函數(shù)的奇偶性課后課時(shí)作業(yè)

222222222222課后課作[組]大連]已知函數(shù)()x時(shí)，()x則f的值為()A．1C．

BD．Bf以f(1)＝－1選B.洛陽]若函數(shù)yf(21)是偶函數(shù)數(shù)y＝(2x是()A．－C．x＝

BD．

Cf(21)是軸x0f(2x1)f)(21)的f(2xx1沈陽]已知函數(shù)f(x，若f(a)＝，則(x＋1()A.C.

BD．Cx＋fx1x1＋f(xf)2

32324f)2f()＝.C.沈陽]已知定義在數(shù)fx)滿f(2)＝f)，x[3,4]時(shí)f)＝ln()A．C．ff

Bf3D．f

Cf(xR上的偶函數(shù)(x[3,4]上是f(x[[∵選浙]已知f)(x在上的x)＝f(xg(x“f(x(x“()的)A．充要條件BC．必D．

Bf)(xff(x)g()＝()h()f(x(x＝(x())h(h()fx(x)若f(x(xhxfxx數(shù)”()”洛陽]已知定義在R上的數(shù)(x)，對(duì)任意x∈f＋4)＝()＋f(2)成立fx＋1)的圖象關(guān)于直線－則(2014)的值為()A．C．

B2014D．4

C

222222222222

f)對(duì)數(shù)yfxf(－24)＝2)f(2)f＝f(2)f以f(2)＝0fx4)fxx以為周ff5032)f(2)C.知()在上于x≥有(2)＝f(x且當(dāng)[0,2]時(shí)x)＝－則f＋f(－于)A．1－eC．－1－

BeD．

B(f(x＋f([0,2]時(shí)，f()－1以f＝f(1)ef(－＝(0)，f(2013)f(－2014)e選知f(x)是周期為當(dāng)0<x<1時(shí)，f)設(shè)a＝f＝＝)A．<C．<c

Ba<<cD．<<

Aalglg，bf＝lg1cf1為>，2>lg>lg以c<<b唐山]已知函數(shù))g(x在R

2222222222且x)＝f(x－則f(2015)的值為

gxf(1)由f(xxg()∴fx1)＝f1)f()f(x＋f(f(x4為周期(2015)×－f(－1)g(0)xx＋2(≥0)，若f(3－

2

)>f(2數(shù)的取值范圍(－(x2(x≥0)[0又f(x以f(xR由f(3af(2a)3a>2a即11．已知函f(在R上x＝(1)求證f()為4的(2)若(x(0≤求x∈－4]時(shí)數(shù)f(xf)x有f(1)有f)fx2)數(shù)f(x有f(－x故f(xf(x而f(x4)f2)＝fx即f(x4的周數(shù)f(xRf(0)0.x[x∈[0,1]f(xf)－x.x[1,0]f)

x[54]時(shí)，4[f(xf(x4)x4.x[54]時(shí)數(shù)f(xxf(x(－∞)x2)＝f(x)0≤x≤，f(x(1)求(π)的值；(2)當(dāng)4≤≤求f)與x(3)寫(＋∞)內(nèi)數(shù)()f(x2)f()f(4)＝f[(x2)2]f(xf)f(x以4為周期f(π)fπ)f(π4)fπ)(4π)πf(xf(xf)[(x1)＋2]＝f(x1)＝f[(1)]，f)fxf()稱．0≤x≤fxxf(x則f(x4≤x≤，f(xS則S4

4××2×1

＝數(shù)fx[41,4kkZ[4k1,43](kZ

44[組]江西]已知函數(shù)fx)sin＋|sinx－是()A．f()πBf)在

C．f()D．f()為[－

Cf(xsincos期2π]fx)＝π

5πsinxx，x，

fxππA誤f()，

Cf()π]上的湖北]設(shè)f()在R上的對(duì)x∈有f(x－＝x＋2)，且x∈[－2,0]時(shí)f()＝間(－2,6]內(nèi)關(guān)xf(x)－(x＋＝個(gè)不a則是)A．(1,2)C．4)

B(2，)D．(4，2)

Dfx在R的f(x軸

xR有f(x－2)＝f(xf(x

當(dāng)x[，)1f(x(2,6]內(nèi)2,6]fxlog(>1)af(log(xa得a(42)R上的函數(shù)y＝x∈f(1)＝；yf(x＋1)的圖y軸對(duì)f的，∈[0,1]，<x有fx)>)．f1211f(2)，是________．

f(3)<

<(2)

f(xf1＝f(x(x)fyf(x＋軸對(duì)稱，f(1)f()f()于對(duì)稱f)[0,1]結(jié)(x[f(2f(1)

2222222222[1,2]，以f(1)<f(2)即f(3)<f

<f(2)在上的函數(shù)f()意aR都有f(b)＝(af＋k為數(shù))．(1)判斷k，fxk＝－1，f)是R(4)＝，若不等式f(mx－mx＋對(duì)意R恒成數(shù)m的f()R上f＝令a0f0)f(0)fk以0.由f(abf(afabx則f(xx(xf(－)f(0)