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(1)Simulazione prova d'esame n. 3. ESAME DI STATO DI LICEO SCIENTIFICO CORSO DI ORDINAMENTO Tema di: MATEMATICA ,OFDQGLGDWRVYROJDXQSUREOHPDWUDLGXHSURSRVWLHFLQTXHTXHVLWLWUDLGLHFLSURSRVWL ( DPPHVVRO XVRGHOODVRODFDOFRODWULFHQRQSURJUDPPDELOH  3UREOHPL 3UREOHPD  x +1 y= 2 ( GDWDODIDPLJOLDGLFXUYH x +k  D

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(39) Soluzioni simulazione prova d'esame n. 3 . 62/8=,21('(,'8(352%/(0, . 6ROX]LRQHGHO3ULPR3UREOHPD.  D

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(43) /¶HTXD]LRQHGHOOXRJRGHLSXQWLGLPDVVLPRHGLPLQLPRGHOODIDPLJOLDGLFXUYHVLRWWLHQH • FDOFRODQGR OD GHULYDWD GHOOD IXQ]LRQH GDWD H XJXDJOLDQGROD D ]HUR . § − x 2 − 2x + k · § x +1 · ¸ = 0 D¨ 2 ¸ = D¨¨ 2 2 ¸ x + k © ¹ ¹ © x +k. (. •. • . ). ­− x 2 − 2 x + k = 0 ° PHWWHQGRODSRLDVLVWHPDFRQO HTXD]LRQHGDWD ®  x +1 °y = 2 x +k ¯ 1 ³HOLPLQDQGR LO SDUDPHWUR N´ H VHPSOLILFDQGR VL RWWLHQH  y =  FKH q O¶HTXD]LRQH 2x ULFKLHVWD. Prof. Franco Fusier. Soluzioni - Pag. 1.

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(48) ,OWULDQJRORUHWWDQJRORGLLSRWHQXVD AB = 2 GLSHULPHWURPDVVLPRqO¶LVRVFHOH 6LSXzGLPRVWUDUHLQGLYHUVLPRGLSHUHVHPSLR. ˆB 3RQHQGR CA. = x FRQ 0 ≤ x ≤. π. LOSHULPHWURVDUjGDWRGD 2 p = 2 + 2 cos x + 2 senx  2 GHULYDQGR ( 2 p )' = D 2 (1 + cos x + senx ) = 2 (− senx + cos x ) . (. TXLQGL ( 2 p )' ≥ 0 TXDQGR 0 ≤. x≤. π. 4. 3HUWDQWRLOPDVVLPRDVVROXWRVLKDSHU x. ). . =. π 4. ҏFKHFRUULVSRQGHDOWULDQJRORUHWWDQJRORLVRVFHOH..  E

(49) 3HULQVFULYHUHXQSDUDOOHOHSLSHGRUHWWDQJRORGLEDVHTXDGUDWDQHOODSLUDPLGHGLEDVH$%&H DOWH]]D VC = h ELVRJQDGLVSRUUHLYHUWLFLGHOSDUDOOHOHSLSHGRLQPRGRFKH • XQRFRLQFLGDFRQ& • GXHYHUWLFL'(VLDQRVXLFDWHWL$&H&% • XQYHUWLFH)VLDVXOO¶DOWH]]D&9 • GXHYHUWLFL*H.VLDQRVXOOHIDFFH9&%H9&$ • XQYHUWLFH-VLDVXOODEDVH$%&. Prof. Franco Fusier. Soluzioni - Pag. 2.

(50) Soluzioni simulazione prova d'esame n. 3 • PHQWUHO¶XOWLPRYHUWLFHFKHLQGLFKLDPRFRQ/VLWURYLVXOODIDFFLD9$% 6H  LQGLFKLDPR FRQ [ O¶DOWH]]D GL WDOH SDUDOOHOHSLSHGR H VH]LRQLDPR OD SLUDPLGH FRQ XQ SLDQR SHUSHQGLFRODUHDOODEDVHHSDVVDQWHSHU9HSHULOSXQWRPHGLR+GHOO¶LSRWHQXVD$%WDOHSLDQR SDVVHUjSHUODVLPPHWULDGHOODILJXUDDQFKHSHULOYHUWLFH/GHOSDUDOOHOHSLSHGR 3HUWDQWRGDOODVLPLOLWXGLQHGHLWULDQJROLUHWWDQJROLFKHVLYHQJRQRDIRUPDUHSRVVLDPRVFULYHUH. = CH : CV  VF ⋅ CH 2 (h − x ) =  'DFXLVLULFDYDFKH FL = 2h CV. ODSURSRU]LRQH FL : VF. 0DVHRVVHUYLDPRFKH)/qODGLDJRQDOHGHOODEDVHGHOSDUDOOHOHSLSHGRLOODWR l GHOTXDGUDWRGL EDVHGHOSDUDOOHOHSLSHGRVLRWWHUUjGLYLGHQGRODGLDJRQDOHSHU. 2  l =. FL 2. =. h−x  2h. 2. 3HUWDQWRLOYROXPHGHOSDUDOOHOHSLSHGRVDUjGDWRGD V. §h−x· = x ⋅¨ ¸  © 2h ¹. 3HUWURYDUHSHUTXDOHYDORUHGL[LOYROXPHqPDVVLPRGHULYLDPR9RWWHQHQGR. V '=. h 2 − 4hx + 3 x 2  4h 2. 3RQHQGR 9¶!  H ULVROYHQGR VL WURYD WHQHQGR FRQWR GHOOH FRQGL]LRQL JHRPHWULFKH FKH ULFKLHGRQR 0 <. x < h

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(52) ,OYDORUHGHOYROXPHPDVVLPR V MAX FDOFRODWRDOSXQWRSUHFHGHQWHq 3HUFLz x. =. 1 VH h = 1  27. 1 1 HSXUHLOODWRGLEDVHq ORVLFDOFRODGDOODIRUPXODULFDYDWDSUHFHGHQWHPHQWH 3 3. 3HUWDQWRODWRGLEDVHHDOWH]]DVRQRXJXDOLHLOSDUDOOHOHSLSHGRqXQFXER  G

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(54) 3HUFDOFRODUHO¶DUHDGHOODSDUWHGLSLDQRFRPSUHVDWUDLOJUDILFRGHOODIXQ]LRQHHO¶DVVH[q. 1. ª 1 § x 2 2 x 3 x 4 ·º 1 1 2  VXIILFLHQWHFDOFRODUHO¶LQWHJUDOHGHILQLWLYR ³ (1 − x ) xdx = « ¨¨ − + ¸¸» = 4 3 4 ¹¼ 0 48 0 ¬4 © 2 1. Prof. Franco Fusier. Soluzioni - Pag. 3.

(55)

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