Tập hợp tất cả tham số m để hàm số \(y=x^3+(m+1)x^2+3x+2\) đồng biến trên \(\mathbb{R}\) là
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Lời giải:
Báo saiTXĐ: \(D=\mathbb{R}\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaaqaaaaa % aaaaWdbiaadMhacqGH9aqpcaWG4bWdamaaCaaaleqabaWdbiaaioda % aaGccqGHRaWkcaGGOaGaamyBaiabgUcaRiaaigdacaGGPaGaamiEa8 % aadaahaaWcbeqaa8qacaaIYaaaaOGaey4kaSIaaG4maiaadIhacqGH % RaWkcaaIYaaabaGaamyEaiaacEcacqGH9aqpcaaIZaGaamiEamaaCa % aaleqabaGaaGOmaaaakiabgUcaRiaaikdacaGGOaGaamyBaiabgUca % RiaaigdacaGGPaGaamiEaiabgUcaRiaaiodaaeaacqGHuoarcaGGNa % Gaeyypa0Jaaiikaiaad2gacqGHRaWkcaaIXaGaaiykamaaCaaaleqa % baGaaGOmaaaakiabgkHiTiaaiodacaGGUaGaaG4maiabg2da9iaad2 % gadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaIYaGaamyBaiabgkHi % TiaaiIdaaaaa!64F6! \begin{array}{l} y = {x^3} + (m + 1){x^2} + 3x + 2\\ y' = 3{x^2} + 2(m + 1)x + 3\\ \Delta = {4(m + 1)^2} - 4.3.3 = 4{m^2} + 8m - 32 \end{array}\)
Để hàm số đồng biến trên \(\mathbb{R}\) thì \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcqaaaaaaaaaWdbe % aacqGHuoarcaGGNaGaeyizImQaaGimaiaaykW7caaMc8UaaGPaVlab % gcGiIiaadIhacqGHiiIZcqWIDesOaaa!43F6! y' % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyyzImlaaa!37B9! \ge 0\,\,\,\forall x \in \mathbb{R}\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGHuh % Y2daGabaabaeqabaGaamyyaiabg2da9iaaiodacqGH+aGpcaaIWaGa % aGPaVlaaykW7caGGOaGaamiBaiaadYgacaqGKbGaaiykaaqaaiabgs % 5aejabgsMiJkaaicdaaaGaay5EaaaaqaaaaaaaaaWdbeaacqGHuoar % cqGHKjYOcaaIWaGaeyi1HSTaamyBamaaCaaaleqabaGaaGOmaaaaki % abgUcaRiaaikdacaWGTbGaeyOeI0IaaGioaiabgsMiJkaaicdaaeaa % caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl % aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlabgsDiBlabgkHi % TiaaisdacqGHKjYOcaWGTbGaeyizImQaaGOmaaaaaa!768F! \begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l} a = 3 > 0\,\,(ll{\rm{d}})\\ \Delta \le 0 \end{array} \right.\\ \Delta \le 0 \Leftrightarrow 4{m^2} + 8m - 32 \le 0\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow - 4 \le m \le 2 \end{array}\)
Vậy \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWGTbGaeyicI48aamWaaeaacqGHsislcaaI0aGaai4oaiaaikda % aiaawUfacaGLDbaaaaa!3DA1! m \in \left[ { - 4;2} \right]\)
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