291 Bài trắc nghiệm Hàm số Cực hay có lời giải chi tiết (P3)

Xét hàm số f(x) = abs( x^2 + ax + b). Gọi M là giá trị lớn nhất

35/40

Xét hàm số f(x)=x2+ax+b. Gọi M là giá trị lớn nhất của hàm số trên [-1;3]. Giá trị của biểu thức a + 2b khi M nhỏ nhất là

4

-4

2

3

Giải thích

Chọn B

Ta có S0_WYah8zdQPgYqzY_QtJ7_NPuuVZ88mm0mj2ijPX_QvE-HEOx7RojLShSl4H7bAAv-KH2kdBxz7BBZTy9RiZGy7lqKjOB_P_BZJnfW9A4l101iZSPyQcNLeUuN4IH149iEumFtVoMWCMjmllw.

Dấu = xảy ra khi A=B.

Ta có bUYMhFkxtqp0G9sggAWVcpEs5s0RVwDl_8WNkQGlrs6br181E90XGqM9hmKhd4WOwYPxaHJiYdANBV7xgAWh45Gs6ZgEpKTO21F-A_p8ESBosVXGddZNy6kGy78_vDziN_MkuiRPM-TwfXgvww.

Dấu = xảy ra khi A= -B.

Xét hàm số 6ZQu25W-V0YSvcEDDl5CiX9qLlMhkklR-eqYZR8cfBJgVIbbNJbhymaWd3hENWOhIU0HymCdXnVHZE-lHzUA7gkNv2tojKnZGF9dbCP1Q4YjyAiRPi6um63-bjdcUL3qMKXJIBnxkjXgKpnZqQ, có Bk2eQAL2k-WnFrN2Qk9cOyIHOCLBNmLp9fhkMGoj0IUs0scmyRICjkHzwuugy8a86yXfxfPx-uNxALfBtyGRxuFLkm00CXTAVvUFVYV6iboFmH5LYs6K8r3EyOblZhSh4OySOB6GacqNbXgt6w.

Trường hợp 1: FUTYrkdiiokuYc6YReWBnKw2uYeeQqhzkrmpz6mnjbSvusuUwYZleaMKVJZa3C3sMGqgjp0Ij5KxYfcxMvS4SHa53pm3p35bWW7U1v8AK3dFA7vgTas1334jMlUM28-wR8aYjluQsHJnetBZrgl1yeYpPu_PiSMi8py2ThwBVhT3qwWsoOySpTotKOp9t8jgGTqwL4-zh71rYVmESQPcrP_Lvqq5bkQUr89L6MLD_d1xnUGNG1VdRyi9SelEjnr4vXC7fh4KtrnSHRb3CnedXepft4JfBPbalyZg.

Khi đó d2Xyl0eo68Svs-_Ycvi8H4_RjOzcNgSIJkIWJGeQQXToRExVtDSr_UdlHlOo5h6m6jj9CqZ0P5eLueE4xwFkgUgO4SQz8AM8Ze4SBZoXQHpr7l4XeNnE-mCcLB6VBLUxF81O7dhqJvHOddrrpA.

Áp dụng bất đẳng thức (1) ta có mjeNI34kgBuQCXEaTynChyg8LurSHVATpmxQ58O0K60pLTBU2VoRteBoC6wrNez_TCNNgvoOH6WJ9QunlyE7kI4ATnkncotIleLAUGaRKwjNSqh27Te5QEVqxPoruammI8gdwkL3zfrDujqxmw.

Trường hợp 2: N70LZLLYjdFS39f9bAIA4c1hr2DLIMeKzbN4OQ4DiI_-DiUwQS0T9xBuyZ3bnw7T-AQgqla1WnjrrekH7aa_nZ1smgAkBXimVbB7pmeMtKuNvYZxkXa2W4hbunVHqW--Ps_T8HeTK77CSPe5zQvuBROLY_dy7DQtxOY7EDoT3nPNg0ePqxmIey8tsp0WVTW780dIEEPCNI5yIDhNjLjfuZv_suRUBO97DnfoMaxUGyoLXV8FOamffGUC4fteRpRQO62pWpl0-8KkQXh9gNpGhUB7mYEXUueAb4kw.

Khi đó FygxfW6KpHtsSgqYR-Zb4PAeF8sOp-KXTHCbn-5ymrWQMbnpDmkQq4765ISu-HOGeL2AmFEN6moFP9rZW4DRW0B9MXdvbxKuyI93Np6ishsMbvKMwGGKZW6r6yJhZTc6VypHGpY2GhVKKSpMwA.

Áp dụng bất đẳng thức (1) và(2) ta có

-Ldmn324DWuP7Ok7Emzh95tKMy_T3P0DDNu4yZbLSUkxX-74q5g8plgyJVv26JKr23dvATNb8act2SJBgwFYatcwbPYnslxIzcjGtONN76KeFi2Uha52YIVna3rDCzefEJK4s8xZgKALSQdkAA

_XZFuValtEi62i87VFO2arHx-5PkepBrn-krgCfvDk_VvB0bwd9nOikCPcyojA19P5ZTvIKozH9GdgD9ZNCdvNHFXQ_Aa26iLrhAekaeVCDqagypcPt8P39udKNSiYCcehBkONlOTU632W8NEA

TRSFgV9slD-c9o3lZ5_8dT1sE7O_BNmZRsR_Us97kEkHlYGnF_ALGk1TSiDZQPBB0rFUPkl8f67SVR5-SzVCSx0Xr1jSbEOItAxwHrci0Xwpv2JdCth8AXoarrxFyG3cYr6X8Qho4JsXN1VQbg.

Suy ra 71g3o1uYavSsQJt8Rcsu_grEoU2BDdxXHsszX7-RoEQKOa8JBo3oJEueRbPL-5hKq2JgELCp-WXFmgwQ8KUsC2NhVUQDexnaEf5fy-wZMzfiuAAmp35Xes0A8J3jtSQVAxEjiGA6-ztOodFzYA.

Vậy M nhận giá trị nhỏ nhất fQLidrcDewuArld3LCLTeInxK3ph6Z9nGymxew6QGWI8JpNmyMM_3pUgKZTCktYyAQiT-oYTW2x2hoUIdVcOzxI0f9POS4dDQTMP31Y0ligbMpm402HAjlMeTKs6u1SwogilY0pPVedMApduvQ khi

_fYW1uJ2h6oTSMb0llDvDwrSt-dSbvESUNBWF6eH023WGHgZOGZWifJ1C07XqhGgS932NtKXiLjHWr3h5rvp9ROU-jrcHfYCE4Urv_f82luT8Fe7E-Z1j8rG9nS2bU1bnH47PMLcx8ltUJKEhA

nuicI-3dSDQ-WBMrVFEwDdfSgsdLrjgf-rYEBcm2sJ_A_QDoVN-05saoGdPp1SndCFMqlZptMvIFopRcWCOgkpQD3-e1dqpoXhHbHfnNTbg_ooHPQtiSCu670WmEcP8dp6bkWMULiL7cBIMKGw.

Do đó p09D2IWgwimVgLfqM9aEdgv2Ur2XZzJWzHEvzqa1CgNMtaKzZ2eWRBV_ohPrMquMHjaenabjsatk1n8K43UgfAwkK-PA_S4wWGN_DldqSeWCMzBfF3CKhcmq_RwAL_hu_B7matFD6ac3cMbd7g.