Cho hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9iaaikdacaaIWaGaaGym % aiaaiMdadaahaaWcbeqaaiaadIhaaaGccqGHsislcaaIYaGaaGimai % aaigdacaaI5aWaaWbaaSqabeaacqGHsislcaWG4baaaaaa!448A! f\left( x \right) = {2019^x} - {2019^{ - x}}\). Tìm số nguyên m lớn nhất để \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamyBaaGaayjkaiaawMcaaiabgUcaRiaadAgadaqadaqaaiaa % ikdacaWGTbGaey4kaSIaaGOmaiaaicdacaaIXaGaaGyoaaGaayjkai % aawMcaaiabgYda8iaaicdaaaa!43F1! f\left( m \right) + f\left( {2m + 2019} \right) < 0\)
Hãy suy nghĩ và trả lời câu hỏi trước khi xem đáp án
Lời giải:
Báo saiHàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9iaaikdacaaIWaGaaGym % aiaaiMdadaahaaWcbeqaaiaadIhaaaGccqGHsislcaaIYaGaaGimai % aaigdacaaI5aWaaWbaaSqabeaacqGHsislcaWG4baaaaaa!448A! f\left( x \right) = {2019^x} - {2019^{ - x}}\) xác định với mọi x thuộc R.
Ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaeyOeI0IaamiEaaGaayjkaiaawMcaaiabg2da9iaaikdacaaI % WaGaaGymaiaaiMdadaahaaWcbeqaaiabgkHiTiaadIhaaaGccqGHsi % slcaaIYaGaaGimaiaaigdacaaI5aWaaWbaaSqabeaacaWG4baaaOGa % eyypa0JaeyOeI0YaaeWaaeaacaaIYaGaaGimaiaaigdacaaI5aWaaW % baaSqabeaacaWG4baaaOGaeyOeI0IaaGOmaiaaicdacaaIXaGaaGyo % amaaCaaaleqabaGaeyOeI0IaamiEaaaaaOGaayjkaiaawMcaaiabg2 % da9iabgkHiTiaadAgadaqadaqaaiaadIhaaiaawIcacaGLPaaacqGH % shI3caWGMbWaaeWaaeaacaWG4baacaGLOaGaayzkaaaaaa!5E59! f\left( { - x} \right) = {2019^{ - x}} - {2019^x} = - \left( {{{2019}^x} - {{2019}^{ - x}}} \right) = - f\left( x \right) \Rightarrow f\left( x \right)\) là hàm số lẻ
Mặt khác \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafa % WaaeWaaeaacaWG4baacaGLOaGaayzkaaGaeyypa0JaaGOmaiaaicda % caaIXaGaaGyoamaaCaaaleqabaGaamiEaaaakiGacYgacaGGUbGaaG % OmaiaaicdacaaIXaGaaGyoaiabgUcaRiaaikdacaaIWaGaaGymaiaa % iMdadaahaaWcbeqaaiabgkHiTiaadIhaaaGcciGGSbGaaiOBaiaaik % dacaaIWaGaaGymaiaaiMdacqGH+aGpcaaIWaGaeyiaIiIaamiEaiab % gIGiolabl2riHkabgkDiElaadAgadaqadaqaaiaadIhaaiaawIcaca % GLPaaaaaa!5A96! f'\left( x \right) = {2019^x}\ln 2019 + {2019^{ - x}}\ln 2019 > 0;\forall x \in R \Rightarrow f\left( x \right)\) đồng biến trên R.
Do đó BPT :\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamyBaaGaayjkaiaawMcaaiabgUcaRiaadAgadaqadaqaaiaa % ikdacaWGTbGaey4kaSIaaGOmaiaaicdacaaIXaGaaGyoaaGaayjkai % aawMcaaiabgYda8iaaicdacqGHuhY2caWGMbWaaeWaaeaacaaIYaGa % amyBaiabgUcaRiaaikdacaaIWaGaaGymaiaaiMdaaiaawIcacaGLPa % aacqGH8aapcqGHsislcaWGMbWaaeWaaeaacaWGTbaacaGLOaGaayzk % aaGaeyi1HSTaamOzamaabmaabaGaaGOmaiaad2gacqGHRaWkcaaIYa % GaaGimaiaaigdacaaI5aaacaGLOaGaayzkaaGaeyipaWJaamOzamaa % bmaabaGaeyOeI0IaamyBaaGaayjkaiaawMcaaaaa!6347! f\left( m \right) + f\left( {2m + 2019} \right) < 0 \Leftrightarrow f\left( {2m + 2019} \right) < - f\left( m \right) \Leftrightarrow f\left( {2m + 2019} \right) < f\left( { - m} \right)\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HSTaaG % Omaiaad2gacqGHRaWkcaaIYaGaaGimaiaaigdacaaI5aGaeyipaWJa % eyOeI0IaamyBaiabgsDiBlaad2gacqGH8aapcqGHsislcaaI2aGaaG % 4naiaaiodaaaa!4833! \Leftrightarrow 2m + 2019 < - m \Leftrightarrow m < - 673\)
Đề thi thử tốt nghiệp THPT QG môn Toán năm 2020
Tuyển chọn số 5