Cho hàm số y = f(x). Hàm số y = f'(x) có đồ thị như sau:
Bất phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg6da+iaadIhadaahaaWcbeqa % aiaaikdaaaGccqGHsislcaaIYaGaamiEaiabgUcaRiaad2gaaaa!40D6! f\left( x \right) > {x^2} - 2x + m\) đúng với mọi \(x\in(1;2)\) khi và chỉ khi
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Lời giải:
Báo saiBất phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HSTaam % yBaiabgYda8iaadAgadaqadaqaaiaadIhaaiaawIcacaGLPaaacqGH % sislcaWG4bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaGOmaiaadI % hacqGH9aqpcaWGNbWaaeWaaeaacaWG4baacaGLOaGaayzkaaaaaa!47A6! \Leftrightarrow m < f\left( x \right) - {x^2} + 2x = g\left( x \right)\) đúng với mọi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabgI % GiopaabmaabaGaaGymaiaacUdacaaIYaaacaGLOaGaayzkaaGaaGPa % VlaaykW7caaMc8UaaiikaiaacQcacaGGPaaaaa!42DB! x \in \left( {1;2} \right)\,\,\,(*)\).
Xét \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4zamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9iaadAgadaqadaqaaiaa % dIhaaiaawIcacaGLPaaacqGHsislcaWG4bWaaWbaaSqabeaacaaIYa % aaaOGaey4kaSIaaGOmaiaadIhaaaa!4354! g\left( x \right) = f\left( x \right) - {x^2} + 2x\) với \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabgI % GiopaabmaabaGaaGymaiaacUdacaaIYaaacaGLOaGaayzkaaaaaa!3C33! x \in \left( {1;2} \right)\) ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabm4zayaafa % WaaeWaaeaacaWG4baacaGLOaGaayzkaaGaeyypa0JabmOzayaafaWa % aeWaaeaacaWG4baacaGLOaGaayzkaaGaeyOeI0IaaGOmaiaadIhacq % GHRaWkcaaIYaGaeyypa0JabmOzayaafaWaaeWaaeaacaWG4baacaGL % OaGaayzkaaGaeyOeI0IaaGOmamaabmaabaGaamiEaiabgkHiTiaaig % daaiaawIcacaGLPaaaaaa!4C92! g'\left( x \right) = f'\left( x \right) - 2x + 2 = f'\left( x \right) - 2\left( {x - 1} \right)\).
Với \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabgI % GiopaabmaabaGaaGymaiaacUdacaaIYaaacaGLOaGaayzkaaaaaa!3C33! x \in \left( {1;2} \right)\) thì f'(x) < 0 và \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0IaaG % OmamaabmaabaGaamiEaiabgkHiTiaaigdaaiaawIcacaGLPaaacqGH % 8aapcaaIWaGaeyO0H4Tabm4zayaafaWaaeWaaeaacaWG4baacaGLOa % GaayzkaaGaeyipaWJaaGimamaabmaabaGaeyiaIiIaamiEaiabgIGi % opaabmaabaGaaGymaiaacUdacaaIYaaacaGLOaGaayzkaaaacaGLOa % Gaayzkaaaaaa!4DBA! - 2\left( {x - 1} \right) < 0 \Rightarrow g'\left( x \right) < 0\left( {\forall x \in \left( {1;2} \right)} \right)\).
Do đó hàm số g(x) nghịch biến trên khoảng (1;2).
Khi đó \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % GGQaaacaGLOaGaayzkaaGaeyi1HSTaamyBaiabgsMiJkaadEgadaqa % daqaaiaaikdaaiaawIcacaGLPaaacqGHuhY2caWGTbGaeyizImQaam % OzamaabmaabaGaaGOmaaGaayjkaiaawMcaaaaa!4891! \left( * \right) \Leftrightarrow m \le g\left( 2 \right) \Leftrightarrow m \le f\left( 2 \right)\).
Đề thi thử tốt nghiệp THPT QG môn Toán năm 2020
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