函數(shù)的單調(diào)性重點(diǎn)考 專題fx2x2mx1在區(qū)間1f1的取值范圍是（A．7,

x fx一定是（奇函 B．偶函 C．增函 D．減函是（

C．3,

D．4,fxf(xf(2x，且在區(qū)間[1上單調(diào)遞減.設(shè)af(ln1.1bf20.4cflog25，則（abC．cb

bcD．bafxln1ax（a為常數(shù)），則（1fxg(x)xf(x在[0上是增函數(shù).若ag(log5.1bg(20.8cg(3，a，b，c的大小關(guān)系為ab

cb

ba

bcyf4xx2在區(qū)間12fx的解析式可以為（fx4xC．fx

fx2D．fxfxR上的偶函數(shù)，且在區(qū)間,0aA．,1

2a的取值范圍是（B．3, 2 C．1,3

D．,13,22 2 yf(x的定義域是00，對任意的x1x20x1x2x2fx2x1fx10yfx1的圖象關(guān)于點(diǎn)10f14x2fx4的解集為（A．1,0

已知定義在Rfx在區(qū)間10f4xfx，f2xfx，則（）A．fkkC．f2.5flog

B．f0.9f1.2D．fsin1fln1 2 fx是定義在Rx20fx1xyRfxfyfxy2f02，則（A．f2

B．f4C．fx在R上單調(diào)遞 D．不等式fx23x41的解集,5∪2,Rfxx，y，恒有f(x1]fy1f(xy1f(11x0f(x0，則下列結(jié)論正確的是（f(0)fx的最小值為fxRxf(xf(2x3的解集為(0)(2已知定義0,在上的函數(shù)fx滿足f()yf(x)xf(y),且當(dāng)x>1時(shí)，fx0，則 C．fx22f

B．f2x2fD．fx3fxf2x2可以 f32af4a50，則實(shí)數(shù)a的取值范圍 fxxyRfx十fyfxyx0fx0f11，則不等式fx284的解集 4 求實(shí)數(shù)cf(x)2cos(2xπa2x0πfx1，求afx在區(qū)間0a和2a7π上均單調(diào)遞增，求實(shí)數(shù)a 3 6

a2x2x

,aR當(dāng)a1fx24f(4x1fxg(xmf(2xf(x在區(qū)間0log23上有零點(diǎn)，求正實(shí)數(shù)m的取fx2x2mx1xm因?yàn)楹瘮?shù)在區(qū)間1m1，解得m4【分析】利用函數(shù)單調(diào)性定義即可得到答案【詳解】對于任意兩個(gè)不相等的實(shí)數(shù)xxfx2fx10 x x1x2fx1fx2【分析】由函數(shù)的對稱性、單調(diào)性即可列出不等式求解fx1fxx=?1對稱，fx在1fa1f1，a1111，解得2a2f(x)f(2xx1af(ln1.1f(2ln1.1ff(x)f(2xx1，則af(ln1.1)f(2ln1.1 而120.42ln1.12log1.1log4.4log5fx在[1 f20.4f(2ln1.1flog25，得bac【分析】由函數(shù)奇偶性的性質(zhì)，定義域關(guān)于原點(diǎn)對稱可求得a,進(jìn)而判斷函數(shù)的奇偶性fxln1ax，有1ax0，即1ax1x0fx1 1則定義域?qū)ΨQ，必然有1a0，即a1fxln1xfxfxln1xln1xln10fx為奇函數(shù)

1

1

1fxg(x)xf(xR上的偶函數(shù)，且在[0ag(log25.1)g(log25.1) 20.82，又45.18，則2log5.13，所以即020.8log5.13 g(20.8g(log5.1)g(3，所以bacC．【詳解】因?yàn)閠4xx2開口向下，對稱軸為t2，可知內(nèi)層函數(shù)t4xx2在區(qū)間12上單調(diào)遞增，x1t3x2t4；可知t4xx234yf4xx2在區(qū)間12Afx4xx2在區(qū)間34ABx34fx2x2x在區(qū)間34B22Cx34π3πfxsinx在區(qū)間3422 Dfxx在區(qū)間34D錯(cuò)誤.所

1a11 即a11或a11，解得a1或a3 2 g(x)xf(xg(x的奇偶性和單調(diào)性，結(jié)合函數(shù)的奇偶性和單yf(x1圖象關(guān)于點(diǎn)(10)fx圖象關(guān)于點(diǎn)(00fx為奇函數(shù)g(x)xf(xg(x)xf(x)xf(x)g(xg(x對于xx(0g(x2g(x1)0(xxg(x在(01 x g(x在(0)上單調(diào)遞減f(14g(14g(14x0f(x4xf(x4，即g

x0f(x)4xf(x)4g(x)g(1，解得1x0f(x4的解集為(10)(1g(x)xf(xg(x)Af4xfxfxx2f2xfxfx關(guān)于點(diǎn)10對稱，f4xf2xf2xfx，f4xfxfx的周期為4fxfxfx為偶函數(shù)，fx在區(qū)間10fx的大致圖象如下圖：f41f10，x0f40f0f4f2xfxf2f0，f1f2f3f40，因?yàn)楹瘮?shù)fx的周期為4，所以fkf1f2f3f42f9f100f1f2f20Akf1.1f1.2f0.9f1.20B不正確； 對于Cflog80flog16log5f4log5

log22

log52 f2.5flog280C 1 Cfln2fln2fln2，且0ln20.7314 所以sinπsin1sinπ 20.7，故1sin1ln20 fx在0,1fsin1fln1D 2 Axy2fxf2fxfx是非常值函數(shù)，∴f21ABxy4xf4xfxf2x0，∴f4f01,f4C，f4xfx

f

1Bx24x2，0f4x1，∴x2fx∴xR,fx0

f4

x

fx1fx

21fxfx fx在RC Dxy0f2f0f04，x2,y0f4f2f08，f4xfx1f8

f

1，又∵fx在R上單調(diào)遞減，∴fx23x41x23x48x<?4x1D錯(cuò)誤.Ax1y0，則f(11]f(0)1f(11f(11f(0)0，ABxytf(t1ft1]20f(t1pRfp xf(x1f(xpp1f(xp1fp10f(x)1f(1)1pRfp)1f(x)1，B錯(cuò)誤；C，由[f(x1][fy1f(xy1f(xyf(xfyf(xfy，x1x2Rx1x2x1x20x0f(x0f(x1x20，f(x10f(x1f(x2)f(x1x2x2f(x2)f(x1x2)f(x2)f(x1x2)f(x2)f(x2f(x1x2f(x2f(x1x2f(x1x2f(x210，f(x1f(x2fxR上的增函數(shù)，C正確；Df(2)f(11f212f(13，則f(2x1f(x1f(2)14f(2x)

f(x)

1f(xf(2x3f(x

f(x)

1f(x)

f(x)

23，令tf(x10，由t423，即t25t40解得0t1或t4，即1f(x0f(x)3fxRf(2)3,f(0)0，x0x2f(xf(2x)3的解集為(0)(2，D正確.【分析】應(yīng)用賦值法得到0x1f(x0f(x),f(x2f(x3性質(zhì)判斷即可x0y0fyyfxxfyxy1f(1

f()

1f(1)1f(1)0x1f1yf(11fyfyf1f(x x1f(x0當(dāng)0x111f1f(x0，即0x1,fx0 所以x1fx0，Ax1f202f10f2x2fx不成立，By2 1賦值xy2，得f f(y)yfyyf(y)，解得f(y)yf(y)y yf(x2x1f(x x f(x2x1f(xx0 x fx22f(xx12f(x xx0x12x

20x1f(x0x12f(x0fx22f(x 當(dāng)0x1f(x0x12f(x0fx22f(xC x2

1 xx2,y，得f f(x)f(x)xff(x)xf(x) 1

x xf(x2x1f(xf(x31f(x2x2f(x11x2f(x x 則fx3fxf2x211x2f2xx2 2fxfx fx3fxf2x2fx不恒為0D正確xx3(答案不唯一fx是R上單調(diào)遞增的奇函數(shù)即可符合題意xx3，xx3(答案不唯一a【分析】根據(jù)函數(shù)圖象關(guān)于20f4xfxfx在2上單調(diào)遞減fx0結(jié)合函數(shù)關(guān)于20fx在R上單調(diào)遞減.再利用函數(shù)的單調(diào)性即可求得a的范圍.fx的圖象關(guān)于20f4xfxf32af12af32af4a50f4a5f12a，則4a512a，得a2.a23,xy02f0f0f00yxfxfxfx為奇函數(shù)，x1x2fx1fx2fx1x2，x0fx0fx1x20fx1fx2，fxR上單調(diào)遞增．f11f14 4 2 2 4f1f1f142 2 4 fx284fx28f1x281，解得3x3．17．(1)c0；(3)3【分析】（1）f00，求出c0（2）f20得d4b2fx0x0x2b2b2，解得b3f13b72(b2)2(b2)2

0fxf00，解得c0（2）由（1）fxx3bx2dfx3x22bxf20，即84bd0，故d4b2fx0x0x2bfx在0上是增函數(shù)，在02上是減函數(shù)，所以2b2，解得b3f11bd1b4b23b72（3）因?yàn)棣力拢?fx03個(gè)實(shí)數(shù)根，fxxαxβx2，

所以

d

，所以

(b(b2)28b

(b(b2)2(b2)2Qb3，b25，(b2)225(b2)2169(b(b2)2

{π,ππ π

,22[,3【分析】（1）x的取值范圍，進(jìn)而求出2xπ的取值范圍，利用余弦函數(shù)的單調(diào)性即可求得結(jié)果f(x)2，得到cos(2xπ)1k 結(jié)果fx的單調(diào)遞增區(qū)間，再讓區(qū)間0a和2a7π求得結(jié)果

3

6【詳解】（1）x0π，U2xππ5π

333在[π0)(π5π上單調(diào)遞增，cosU

cosπ1，f

2a21，a1（2）由（1）f(x)2cos(2xπ3，Qf(x)2，cos(2xπ)1 2xπ2kπ2πkZxkππkZxkππkZ xπ,π，當(dāng)k1時(shí)xπ；當(dāng)k0xππ；當(dāng)k1x5πx的取值集合為{πππ5π 22

（3）由2kππ2xπ2kπ(kZ，得kππxkππ(kZ) fx的單調(diào)遞增區(qū)間為[kππkππ](kZ 當(dāng)k

fx的單調(diào)遞增區(qū)間為ππ36當(dāng)k1fx的單調(diào)遞增區(qū)間為27 又函數(shù)f(x)在區(qū) 和[2a,

0a

，解得πaπ π的取值