Gọi m, M lần lượt là giá trị nhỏ nhất và giá trị lớn nhất của hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9iaaikdacaWG4bGaey4k % aSIaci4yaiaac+gacaGGZbWaaSaaaeaacqaHapaCcaWG4baabaGaaG % Omaaaaaaa!435F! f\left( x \right) = 2x + \cos \frac{{\pi x}}{2}\) trên đoạn [-2;2]. Giá trị của m + M bằng:
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Lời giải:
Báo saiTa có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9iaaikdacaWG4bGaey4k % aSIaci4yaiaac+gacaGGZbWaaSaaaeaacqaHapaCcaWG4baabaGaaG % OmaaaacqGHshI3caWGMbGaai4jamaabmaabaGaamiEaaGaayjkaiaa % wMcaaiabg2da9iaaikdacqGHsisldaWcaaqaaiabec8aWbqaaiaaik % daaaGaci4CaiaacMgacaGGUbWaaSaaaeaacqaHapaCcaWG4baabaGa % aGOmaaaaaaa!556E! f\left( x \right) = 2x + \cos \frac{{\pi x}}{2} \Rightarrow f'\left( x \right) = 2 - \frac{\pi }{2}\sin \frac{{\pi x}}{2}\)
Vì \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0IaaG % ymaiabgsMiJkGacohacaGGPbGaaiOBamaalaaabaGaeqiWdaNaamiE % aaqaaiaaikdaaaGaeyizImQaaGymaiabgsDiBlabgkHiTmaalaaaba % GaeqiWdahabaGaaGOmaaaacqGHKjYOdaWcaaqaaiabec8aWbqaaiaa % ikdaaaGaci4CaiaacMgacaGGUbWaaSaaaeaacqaHapaCcaWG4baaba % GaaGOmaaaacqGHKjYOdaWcaaqaaiabec8aWbqaaiaaikdaaaGaeyO0 % H4TaaGimaiabgYda8iaaikdacqGHsisldaWcaaqaaiabec8aWbqaai % aaikdaaaGaeyizImQaaGOmaiabgkHiTmaalaaabaGaeqiWdahabaGa % aGOmaaaaciGGZbGaaiyAaiaac6gadaWcaaqaaiabec8aWjaadIhaae % aacaaIYaaaaiabgsMiJkaaikdacqGHsisldaWcaaqaaiabec8aWbqa % aiaaikdaaaaaaa!7144! - 1 \le \sin \frac{{\pi x}}{2} \le 1 \Leftrightarrow - \frac{\pi }{2} \le \frac{\pi }{2}\sin \frac{{\pi x}}{2} \le \frac{\pi }{2} \Rightarrow 0 < 2 - \frac{\pi }{2} \le 2 - \frac{\pi }{2}\sin \frac{{\pi x}}{2} \le 2 - \frac{\pi }{2}\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % OzaiaacEcadaqadaqaaiaadIhaaiaawIcacaGLPaaacqGH+aGpcaaI % WaGaeyiaIiIaamiEaiabgIGiopaadmaabaGaeyOeI0IaaGOmaiaacU % dacaaIYaaacaGLBbGaayzxaaGaeyO0H4naaa!48F3! \Rightarrow f'\left( x \right) > 0\forall x \in \left[ { - 2;2} \right] \Rightarrow \) hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9iaaikdacaWG4bGaey4k % aSIaci4yaiaac+gacaGGZbWaaSaaaeaacqaHapaCcaWG4baabaGaaG % Omaaaaaaa!435F! f\left( x \right) = 2x + \cos \frac{{\pi x}}{2}\) là hàm đồng biến trên [-2;2]
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGHsh % I3caWGMbWaaeWaaeaacqGHsislcaaIYaaacaGLOaGaayzkaaGaeyiz % ImQaamOzamaabmaabaGaamiEaaGaayjkaiaawMcaaiabgsMiJkaadA % gadaqadaqaaiaaikdaaiaawIcacaGLPaaacqGHaiIicaWG4bGaeyic % I48aamWaaeaacqGHsislcaaIYaGaai4oaiaaikdaaiaawUfacaGLDb % aaaeaacqGHshI3daGabaabaeqabaGaamytaiabg2da9maaxababaGa % ciyBaiaacggacaGG4baaleaadaWadaqaaiabgkHiTiaaikdacaGG7a % GaaGOmaaGaay5waiaaw2faaaqabaGccaWGMbWaaeWaaeaacaWG4baa % caGLOaGaayzkaaGaeyypa0JaamOzamaabmaabaGaaGOmaaGaayjkai % aawMcaaiabg2da9iaaiodaaeaacaWGTbGaeyypa0ZaaCbeaeaaciGG % TbGaaiyAaiaac6gaaSqaamaadmaabaGaeyOeI0IaaGOmaiaacUdaca % aIYaaacaGLBbGaayzxaaaabeaakiaadAgadaqadaqaaiaadIhaaiaa % wIcacaGLPaaacqGH9aqpcaWGMbWaaeWaaeaacqGHsislcaaIYaaaca % GLOaGaayzkaaGaeyypa0JaeyOeI0IaaGynaaaacaGL7baaaeaacqGH % shI3caWGnbGaey4kaSIaamyBaiabg2da9iaaiodacqGHRaWkcaGGOa % GaeyOeI0IaaGynaiaacMcacqGH9aqpcqGHsislcaaIYaaaaaa!889A! \begin{array}{l} \Rightarrow f\left( { - 2} \right) \le f\left( x \right) \le f\left( 2 \right)\forall x \in \left[ { - 2;2} \right]\\ \Rightarrow \left\{ \begin{array}{l} M = \mathop {\max }\limits_{\left[ { - 2;2} \right]} f\left( x \right) = f\left( 2 \right) = 3\\ m = \mathop {\min }\limits_{\left[ { - 2;2} \right]} f\left( x \right) = f\left( { - 2} \right) = - 5 \end{array} \right.\\ \Rightarrow M + m = 3 + ( - 5) = - 2 \end{array}\)
Đề thi thử tốt nghiệp THPT QG môn Toán năm 2020
Tuyển chọn số 4