2022屆高考數(shù)學(xué)一輪復(fù)習(xí)24配套練習(xí)

【贏在高考】2022屆高考數(shù)學(xué)一輪復(fù)習(xí)配套練習(xí)1若10x310y4則10xy的值為A3B443C3D223【答案】A【解析】10xy10x310y42下列等式:36a32a;326(2)2;3424(3)42中一定建立的有個(gè)個(gè)個(gè)個(gè)【答案】A【解析】36a336a2a;323206(2)2622320∴326(2)2;34204(3)420∴3424(3)423函數(shù)f(x)axb的圖象如右圖,其中a、b為常數(shù),則下列結(jié)論正確的選項(xiàng)是>1,b1,b>00yax0,即ba51f(x)axfn,則m,n的關(guān)系為2n0>n0511f(x)ax(51)xfn,∴22my(1)x3x8y(1)xy3xy(1)x3x8113a6a21a(a)2(a0)a3a333333(a)34(b)3(ab0)y4x32x3(0](01][24](0][12]4b)a3(22)[17](23)2[125]23[51][15]y(2x232xx24xx33244222222x[11][24]x(0][12]yaxb1(aa1)0>1且b>01且byaxb1a0b10x[12]f(x)2x14af(2)bf(3)f(x)f(x)2xaf(1)f(3)bf(4)cf(1)2xa2a(1]21)(01]22321](01]21)(02]21)[02]2xa2a(1]2x(02]0a2a22aa02a10a13a3nan(n1n1(3x7)0x2(a2)2nf(x)(x2)2a2a2x72x163y130a30x20x2x72x163y133(a2)23273x70327An(nan)(n)yax(a0a1)a3a72a5a3a72a5a3a72a5a3a72a5a3a72a5An(nan)(n)yax(0a1)的圖象上,所以有aann故a3a7a3a7由基本不等式得a3a72a3a72a5(因?yàn)閍0a1進(jìn)而基本不等式的等號(hào)不建立,又2a52a5故選A8已知函數(shù)()2x1對(duì)于滿足fx0x1x2x1x2(1)(x2x1)[f(x2)f(x1)]0(2)x2f(x1)x1f(x2)(3)f(x2)f(x1)x2x1(4)f(x1)f(x2)f(x1x2)f(x1)0f(x2)0(x1f(x1))(x2f(x2))3a022x10x20618a[kk1)k31130110x[20]y3x12[51]x[20]y3x1233321015xa131時(shí),則f在[1,2]上是增函數(shù),32y32y1yf(x)a(a0a1)3222∴f2-f(1)a∴a322②當(dāng)0f(2)aa1a13ya2x2ax1(a0a1)x[11]axt0,則2222yt22t1(t1)22其對(duì)稱軸為t=-1該二次函數(shù)在[1)上是增函數(shù)①若a>1,∵x[11]∴tax[a1a]故當(dāng)t=a,即=1時(shí)ymaxa22a114解得a=3a=-5舍去②若0x[11]tax[a1]t1ymax(a11)2214a11(1y2x23x4aa353x23x40x23x404x14x1tx23x4tx23x40x23x4x3t2[43]f(x)2

(x3)2254x1tmax4x3tmin00t2524252445y(1x23x4[21]tx23x4(x3225(4x1)22)82)4x23x43x1tx23x4y(1)x23x4[43][31][31]22222a2xb3xab00,判斷函數(shù)f的單一性;2若abf時(shí)的的取值范圍【解】1當(dāng)a>0,b>0時(shí),因?yàn)閥a2x、yb3x都單一遞增,所以函數(shù)f單一遞增;當(dāng)