专题03 导数多选题 (解析版)新课标试卷

fxfxf00f(x)cosx，f(x)sinx0下2()6ffln0．f．36243f．f6．f2f343f(x)g(x)，x，2xf(xxf(xx则g(x)，cos2xf(x)cosxf(x)sinx0，f(xxf(xxg(x)0在cos2x2f(x)g(x)在x2ff664gg6ffA4642coscos64ff(x)f00g0()gx0又在x20Bf323ff又g663gf3fC363coscos63ff43g2f又g4fD343coscos43x2xax1f(x)a23）32201x2xax1f(x)322f(x)3x4xa22x2xax1f(x)322则f(x)3x4xa与x22233233即443a0a22ABABfxfx1xfxxfxlnx，f1）220,xfx0,xfx．．在．．在0,1120,xfxxfx在在2f得，则），x，x设＞x即）0得0得0，xfx0,1即在121）．，2,2fxxsinxxcosx）fx．．．．fx0,在fx4fx4fx2,2fxfxxsinxxcosxfx1cosxcosxxsinx1xsinx，fx0fxx0,0,在当.f00fx0x1，x12,2sinxy，yx2,24fx2,2fx2..xf(x)）x21fxe．．在x2efxf()f(．f(f(x)1ekk在(0,)x22【解析】函数定义域为(0,1x)，f'(x),x3f'(x)f'(x)0e)，f(x)单调递增，当x(e,)当x在x，f(x)f(x)1f(e)eAef(1)0f(x)0x0x11f(x)0Bf(x)1212223f()f(f()f(e，221ln424，1xx，x(e,)h'(x)0h(x)e4h(4)∴h)，设h(x)h'(x)xx2lnln4ln2(f()，即∴f42(f()f(即fCx11x11令g(x)x2x20则g'(x))'()0gx(,)x（xx，x3ee11eg'(x)0，g(x)g()在x，2ee1e(x)k(0,)k∴f在xD22x．(0,)fx122xfxfxxfxx对）且x(0,)2f23f15．11f12,x1则fxxx222f32f17．1212f12,0x1x则fxx2fxxx1gx2，221fxxxfxxxfxfx1xx2gx2则2x1x12'0gx,x1fxfxx2x2在(0,)gxg123gg22315ff故故A.,，f32f17121gxg)0x1若f12gx(0,在当即112fxxx12xx即fx故DB.222，aa62xax(a0)(,f(x)在)a3223634．．5．．ax0x(af(x)2x(3xa)，，132aax0f(x)0x当或x0f(x)0，33a,,0，a,则f(x)33aaf()a3f(x)在x，33aaaaa3)(2x)0x(x由f(x)x或，23336aa6又f(x)在(,)23aa6aa4，336fx2a0fx有4a4e123{x|x0}f(x)f(x)fxfx有2x0有4a0，20当xfx2e2a1e2eex则f(x),f(x)x0,f(x)当xeee由f(x)0得xf(x)00xf(x)0，xaaaeef()4f()0fx即aaeaee1a)0a22211，af()ae，a1a2,3,4eaxxx，xx(f(x))x1212aex2x．．12xx1xx2x．f(x)x．012120eaxf(x)ea，f(x)xxfx当f(x)ea0在Ra0x0xlnaxlnaf(x)ea0，当af(x)ea0xxfx(,lna)(lna,)在eaxf(x)且xx，x,xx1212ealnaaalnaa(1lna)0aa0，则f(lna)ae所以1lna0Axln(axx)axx)2xx)x，212121212e2fe2a0xf10，取a2220x1x2x1B12100x1xx1C12由f(0)1fx(,lna)(lna,)在xlna0xx2x2lnaD120）ca，，bc74aa，cb54aacbab22,22ln34，c4ln5，bcbaa35bb1b0aaa1对，cba4a5(a102cba4a7cba4a52bc22123cba，①②b22a2b1a2．1aa(a2，2)2b1