Cho hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9maaceaabaqbaeaabiWa % aaqaamaalaaabaWaaOaaaeaacaaIYaGaamiEaiabgUcaRiaaiIdaaS % qabaGccqGHsislcaaIYaaabaWaaOaaaeaacaWG4bGaey4kaSIaaGOm % aaWcbeaaaaaakeaacaqGRbGaaeiAaiaabMgaaeaacaWG4bGaeyOpa4 % JaeyOeI0IaaGOmaaqaaiaaicdaaeaacaqGRbGaaeiAaiaabMgaaeaa % caWG4bGaeyypa0JaeyOeI0IaaGOmaaaaaiaawUhaaaaa!512F! f\left( x \right) = \left\{ {\begin{array}{*{20}{l}} {\frac{{\sqrt {2x + 8} - 2}}{{\sqrt {x + 2} }}}&{{\rm{khi}}}&{x > - 2}\\ 0&{{\rm{khi}}}&{x = - 2} \end{array}} \right.\) . Tìm khẳng định đúng trong các khẳng định sau:

\((I)\) \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaCbeaeaaci % GGSbGaaiyAaiaac2gaaSqaaiaadIhacqGHsgIRdaqadaqaaiabgkHi % TiaaikdaaiaawIcacaGLPaaadaahaaadbeqaaiabgUcaRaaaaSqaba % GccaWGMbWaaeWaaeaacaWG4baacaGLOaGaayzkaaGaeyypa0JaaGim % aaaa!456E! \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} f\left( x \right) = 0\) .

\((II)\) \(f(x)\) liên tục tại \(x=-2\).

\((III)\) \(f(x)\) gián đoạn tại \(x=-2\).

A. Chỉ (III)

B. Chỉ (I)

C. Chỉ (I) và (II)

D. Chỉ (I) và (III)

Suy nghĩ trả lời câu hỏi trước khi xem đáp án

Chủ đề: Đề thi THPT QG

Môn: Toán

Lời giải:

Báo sai

Hàm số \(f(x)\) xác định trên nửa khoảng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaKGeaeaacq % GHsislcaaIYaGaai4oaiabgUcaRiabg6HiLcGaay5waiaawMcaaaaa % !3C81! \left[ { - 2; + \infty } \right)\).

Ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaCbeaeaaci % GGSbGaaiyAaiaac2gaaSqaaiaadIhacqGHsgIRdaqadaqaaiabgkHi % TiaaikdaaiaawIcacaGLPaaadaahaaadbeqaaiabgUcaRaaaaSqaba % GccaWGMbWaaeWaaeaacaWG4baacaGLOaGaayzkaaGaeyypa0ZaaCbe % aeaaciGGSbGaaiyAaiaac2gaaSqaaiaadIhacqGHsgIRdaqadaqaai % abgkHiTiaaikdaaiaawIcacaGLPaaadaahaaadbeqaaiabgUcaRaaa % aSqabaGcdaWcaaqaamaakaaabaGaaGOmaiaadIhacqGHRaWkcaaI4a % aaleqaaOGaeyOeI0IaaGOmaaqaamaakaaabaGaamiEaiabgUcaRiaa % ikdaaSqabaaaaaaa!56EF! \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} \frac{{\sqrt {2x + 8} - 2}}{{\sqrt {x + 2} }}\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0ZaaC % beaeaaciGGSbGaaiyAaiaac2gaaSqaaiaadIhacqGHsgIRdaqadaqa % aiabgkHiTiaaikdaaiaawIcacaGLPaaadaahaaadbeqaaiabgUcaRa % aaaSqabaGcdaWcaaqaaiaaikdacaWG4bGaey4kaSIaaGioaiabgkHi % TiaaisdaaeaadaGcaaqaaiaadIhacqGHRaWkcaaIYaaaleqaaOWaae % WaaeaadaGcaaqaaiaaikdacaWG4bGaey4kaSIaaGioaaWcbeaakiab % gUcaRiaaisdaaiaawIcacaGLPaaaaaaaaa!4FC6! = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} \frac{{2x + 8 - 4}}{{\sqrt {x + 2} \left( {\sqrt {2x + 8} + 4} \right)}}\) \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0ZaaC % beaeaaciGGSbGaaiyAaiaac2gaaSqaaiaadIhacqGHsgIRdaqadaqa % aiabgkHiTiaaikdaaiaawIcacaGLPaaadaahaaadbeqaaiabgUcaRa % aaaSqabaGcdaWcaaqaaiaaikdadaGcaaqaaiaadIhacqGHRaWkcaaI % YaaaleqaaaGcbaWaaOaaaeaacaaIYaGaamiEaiabgUcaRiaaiIdaaS % qabaGccqGHRaWkcaaI0aaaaiabg2da9iaaicdaaaa!4BB1! = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} \frac{{2\sqrt {x + 2} }}{{\sqrt {2x + 8} + 4}} = 0\)

Khẳng định (I) đúng.

Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaCbeaeaaci % GGSbGaaiyAaiaac2gaaSqaaiaadIhacqGHsgIRdaqadaqaaiabgkHi % TiaaikdaaiaawIcacaGLPaaadaahaaadbeqaaiabgUcaRaaaaSqaba % GccaWGMbWaaeWaaeaacaWG4baacaGLOaGaayzkaaGaeyypa0ZaaCbe % aeaaciGGSbGaaiyAaiaac2gaaSqaaiaadIhacqGHsgIRdaqadaqaai % abgkHiTiaaikdaaiaawIcacaGLPaaadaahaaadbeqaaiabgkHiTaaa % aSqabaGccaWGMbWaaeWaaeaacaWG4baacaGLOaGaayzkaaGaeyypa0 % JaamOzamaabmaabaGaeyOeI0IaaGOmaaGaayjkaiaawMcaaiabg2da % 9iaaicdaaaa!595D! \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ - }} f\left( x \right) = f\left( { - 2} \right) = 0\) , theo định nghĩa hàm số liên tục trên một đoạn thì hàm số liên tục tại x = -2 . Khẳng định (II) đúng, khẳng định (III) sai.

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