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函數(shù)(中)

2023-07-17 17:23 作者:24bs  | 我要投稿

%5Cexists%20x%5Cin%20%5B0%2C2%5D%5C%20s.t.%20(e-1)%5Cln%20a%5Cge%20ae%5E%7B1-x%7D%2Be(x-1)-x,求?a?取值范圍。

%5Ccdots%5CLeftrightarrow%20e(%5Cln%20a%2B1-x)%2B1%5Cge%20e%5E%7B%5Cln%20a%2B1-x%7D%2B%5Cln%20a%2B1-x

令?t%5Ctriangleq%20%5Cln%20a%2B1-x%5Cin%20%5B%5Cln%20a-1%2C%5Cln%20a%2B1%5D,則?et%2B1%5Cge%20e%5Et%2Bt。研究?f(t)%3De%5Et%2B(1-e)t?即可。

a%5Cin%20%5B0%2C4%5Df(x)%3D%5Cdfrac%7B4x-a%7D%7Bx%5E2%2B1%7D。(1) 當(dāng)?x%3E0?時(shí),證明:f(x)%5Cle%20%5Cdfrac%7Ba%7D%7B2%7Dx-a%2B2;(2) 設(shè)?x_1%2Cx_2%5Cin%20%5Cmathbb%20R,若?m%5Cin%20%5Cmathbb%20R?滿足?f(x_1)f(x_2)%3D-m%5E2,證明:f(m-a)-f(1)%3C%5Cdfrac18

(1) 法一:直接移項(xiàng)后因式分解;法二:以?a?為主元,然后因式分解

(2) 令t%5Ctriangleq%204x-a,則?f(x)%3D%5Cdfrac%7B16t%7D%7Bt%5E2%2B2at%2Ba%5E2%2B16%7D,故?f(x)%5Cin%20%5B%5Cdfrac%7B8%7D%7Ba-%5Csqrt%7Ba%5E2%2B16%7D%7D%2C%5Cdfrac%7B8%7D%7Ba%2B%5Csqrt%7Ba%5E2%2B16%7D%7D%5D。故?-m%5E2%5Cge%20f_%7B%5Cmin%7D%5Ccdot%20f_%7B%5Cmax%7D%3D-4,即?m%5Cin%20%5B-2%2C2%5D。之后分?m-a%5Cle%20(%5Ctext%7Bor%7D%20%3E)%5C%20%200?討論即可。(其中需要使用 (1) 中結(jié)論)

偶函數(shù)?f(x)?定義域?yàn)?%5Cmathbb%20R%5Cforall%20x%5Cin%20%5B0%2C%2B%5Cinfty),2f(x)%2Bxf'(x)%3E0。判斷以下結(jié)論正確性:

......

對(duì)于此類題目,參見?https://www.bilibili.com/video/BV1UA4y1f7TZ/

a%2Cb%3E0,ae%5E%7Ba%2B1%7D%2Bb%3Cb%5Cln%20b,證明:b%3Ee%5E%7Ba%2B1%7D

%5Ccdots%5CLeftrightarrow%20e%5Ea%5Cln%20e%5Ea%3C%5Cdfrac%7Bb%7D%7Be%7D(%5Cln%20b-1)%3D%5Cdfrac%7Bb%7D%7Be%7D%5Cln%20%5Cdfrac%7Bb%7D%7Be%7D,對(duì)?f(x)%3Dx%5Cln%20x?進(jìn)行簡(jiǎn)單分析即可。

f(x)%3Dx%5E2-1?與?g(x)%3Da%5Cln%20x-1?的圖像存在公切線,求?a%5Cin%20%5Cmathbb%20R_%2B?取值范圍。

設(shè)公切線與兩圖像的切點(diǎn)分別為?(x_1%2Cx_1%5E2-1),(x_2%2Ca%5Cln%20x_2-1)f'(x)%3D2x,g'(x)%3D%5Cdfrac%7Ba%7D%7Bx%7D,故公切線既為直線?y%3D2x_1(x-x_1)%2Bx_1%5E2-1,也為直線?y%3D%5Cdfrac%7Ba%7D%7Bx_2%7D(x-x_2)%2Ba%5Cln%20x_2-1。對(duì)比系數(shù)得?2x_1%3D%5Cdfrac%7Ba%7D%7Bx_2%7D,-x_1%5E2-1%3D-a%2Ba%5Cln%20x_2-1,故?%5Cdfrac%7Ba%7D%7B4%7D%3Dx_2%5E2(1-%5Cln%20x_2)。對(duì)?h(x)%3Dx%5E2(1-%5Cln%20x)?進(jìn)行簡(jiǎn)單分析即可。

f(x)?定義域?yàn)?%5Cmathbb%20R,f(x%2By)%2Bf(x-y)%3Df(x)f(y)f(1)%3D1,求?%5Csum%5Climits_%7Bk%3D1%7D%5E%7B22%7Df(k)。

  • 法一:注意到?2%5Ccos(x%2By)%2B2%5Ccos(x-y)%3D2%5Ccos%20x%5Ccdot2%5Ccos%20y,故可設(shè)?f(x)%3D2%5Ccos%20%5Cdfrac%7B%5Cpi%20x%7D3。后略。

  • 法二:令?y%3D1?得?f(x%2B1)%2Bf(x-1)%3Df(x),即?f(x%2B2)-f(x%2B1)%3D-f(x)。同理?f(x%2B2)%3Df(x%2B3)%2Bf(x%2B1),故?f(x%2B3)%3D-f(x),則?f(x)?周期為?6。后略。

f(x)%3D(1-x%5E2)(x%5E2%2Bax%2Bb)?的圖像關(guān)于直線?x%3D-2?對(duì)稱,求?f(x)?最大值。

利用必要性探路,通過代入來求出?a%2Cb。

f(x)%2Cg(x)?定義域?yàn)?%5Cmathbb%20Rf(x)%2Bg(2-x)%3D5,g(x)-f(x-4)%3D7。若?g(x)?圖像關(guān)于直線?x%3D2?對(duì)稱,g(2)%3D4,求?%5Csum%5Climits_%7Bk%3D1%7D%5E%7B22%7Df(k)。

可知?f(x)?周期為?4,隨后玩一玩就能得到答案。


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