دورة الؼام اهتحا ات الشهادة الثا ىية الؼاهة وزارة التربية والتؼلين الؼالي اإلنث يي 07 حسيراى 7102 فرع: الؼلىم الؼاهة

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1 70 الؼادي ة دورة الؼام اهتحا ات الشهادة الثا ىية الؼاهة وزارة التربية والتؼلين الؼالي اإلنث يي 07 حسيراى 70 فرع: الؼلىم الؼاهة الوديرية الؼاهة للتربية دائرة االهتحا ات الرسوية االسن: هسابقة في هادة الرياضيات ػدد الوسائل: سث الرقن: الودة: أربغ ساػات هالحظة: - يسوح باستعوال آلت حاسبت غيز قابلت للبزهجت او اختزاى الوعلوهاث او رسن البياناث. - يستطيع الوزش ح اإلجابت بالتزتيب الذي يناسبه ( دوى االلتزام بتزتيب الوسائل الواردة في الوسابقت(. I- ( points) In the table below, only one among the proposed answers to each question is correct. Write the number of each question and give, with justification, the answer that corresponds to it. Questions Answers a b c d f is the function defined over 5 5 ; as f() = 5 An antiderivative of f is:. arcsin 5 arcsin arcsin 5 arcsin 5 5 If T 3ln t dt e with > 0, then T' 3 z and z' are two comple numbers. 3 If z i z' iz with z i, then z In the comple plane referred to an orthonormal system, M and M' are two points with respective non-zero affies z and z'. right isosceles equilateral right isosceles If z' iz, then triangle OMM' is: Page of

2 II- (.5 points) In the space referred to a direct orthonormal systemo;i, j,k, consider the following points: A ( ; ; ), B ( ; ; 3) and E ( ; ; 3 ). Let (P) be the plane with equation y z 5 0. Let be the perpendicular bisector of [AB] in (P). ) Verify that the points A and B are in the plane (P). ) a- Verify that V ; ; is a direction vector to. b- Write a system of parametric equations of the line. 3) Let I be a point on such that I > 0. Consider, in the plane (P), the circle (C) with center I and radius 3 that is tangent to (AB). a- Determine the coordinates of I. b- Verify that E is on the circle (C). ) Denote by (D) the line defined as t y t 3 z t Show that the line is tangent to the circle (C) at E. where t. III- (.5 points) Consider two urns U and U. U contains two red balls and one green ball. U contains four red balls and three green balls. Each red ball holds the number and each green ball holds the number. One ball is randomly selected from U. If this ball is red, then one ball is randomly selected from U. (Hence, we get two balls) If it is green, then two balls are randomly and simultaneously selected from U. (Hence, we get three balls) Consider the following events : R : «One red ball is selected from U», R : «One red ball is selected from U», D: «The selected balls have the same color». ) Calculate the probability P(R R ). ) Verify that PD. 7 3) Let S be the sum of numbers on the selected balls. a- Verify that the possible values of S are : 3 ; ; 0 ; ;. b- Calculate P(S < 0). c- Knowing that S < 0, calculate the probability that the selected balls don t have the same color. Page of

3 IV- (3 points) In the plane referred to an orthonormal system O;i, j, consider the point E(;0) and the two variable points M(m ; 0) and N(0 ; n) such that OM EN with m and n being two real numbers (m or m ). Let P be the point defined as NP OM. Part A ) Verify that m n. ) a- Find the coordinates of P in terms of m and n. b- Show that P moves on the hyperbola (H) with equation y. 3) Denote by A and A' the vertices of (H), and by F and F' its foci. a- Find the coordinates of A, A', F and F' ( A 0 and F 0 ). b- Write the equations of the asymptotes of (H) and draw (H). Part B Let (E) be the ellipse so that A, A' and B(0;) are three of its vertices. ) Draw (E) in the system O;i, j. ) The tangent at B to (E) intersects (H) at L with L 0. a- Show that OFLB is a rectangle. b- Calculate the area of the region interior to quadrilateral OALB and eterior to (E). 3) Let G be the point defined as OG OF. Show that the line (LG) is tangent to (H). 5 V- (3 points) In the figure to the right: DICE and JIKF are two direct squares with centers G and E respectively. A is the symmetric of C with respect to I. O is the symmetric of E with respect to D. Let S be the direct plane similitude that maps A onto I and I onto E. Part A ) a- Show that the ratio of S is equal to and that is an angle of S. b- Determine S(C). ) a- S S is a similitude. Find an angle of S S and calculate its ratio. b- Find S S(A) and deduce that O is the center of S. 3) The two straight lines (OC) and (AD) intersect at L. Let L = S(L). Prove that the three points I, D and L are collinear. Part B The plane is referred to a direct orthonormal system O;OA,OD. ) Write the comple form of S and determine the affi of G' such that G'= S(G). ) Let (T) be the ellipse with center I. The points O and G are two of its vertices. Denote by (T) be the image of (T) under S. Write an equation of (T). Page 3 of

4 VI- (7 points) Part A Consider the differential equation (E) : y y e. Let y z e. Part B ) Form the differential equation (E) satisfied by z. ) Find the particular solution of (E) whose representative curve in an orthonormal system passes through the point A( ; ). Consider the function f defined on representative curve in an orthonormal system O; i, j. ) a- Determine b- Determine lim f (). lim f (). Deduce an asymptote (d) to (C). ) a- Calculate f (), and set up the table of variations of f. b- Show that the equation f() = 0 has two roots and 0. Verify that.6.5. as f () ( )e. Denote by (C) its 3) a- Show that (C) has an inflection point whose coordinates are to be determined. b- Write an equation of ( ), the tangent to (C) at its inflection point. ) Let (d') be the line with equation y =. a- Verify that f () e. b- Study, according to the values of, the relative positions of (d') and (C). 5) Draw (d), ( ), (d) and (C). 6) a- Use the differential equation (E) to find an antiderivative of f. b- Deduce the area of the region bounded by (C), (d) and the two lines with equations = α and = 0. 7) Let g be the function defined as g() = ln( f()). Determine the domain of definition of g. Page of

5 وزارة التربية والتعليم العالي المديرية العامة للتربية دائرة االمتحانات الرسمية عدد المسائل: ست امتحانات الشهادة الثانوية العامة فرع: العلوم العامة أسس التصحيح مادة الرياضيات دورة العام 07 العادي ة اإلثنين حزيران 07 QI Answers M d d arcsin 5 5 ( / 5) 5 c e T'() = 3(ln) ; T' ( ) = 3(lne) = = d 3 z i z i z' a iz iz z' e z π i ; then triangle OMM is isosceles at O b QII Answers M A + y A + z A 5 = 0 and B + y B + z B 5 = 0 a VAB 0 and VNP 0 So AB NP / / V b J(0 ; 0 ; 5 k ) midpoint of [AB] ; ( ) y k 5 z k where k R 3a IJ = 3 ; IJ = 9 ; k + k + k = 9 ; k = ; k = since I > 0; then I( ; ; ) 3b E + y E + z E 5 = 0, So E (P) ; IE = + + = 9, IE = 3 = R t = - ; t + = ; -t + 3 = 3 gives t = 0, then E (D). VD IE 0 then (D) (IE) (t-) + (t + ) + (-t + 3 ) 5 = 0, so (D) (P) Thus (D) is tangent to (C) at E. Page of

6 QIII Answers M P(R R ) = C3 3 P(D) = P(R R ) + P(V V) =, then PD 3 C a 3 (V, V) ; (V and (R, V)) ; 0 (R and V) ; (V, R) ; (R and R ) P(S < 0) = P(S = 3) + P(S = ) = P(V, V) + P(V, (R V)) = 3b C C 3 C 7 7 3c P D PDS 0 PS S 0 PS 0 PS QIV Answers M A EN OE ON then m n m Aa P ;n y Ab m and n y thus y or A3a A(;0)and A ( ;0) ; F( 5;0) and F ( 5;0) A3b Asymptotes: y and y. Drawing of (H) Tangent at B: y. B Ba Bb B3 6 0 ; 5 ; L( 5;) ; F L ; BL OF and BOF ˆ 90 So OFLB is a rectangle. ( 5) Area of OALB ( 5) Area = ( 5) Area of (E) ( 5) units of area 5 G ;0 5 ; L( 5;) ; slope of (GL) 5 ; y So 8 yy and y. y 5 yl 5 slope of (GL). Page of

7 QV Answers M S: A I I E Aa IE IE. Angle of S = AI;IE IC;IE. AI IC Ab S(C) = F since C is the symmetric of A with respect to I then S(C) is the symmetric of I of with respect to E. Aa SS is a similitude with ratio and angle. SS(A) = S (I) = E, and we have OE = OA ; Ab OA;OE ;.5 therefore O is the center of SS, hence O is the center of S. A3 B B S(A) = I, then S((AD)) is a line passing through I making an angle π with (AD), then it is line (ID). L (AD),thus S(L) L' (ID). z = az + b, S has O as a center, then b = 0, thus z = a z. i a = e i ; thus z = ( + i) z. 3 3 zg i then z G' ( i) i i. (T) has I as a center and O and G as vertices; then (T ) has S(I) = E as a center and S(O) = O and S(G) = G as vertices. Therefore, the focal ais of (T ) is (OE) / / to the ais of ordinates, E (0 ; ), (y) a = OE = and b = EG =. Thus, an equation of (T ) is. QVI Answers Note A y y e ; y z e ; y z (e e ) ; (E ):zz 0 The general solution of (E ) is: z ke ; the general solution of (E) is: A Ba y ke e. y()= (k + )e + =, then k =. f() = ( + )e - The particular solution of (E) is: y( ), then k ; thus, y ()e. lim f() Bb lim f() ; (d): y horizontal asymptote. Page 3 of

8 Ba Bb f () ( )e f () 0 f() e 0.7 On ], -[ : f is continuous and strictly decreasing from to 0.7, then the equation f() = 0 has one unique solution α ], -[ and f(.6) f(.5) , then -.6 < α < -.5. Moreover f(0) 0. f ''() = e ( ) e e B3a f ''() = 0 for = 0 and changes its sign from positive to negative, then O(0,0) is an inflection point of (C). B3b f (0) ; y0 ( 0) ; ( ): y is tangent to (C). Ba f() ( )e ( )( e )..5 Bb e - 0 f() position (C) is above (d') (C) cuts (d') in (-;) (C) is below (d') f() lim asymptotic direction parallel to y y. (C) cuts (d') in (0;0) (C) is above (d').5 B5.5 B6a f '() + f() = an antiderivative of f is e ; f() = e f '() ; f()d e f() c e ( )e ( 3)e., then B6b α A [ f()]d ( 3)e ] 3 α (α3)e 0 0 α α units of area..5 α B7 f() 0; f() 0; Using part B--b, 0 OR graphically. Page of