Cho hàm số \(f(x)\). Hàm số \(y=f'(x)\) có bảng xét dấu như sau

\(\begin{array}{c|c} x & -\infty\,\,\,\,\,\,\,\,\,\,\,\,-2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\ 3 \,\,\,\,\,\,\,\,\,\,\,\,+\infty \\ \hline f'(x) &\,\,\,\, - \,\,\,\,\,\,0 \,\,\,+\,\,\,\,\,\, 0 \,\,\,\,\,+\,\,0 \,\,\,\,\,\,- \end{array}\)

Số điểm cực tiểu của hàm số \(y=f(x^2+3x)\) là

A. \(5\)

B. \(4\)

C. \(3\)

D. \(2\)

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Chủ đề: Đề thi THPT QG

Môn: Toán

Lời giải:

Báo sai

Xét hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWG5b % Gaeyypa0JaamOzaiaacIcacaWG4bWaaWbaaSqabeaacaaIYaaaaOGa % ey4kaSIaaG4maiaadIhacaGGPaaabaGaamyEaiaacEcacqGH9aqpca % GGOaGaaGOmaiaadIhacqGHRaWkcaaIZaGaaiykaiaadAgacaGGNaGa % aiikaiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaIZaGaam % iEaiaacMcaaaaa!4DA9! \begin{array}{l} y = f({x^2} + 3x), \,\mathrm{ta\,có}\,y' = (2x + 3)f'({x^2} + 3x) \end{array}\)

cho \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpcaaIWaGaeyi1HS9aamqaaqaabeqaaiaaikdacaWG4bGa % ey4kaSIaaG4maiabg2da9iaaicdaaeaacaWGMbGaai4jaiaacIcaca % WG4bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaG4maiaadIhacaGG % PaGaeyypa0JaaGimaaaacaGLBbaacqGHuhY2daWabaabaeqabaGaam % iEaiabg2da9maalaaabaGaeyOeI0IaaG4maaqaaiaaikdaaaaabaGa % amOzaiaacEcacaGGOaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgU % caRiaaiodacaWG4bGaaiykaiabg2da9iaaicdaaaGaay5waaaaaa!5C1F! y' = 0 \Leftrightarrow \left[ \begin{array}{l} 2x + 3 = 0\\ f'({x^2} + 3x) = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \frac{{ - 3}}{2}\\ f'({x^2} + 3x) = 0 \end{array} \right.\)

Dựa vào bảng biến thiên của \(f'(x)\):

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacE % cacaGGOaGaamiEaiaacMcacqGH9aqpcaaIWaGaeyi1HS9aamqaaqaa % beqaaiaadIhacqGH9aqpcqGHsislcaaIYaaabaGaamiEaiabg2da9i % aaigdaaeaacaWG4bGaeyypa0JaaG4maaaacaGLBbaaaaa!4828! f'(x) = 0 \Leftrightarrow \left[ \begin{array}{l} x = - 2\\ x = 1\\ x = 3 \end{array} \right.\)

suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacE % cacaGGOaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaioda % caWG4bGaaiykaiabg2da9iaaicdacqGHuhY2daWabaabaeqabaGaam % iEamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaiodacaWG4bGaeyyp % a0JaeyOeI0IaaGOmaaqaaiaadIhadaahaaWcbeqaaiaaikdaaaGccq % GHRaWkcaaIZaGaamiEaiabg2da9iaaigdaaeaacaWG4bWaaWbaaSqa % beaacaaIYaaaaOGaey4kaSIaaG4maiaadIhacqGH9aqpcaaIZaaaai % aawUfaaiabgsDiBpaadeaaeaqabeaacaWG4bGaeyypa0JaeyOeI0Ia % aGymaaqaaiaadIhacqGH9aqpcqGHsislcaaIYaaabaGaamiEaiabg2 % da9maalaaabaGaeyOeI0IaaG4maiabgglaXoaakaaabaGaaGymaiaa % iodaaSqabaaakeaacaaIYaaaaaqaaiaadIhacqGH9aqpdaWcaaqaai % abgkHiTiaaiodacqGHXcqSdaGcaaqaaiaaikdacaaIXaaaleqaaaGc % baGaaGOmaaaaaaGaay5waaaaaa!7122! f'({x^2} + 3x) = 0 \Leftrightarrow \left[ \begin{array}{l} {x^2} + 3x = - 2\\ {x^2} + 3x = 1\\ {x^2} + 3x = 3 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = - 1\\ x = - 2\\ x = \frac{{ - 3 \pm \sqrt {13} }}{2}\\ x = \frac{{ - 3 \pm \sqrt {21} }}{2} \end{array} \right.\)

trong đó \(x = \frac{{ - 3 \pm \sqrt {13} }}{2}\) là nghiệm kép

từ bảng xét dấu của \(f'(x)\) ta có thể suy ra:\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacE % cadaqadaqaaiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaI % ZaGaamiEaaGaayjkaiaawMcaaiabgYda8iaaicdacqGHuhY2daWaba % abaeqabaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaioda % caWG4bGaeyipaWJaeyOeI0IaaGOmaaqaaiaadIhadaahaaWcbeqaai % aaikdaaaGccqGHRaWkcaaIZaGaamiEaiabg6da+iaaiodaaaGaay5w % aaGaeyi1HS9aamqaaqaabeqaaiabgkHiTiaaikdacqGH8aapcaWG4b % GaeyipaWJaeyOeI0IaaGymaaqaaiaadIhacqGH8aapdaWcaaqaaiab % gkHiTiaaiodacqGHsisldaGcaaqaaiaaikdacaaIXaaaleqaaaGcba % GaaGOmaaaaaeaacaWG4bGaeyOpa4ZaaSaaaeaacqGHsislcaaIZaGa % ey4kaSYaaOaaaeaacaaIYaGaaGymaaWcbeaaaOqaaiaaikdaaaaaai % aawUfaaaaa!67F2! f'\left( {{x^2} + 3x} \right) < 0 \Leftrightarrow \left[ \begin{array}{l} {x^2} + 3x < - 2\\ {x^2} + 3x > 3 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} - 2 < x < - 1\\ x < \frac{{ - 3 - \sqrt {21} }}{2}\\ x > \frac{{ - 3 + \sqrt {21} }}{2} \end{array} \right.\)

Ta có bảng xét dấu của \(y'\)

Quan sát bảng biến thiên ta thấy hàm số \(y=f(x^2+3x)\) có các điểm cực tiểu là

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadIhacqGH9aqpcqGHsislcaaIYaaabaGaamiEaiabg2da9maa % laaabaGaeyOeI0IaaG4maiabgUcaRmaakaaabaGaaGOmaiaaigdaaS % qabaaakeaacaaIYaaaaaaacaGL7baaaaa!41B7! \left\{ \begin{array}{l} x = - 2\\ x = \frac{{ - 3 + \sqrt {21} }}{2} \end{array} \right.\)

Vậy hàm số có hai điểm cực tiểu

Câu hỏi này thuộc đề thi trắc nghiệm dưới đây, bấm vào Bắt đầu thi để làm toàn bài