Cho hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maalaaabaGaaGOmaiaadIhacqGHsislcaaIZaaabaGaamiEaiab % gkHiTiaaikdaaaaaaa!3E10! y = \frac{{2x - 3}}{{x - 2}}\) có đồ thị (C). Gọi I là giao điểm của các đường tiệm cận của (C). Biết rằng tồn tại hai điểm M thuộc đồ thị (C) sao cho tiếp tuyến tại M của ( C) tạo với các đường tiệm cận một tam giác có chu vi nhỏ nhất. Tổng hoành độ của hai điểm M là

A. 4

B. 0

C. 3

D. 1

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Chủ đề: Đề thi THPT QG

Môn: Toán

Lời giải:

Báo sai

HD:

Gọi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamytamaabm % aabaGaamyyaiaacUdadaWcaaqaaiaaikdacaWGHbGaeyOeI0IaaG4m % aaqaaiaadggacqGHsislcaaIYaaaaaGaayjkaiaawMcaaiabgIGiop % aabmaabaGaam4qaaGaayjkaiaawMcaaiabgkDiElqadMhagaqbamaa % bmaabaGaamyyaaGaayjkaiaawMcaaiabg2da9iabgkHiTmaalaaaba % GaaGymaaqaamaabmaabaGaamyyaiabgkHiTiaaikdaaiaawIcacaGL % PaaadaahaaWcbeqaaiaaikdaaaaaaaaa!5148! M\left( {a;\frac{{2a - 3}}{{a - 2}}} \right) \in \left( C \right) \Rightarrow y'\left( a \right) = - \frac{1}{{{{\left( {a - 2} \right)}^2}}}\) ; tâm I(2;2).

Phương trình tiếp tuyến tại M là: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iabgkHiTmaalaaabaGaaGymaaqaamaabmaabaGaamyyaiabgkHi % TiaaikdaaiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaaaaOGaai % OlamaabmaabaGaamiEaiabgkHiTiaadggaaiaawIcacaGLPaaacqGH % RaWkdaWcaaqaaiaaikdacaWGHbGaeyOeI0IaaG4maaqaaiaadggacq % GHsislcaaIYaaaaaaa!4A92! y= - \frac{1}{{{{\left( {a - 2} \right)}^2}}}.\left( {x - a} \right) + \frac{{2a - 3}}{{a - 2}}\)

Tiếp tuyến d cắt x = 2 tại \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqamaabm % aabaGaaGOmaiaacUdacaaIYaGaey4kaSYaaSaaaeaacaaIYaaabaGa % amyyaiabgkHiTiaaikdaaaaacaGLOaGaayzkaaGaeyO0H4Taamysai % aadgeacqGH9aqpdaWcaaqaaiaaikdaaeaadaabdaqaaiaadggacqGH % sislcaaIYaaacaGLhWUaayjcSdaaaaaa!4A2A! A\left( {2;2 + \frac{2}{{a - 2}}} \right) \Rightarrow IA = \frac{2}{{\left| {a - 2} \right|}}\)
Tiếp tuyến d cắt y = 2 tại .\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOqamaabm % aabaGaaGOmaiaadggacqGHsislcaaIYaGaai4oaiaaikdaaiaawIca % caGLPaaacqGHshI3caWGjbGaamOqaiabg2da9iaaikdadaabdaqaai % aadggacqGHsislcaaIYaaacaGLhWUaayjcSdaaaa!486E! B\left( {2a - 2;2} \right) \Rightarrow IB = 2\left| {a - 2} \right|\)

Do đó IA.IB = 4 mà \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4qamaaBa % aaleaacqqHuoarcaWGjbGaamyqaiaadkeaaeqaaOGaeyypa0Jaamys % aiaadgeacqGHRaWkcaWGjbGaamOqaiabgUcaRiaadgeacaWGcbGaey % ypa0JaamysaiaadgeacqGHRaWkcaWGjbGaamOqaiabgUcaRmaakaaa % baGaamysaiaadgeadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaWGjb % GaamOqamaaCaaaleqabaGaaGOmaaaaaeqaaaaa!4E1C! {C_{\Delta IAB}} = IA + IB + AB = IA + IB + \sqrt {I{A^2} + I{B^2}} \)

Áp dụng bất đẳng thức AM – GM, ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadMeacaWGbbGaey4kaSIaamysaiaadkeacqGHLjYScaaIYaWa % aOaaaeaacaWGjbGaamyqaiaac6cacaWGjbGaamOqaaWcbeaakiabg2 % da9iaaisdaaeaacaWGjbGaamyqamaaCaaaleqabaGaaGOmaaaakiab % gUcaRiaadMeacaWGcbWaaWbaaSqabeaacaaIYaaaaOGaeyyzImRaaG % OmaiaadMeacaWGbbGaaiOlaiaadMeacaWGcbGaeyypa0JaaGioaaaa % caGL7baacqGHshI3caWGdbWaaSbaaSqaaiabfs5aejaadMeacaWGbb % GaamOqaaqabaGccqGHLjYScaaI0aGaey4kaSIaaGOmamaakaaabaGa % aGOmaaWcbeaaaaa!5D90! \left\{ \begin{array}{l} IA + IB \ge 2\sqrt {IA.IB} = 4\\ I{A^2} + I{B^2} \ge 2IA.IB = 8 \end{array} \right. \Rightarrow {C_{\Delta IAB}} \ge 4 + 2\sqrt 2 \)

Dấu bằng xảy ra khi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamysaiaadg % eacqGH9aqpcaWGjbGaamOqaiabg2da9iaaikdacqGHuhY2daabdaqa % aiaadggacqGHsislcaaIYaaacaGLhWUaayjcSdGaeyypa0JaaGymai % abgsDiBpaadeaaeaqabeaacaWGHbGaeyypa0JaaGymaaqaaiaadgga % cqGH9aqpcaaIZaaaaiaawUfaaaaa!4E60! IA = IB = 2 \Leftrightarrow \left| {a - 2} \right| = 1 \Leftrightarrow \left[ \begin{array}{l} a = 1\\ a = 3 \end{array} \right.\). Vậy \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaabqaeaaca % WGHbGaeyypa0JaaGinaaWcbeqab0GaeyyeIuoaaaa!3AB1! \sum {a = 4} \).

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