameter). z λ a- Prove that the line (D) is a bisector of one of the angles formed by (d) and (d'). b- E ( - ; - ; ) is a point on (D) ; designate by (C) the circle in plane (P), with center E, that is tangent to (d) at T and to (d') at S. Determine the nature of the quadrilateral ATES and calculate the length AT. c- Write an equation of the mediator plane of [AE] and deduce a system of parametric equations of the straight line (TS).

2 III - (.5 points) A tourist agency offers its customers two choices of 7-day voyages: full-board or half-board. The agency published the following advertisement: Destination Choice Full-board Half-board France 5 LL LL Italy 5 LL LL Turkey 8 LL 7 LL This agency estimates that 5 % of its customers choose France, 5 % choose Italy, and the others choose Turkey, and that out of the customers to any destination, 6 % choose full-board. A customer is questioned at random. Consider the following events: F : «the questioned customer has chosen France». I : «the questioned customer has chosen Italy». T : «the questioned customer has chosen Turkey». : «the questioned customer has chosen full-board». ) a- Calculate the following probabilities : P(F) ; P(I) ; P(T) and P(). b- The questioned customer had chosen full-board, what is the probability that he chose Italy? ) Let X be the random variable that is equal to the amount paid to the agency by a voyager. a- Determine the probability distribution of X. b- Calculate the mean (epected value) E(X). What does the number thus obtained represent? c- Estimate the sum received by the agency when it serves voyagers.

3 IV - ( points) C In the adjacent figure, AC, AD and CDE are three direct equilateral triangles π such that ( A;AC) π. Designate by I the midpoint of [ A ]. ) Show that AE = A. A I D E Let S be the direct similitude, with center W, ratio k and angle that transforms A onto and E onto D. π ) Determine k and verify that θ π. ) Designate by (T) the circle circumscribed about triangle ACE. Prove that the image of (T), under S, is the circle (T ') of diameter [D] and deduce that the image of point C under S is point J, the midpoint of [DE]. ) The comple plane is referred to a direct orthonormal system (A; u, v) such that u AI. a- Determine the affies of the points, C, D and E. b- Find the comple form of S and specify the affi of its center W. π 5) Let S' be the direct similitude with center W, ratio and angle. a- Determine the nature and the elements of the transformation S'oS. b- Calculate the affi of point A', the image of A under S'oS. V - (.5points) In the plane referred to an orthonormal system (O; i, j ) (unit cm), consider the parabolas (P) and (P') of equations y = and = y respectively. ) Determine the verte, the focus and the directri of each of these two parabolas. ) Verify that the point A ( ; ) is common to (P) and (P'), and prove that (OA) is a common tangent to these two parabolas. ) Prove that the line (d), perpendicular to (OA) at O, is a common tangent to (P) and (P'). ) Draw (d), (P) and (P'). 5) The area of the region bounded by (P), the ais of abscissas and the line of equation = is equal to cm. Deduce the area, in cm, of the region bounded by (P), (P'), the ais of abscissas and the ais of ordinates.

4 VI - ( 7 points) A- Consider the differential equation (E) : y '' +y' +y =. Let z = y. ) Form a differential equation (E ) satisfied by z, and solve (E ). ) Deduce the general solution of (E) and find the particular solution of (E) whose representative curve, in an orthonormal system (O ; i, j ), is tangent at O to the ais of abscissas. - Let f be the function defined, on IR, by f() e e representative curve in the system (O; i, j ). and (C) be its ) Calculate lim f() ) Calculate lim f() and deduce an asymptote (d) of (C). f() and lim. ) Find f '() and set up the table of variations of f. ) Prove that (C) has a point of inflection I whose coordinates are to be determined. 5) Determine the coordinates of the point of intersection of (C) with its asymptote (d). 6) Draw (d) and (C). 7) Calculate the area of the region bounded by the curve (C), its asymptote (d) and the ais of ordinates. 8) Let g be the function given by g() = ln(f()), and let (G) be its representative curve. a- Justify that the domain of the definition of g is ; ; its table of variations., and set up b- Prove that the line (D) of equation y = is an asymptote of (G). c- Solve each of the equations g() = and g() =. d- Draw (D) and (G) in a new system of aes.

5 GENERAL SCIENCES MATH st SESSION() Q Short answers M f () = 9 f () f () y O I (E) : y = y, y = (9 ) 9 9 or y = 9 9 Thus (C) is the part of (E) located above the ais of abscissas. (E) =(C) (C ) where (C ) is the symmetric of (C) with respect to the ais of abscissas. y O π A = f() d = sin θdθ = 6 sin = 6 u π = 6 ( cos θ)dθ 5

6 A is the point of (d) corresponding to t =. A is the point of (d ) corresponding tom =. V ( ; ; ) and d V d ( ; ;) ; d V. V =, thus (d) M(,y,z) is a point on (P) iff AM.(Vd Vd ) ; So (P) : + y z + = d (d ) A is a point on (D) corresponding to =. -a cos( V d ; Vd.VD V D ) = =. Vd VD Thus one of the angles formed by (d) and (D) is equal to 5 o. Or : V D ( ; ;) ; V D = V d + V d where V d = V d = Or : Let E( ; ; ) be a point on (D) ; d(e ;(d)) = d(e,(d )) =. (D) (d ) II S E -b A T (d) -c  Tˆ AE AT =. Or AT = ES =. Let L be the mid point of [AE] ; L( ; ; ) o Ŝ 9 and ES = ET,then ATES is a square, consequently (Q) : LM.AE where AE( ; ; ) (Q) : + z = T and S are two points belonging to (Q) and to(p), thus (TS) is the line of intersection of (Q) and (P). + z = (TS) : + y z + = (TS) : = +, y = 5, z =. 6

7 F I.6. III. T P( F) =.5.6 =.5 ; P( I) =.5.6 =. -a P( T) =..6 =.; P() = P( F)+P( I)+P( T) =.6 P() =.6 because 6% of customers choose full-board. P(I ). -b P(I/) = = =.5. P( ) a i p i b -c E(X) = 75 5,75,5LL is the average amount paid by a voyager to the agency. If the agency serves voyagers, it estimates to receive: E(X) = 5,, LL C A I E IV (CI) : bisector of A Ĉ then I Ĉ = o. I Ĉ E = 6 o then (C) is a bisector of I Ĉ E. Similarly (D) : bisector of I Dˆ E, then is the center of the equilateral triangle DEC ; E = I = A and consequently AE = A. D A S(A) = and S(E) = D then k = = AE AE and = (AE,D) () = (E, D) () = D J () IV The triangle ACE is right at C, the circle (T) has as diameter [AE], thus (T ) is the circle of diameter [D] = S( [AE] ). J: mid point of [ED] and DE is isosceles then Ĵ D = 9 o and J(T ). 7

8 AEC is a direct semi - equilateral triangle and DJ is direct semiequilateral triangle, thus S(C) = J. -a z = ; z C = + i ; z D = i and z E =. -b i z = az + b = e z + b = ( i ) z + b S(A) = then b = so z = ( i ) z + Or : S(A) = and S(E) = D give = + b and i = a() + b b = and a = ( i ). b z W = a (5 7 i ) S os is similitude of center W, of ratio = and angle 5-a =. S os is a central symmetry with center W. 5-b z A = z W = (5 i ) 7 (P) : y = = ( ) and (P ) : =y. V (P) has verte S(,), focus F( ;) and diretri (y y). (P ) has verte S (, ), focus F ( ;) and directri ( ) The coordinates of A satisfy the equations of (P) and of (P ), thus A is a common point to these parabolas. (OA) : y = (OA) (P) : = ; ( ) =, = = (double root) (OA) is tangent to (P) at A. (OA) (P ) : y = y ; (y ) =, y = y = (double root) (OA) is tangent to (P ) at A. Or : yy =, y = y and ya = ; the equation of the tangent at A to (P) is y = ( ) ; y = which is (OA) Similarly for (P ). Notice that (OA) is the bisector of F ÂF'. (d) : y =. ( d) (P) : ; double root ' " A ( d) (P') : y y ; double root y' y" ya Or (d) is the symmetric of (OA) with respect to the focal ais ( ais of symmetry) of (P) thus (d) is tangent to (P).Similarly (d) is the symmetric of (OA) with respect to the focal ais (y y) of (P ). 8

9 y F A V O F (d) 5 Let S be the required area. The area of the domain bounded by (P), ( ) and the line = is equal to the area limited by (P ), (y y) and the line y =. S = (area of square OFAF ) = 9 6 = cm VI A- A- - y + y + y = ; z = y (E ) : z + z + z = Characteristic equation of (E ) : r +r + = ; r = or r = Then z = C e - +C e - y = z + = C e - +C e - + y() = and y () = give C + C = and C C = ; C = and C = Then y = e - + e - + f() = e - e - + lim f (), the line (d) of equation y = is an asymptote of (C). 9

10 lim f () lim e (e e ) f ( ) e ( e e lim lim y y an asymptotic direction. f () = -e - + e - = e - ( e - ), f () f () f () = e - e - = e - ( e - ) ) ( ) + f () vanishes at = ln and changes sign ;moreover f(ln) = Then the point W(ln ; ) is a point of inflection of (C). (C) cuts (d) ; e - e - + = ; e - (e - ) = ; e - = ; = ln (C) cuts (d) at ( ln ; ). + + y VI -6 ln O W -7 S = ( f ())d = ( e e )d = e e = u ln ln ln g() = ln(f()) g is defined for f() > which corresponds to ] ;[ ]; [ -8 a g() = ln(f()) and ln is strictly increasing, therefore g and f have the same sense of variation. + g() + f '() Or g'() then g () has the same sign as f () on D g. f () And lim f ( ) then lim g( ) lim f ( ) then lim g( ) D g

11 -8 b -8 c g() ( ) ln(e e ) ln(e ) ln (- e + e ) lim [g() ] = ln = ; (D) is an asymptote of(g) at. g() = is equivalent to ln(f()) = ; f() = ; = ln. g() = is equivalent to ln( e + e ) = ; e + e = ; e = ; = ln y VI ln O ln -8 d