高二導(dǎo)數(shù)練習(xí)題及答案

高二數(shù)學(xué)導(dǎo)數(shù)專題訓(xùn)練1程為BS=1+t+t2f其((tf的(x單)位是秒,那么物體在3秒末.,1)(1,1)C((1,)〕AB.C.D.abA.f(x)〉0B.f(x)〈0C.f(x)=0D.無(wú)法確定A.充分不必要條件B?必要不充分條件C.充要條件.非充分非必要條件yxpA(1,0)B(2,8)C(1,0)和(1,4)D&函數(shù)y1(2,8)和(1,4)A.極小值-1，極大值1B.極小值-2，極大值3C.極小值-1，極大值3D.極小值-2，極大值29.對(duì)于R上可導(dǎo)的任意函數(shù)f(x)，假設(shè)滿足(x1)f'(x)0,那么必有〔〕Af(0)f(2)2f(1)Bf(0)f(2)2f(1)Cf(0)f(2)2f(1)Df(0)f(2)2f(1)的值為〔〕C.C11.函數(shù)yx3x2x的單調(diào)區(qū)間為_(kāi)_____________________________________________.13.曲線yx34x在點(diǎn)(1,3)處的切線傾斜角為_(kāi)_________________.n1fxaxbxc1)，且在x1處的切線方程是yx2，請(qǐng)解答以下問(wèn)題：18.函數(shù)f(x)ax3bx2(c3a2b)xd的圖象如以下圖.fx1〔III]在〔II〕的條件下，函數(shù)yf(x)與yf(x)5xm的圖象3有三個(gè)不同的交點(diǎn)，求m的取值范圍.19．函數(shù)f(x)ln(x1)k(x1)1．kxyfx3m求m的取值范圍.y參考答案AABCBACCDB12.(,0)1333y2nn22nxy2nn22nxoaan1&P三、解答題:15.解:設(shè)切點(diǎn)為P〔a15.解:設(shè)切點(diǎn)為P〔a,b〕，函數(shù)yx32n2nx6x切線的斜率ky'|xa3a26a3，得a1，代入到得b3，即P(1,3),y33(x1),3x16.解:設(shè)小正方形的邊長(zhǎng)為x厘米，那么盒子底面長(zhǎng)為82x,寬為V(82x)(52x)x4x26x40x32,令,得或103x25VV17.解：〔1〕f(x)ax4bx2f(x)4ax12bx,k1f(1)4a2b1,ax4bx2c55223.10f(x)-x423x0,或xxx922fd3d3得3a2bc3a2b0c0〔II〕依題f'(2)3且f(2)512a12a4b3a2b38a4b6a4b35所以f(x)x36x29x3?……〔8分〕〔III〕f(x)3x212x9?可轉(zhuǎn)化為：3x6x29x3x35xm有二gxxx3x2x4,223040x4,+增,43-減2+增當(dāng)且僅當(dāng)g當(dāng)且僅當(dāng)g27mm2xx???當(dāng)x(1,2)時(shí)，f(x)0,當(dāng)x(2,)時(shí),f(x)0,xx1 x1k1,■/kkf(x)0,kf(x)0,x(11kx 1 11,)時(shí),f(x)0,1...f(x)在(1,1-)內(nèi)是增函數(shù),kk1...f(x)的最大值是f(1—)Ink,k fx所以f(1)0,即3m6(m1)n0，所以n3m62111,00極大值單調(diào)遞減0m012m0xf(x)f(x)2故有上表知，當(dāng)m0時(shí)，f(x)在,1單調(diào)遞減,m2在(1,1)單調(diào)遞增，在(1,)上單調(diào)遞減m〔3〕由得f(x)3m，即mx22(m1)x202又m0所以x(m1)xm