{"id":4220,"date":"2015-03-09T04:32:07","date_gmt":"2015-03-09T02:32:07","guid":{"rendered":"https:\/\/kamerpower.com\/?p=4220"},"modified":"2025-01-29T19:10:46","modified_gmt":"2025-01-29T18:10:46","slug":"epreuve-de-mathematiques-douanes-2013-concours-controleurs-adjoints-des-douanes-c-cameroun","status":"publish","type":"post","link":"https:\/\/kamerpower.com\/fr\/epreuve-de-mathematiques-douanes-2013-concours-controleurs-adjoints-des-douanes-c-cameroun\/","title":{"rendered":"\u00c9preuve de Math\u00e9matiques DOUANES 2013 concours: controleurs adjoints des DOUANES C Cameroun"},"content":{"rendered":"

REPUBLIQUE DU CAMEROUN<\/span> — REPUBLIC OF CAMEROON<\/span><\/span>
\nPaix -Travail- Patrie<\/span> — Peace -Work- Fatherland<\/span><\/span>
\nMINISTERE DE LA FONCTION PUBLIQUE<\/span> — MINISTRY OF THE PUBLIC SERVICE<\/span><\/span>
\nET DE LA REFORME ADMINSTRATIVE<\/span> — AND ADMINISTRATIVE REFORM<\/span><\/span><\/p>\n

CONCOURS DIRECTS DU 14 DECEMBRE 2013 POUR LE RECRUTEMENT DANS LE CORPS DE FONCTIONNAIRES DES REGIES FINANCIERES ( DOUANES<\/a> )<\/span><\/h4>\n

CONTROLEURS ADJOINTS DES DOUANES C<\/span><\/p>\n

\u00c9PREUVE\u00a0DE\u00a0MATH\u00c9MATIQUES<\/span><\/strong><\/a>
\nDur\u00e9e : 3h<\/span>
\nCoeff .4<\/span><\/p>\n

KA<\/span>M<\/span>ER<\/span>POWER<\/span>.COM<\/span>
\nPas de calculatrice<\/span><\/p>\n

A – TRAVAUX NUMERIQUES<\/strong>:<\/span> 7,5 pts<\/span><\/p>\n

On donne
\n<\/span><\/p>\n

A = 7\/18 x 2\/7 – (5\/3 – 2)^2
\nB = 2\u221a5(2 – 5\u221a5) – (1 + \u221a5)(3 + 2\u221a5)<\/span><\/p>\n

1.<\/span> Ecrirc A sous forme de fraction irr\u00e9ductible. (0,7.5 pt)<\/span>
\n2.<\/span> D\u00e9velopper et r\u00e9duire B. (0,75 pt)<\/span><\/p>\n

L-Soit A = 12x^2 – 3 + (5-4x)(1-2x)\u00a0<\/span><\/p>\n

1.<\/span> Factoriser 12×2 – 3 puis en d\u00e9duire une factorisation de A. (0,5pt + 0,75pt)<\/span>
\n2.<\/span> R\u00e9soudre l’\u00e9quation (2 – 5x)(x + 2) = 0. (0,5 pt)<\/span><\/p>\n

II.<\/span>– R\u00e9soudre dans R le syst\u00e8me d’in\u00e9quations:<\/span><\/p>\n

{ 2x + 1 < 3x – 1
\n3x + 1 <= 2x + 4<\/p>\n

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(0,75 pt)<\/span><\/p>\n

III.<\/span>– On donne l’\u00e9quation\u00a0<\/span><\/p>\n

q = (2\u221a5)\/(\u221a5+3)<\/p>\n

\n

1.<\/span> Montrer que\u00a0<\/span><\/p>\n

q = \u00a0(-7+5\u221a5)\/4<\/p>\n

\n

2. <\/span>Sachant que 2,236 < \u00a0< 2,237 donner une encadrement de q \u00e0 \u00a0pr\u00e8s.<\/span>
\n(0,7spt)<\/span><\/p>\n

IV-<\/span> x enfants d\u00e9cident de donner 300 F chacun pour acheter un ballon. Mais l’un des\u00a0<\/span>enfants doit renoncer \u00e0 cet achat. Du coup chacun des autres devra donner 20 F de\u00a0<\/span>plus.<\/span><\/p>\n

l.<\/span> Traduire le prix P du ballon dans chaque cas. (0,5pt x 2)<\/span>
\n2.<\/span> D\u00e9terminer alors le prix du ballon, (l pt)<\/span><\/p>\n

B – TRAVAUX GEOMETRIQUES<\/strong>:\u00a0<\/span>6,5 pts<\/span><\/p>\n

On consid\u00e8re la figure ci-\u00a0<\/span>contre o\u00f9 (C) est le cercle de centre O et\u00a0<\/span>de rayon 4 cm. [OA] et [OB] sont des\u00a0<\/span>rayons. On donne OH = 1,6 cm, les\u00a0<\/span>droites (BH) et AO) sont\u00a0<\/span>perpendiculaires la droite\u00a0\u00a0<\/span>est tangente \u00e0 (C) en A. (T) et (OB) se\u00a0<\/span>coupent en\u201c E.<\/span><\/p>\n

\"TRAVAUX<\/p>\n

1.<\/span> Calculer sinOBH puis d\u00e9terminer mes OBH au degr\u00e9 pr\u00e8s. (0,75pt + 0,25pt)<\/span>
\n2.a)<\/span> Montrer que (BH) et (T) sont parall\u00e8les. \u00a0(0,5 pt)<\/span><\/p>\n

2.b)<\/span> Quelle est la nature de AEOBH ? Calculer \u00a0puis en d\u00e9duire OE. (0,25+0,75pt)<\/span><\/p>\n

II-<\/span> ABC est un triangle quelconque. I\u00a0est le milieu de [AB].<\/span><\/p>\n

1.<\/span> Construire les points D, M et N tels que AD = 2AB ; CM = AD; NB = CA\u00a0.<\/span>\u00a0(0,25pt x 3)<\/span><\/p>\n

2. a)<\/span> Justifier que AB = CN\u00a0 et que CM = 2AB\u00a0\u00a0(0,25pt x 2)<\/span><\/p>\n

b)<\/span> En d\u00e9duire que N est le milieu du segment [CM]. (0,25 pt)<\/span><\/p>\n

3- a)<\/span> Exprimer CM\u00a0\u00a0en fonction de IA. (0,5 pt)\u00a0<\/span><\/p>\n

b)<\/span> Que peut-on dire des vecteurs \u00a0CM\u00a0 et \u00a0IA\u00a0? (0,25 pt)\u00a0<\/span><\/p>\n

III.-<\/span> On consid\u00e8re le c\u00f4ne de r\u00e9volutiOn dont une esquisse est donn\u00e9e ci-contre, \u00a0<\/span>On donne OA=2cm; OB=3cm.<\/span><\/p>\n

\"TRAVAUX
\n1.<\/span> Calculer la longueur de la g\u00e9n\u00e9ratrlce AB en valeur exacte. (0,75 pt)\u00a0<\/span>
\n2.<\/span> Calculer le volume de ce cone en valeur exacte et arrondie au dixierne pr\u00e8s.\u00a0<\/span>(0,75pt x 2)<\/span><\/p>\n

C – PROBLEME<\/strong>:<\/span> l6 pts<\/span><\/p>\n

Le plan est muni du rep\u00e8re orthonorm\u00e9 (o, i, j). \u00a0<\/span><\/p>\n

1.<\/span> Placer dans ce rep\u00e8re les points A(6 ,5) ; 3(2, -3) et C(-4, O) (l pt)<\/span>
\n2.<\/span> Calculer les distances AB ; BC et CA en valeur exacte, (0,5pt x 3)<\/span>
\n3.<\/span> En d\u00e9duire la nature du triangle ABC et calculer son aire. (0,5pt + 0,5pt)<\/span><\/p>\n

4.<\/span> Calculer le p\u00e9rim\u00e8tre du triangle ABC sous la forme \u00a0\u00a0(0,5 pt)<\/span><\/p>\n

5.<\/span> Soit (C) le cercle circonscrit au triangle ABC.<\/span>
\na)<\/span> D\u00e9terminer les coordonn\u00e9es de E centre du cercle (C). (0,5 pt)<\/span>
\nb)<\/span> Calculer la valeur exacte du rayon de (C). (0,5 pt)<\/span><\/p>\n

6.<\/span> Calculer tanACB\u00a0en valeur exacte puis en d\u00e9duire une valeur approch\u00e9e du\u00a0<\/span>degr\u00e9 pr\u00e8s de mes ACB\u00a0 (0,25 pt\u00a0x 2)<\/span><\/p>\n

7.<\/span> D\u00e9terminer les coordonn\u00e9es du point D tel que ABCD soit\u00a0<\/span>un parall\u00e9logramme. (0,5pt)<\/span><\/p>\n

\u00a0\u00c9preuve de Math\u00e9matiques DOUANES 2013 concours: controleurs adjoints des DOUANES C Cameroun<\/span><\/h2>\n","protected":false},"excerpt":{"rendered":"

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