07－08學年（1）本科《微積分》第一章統(tǒng)測試卷答案（演示）

1+x21+x2一．單項選擇題(每題2分，共20分)一．單項選擇題(每題2分，共20分)A．A?BB．A?BCAnBBDAnBBAnB=0[,3]sinx≤eeA．sinx≤eeC．周期函數(shù)D．單調(diào)函數(shù)A．奇函數(shù)B．偶函數(shù)C．單調(diào)函數(shù)D．有界函數(shù)1+2x+x211+2x+x21+x2=1+2x2=1+x2fxx，函數(shù)g(x)≥f(x)，A．f[f(x)]≥g[g(x)]B．f[f(x)]≤g[g(x)]C．f[f(x)]>g[g(x)]D．f[f(x)]<g[g(x)]f[f(x)]≤f[g(x)]≤g[g(x)]A．f[?(x)]，?[f(x)]都是奇函數(shù)B．f[?(x)]，?[f(x)]都是非奇非偶函數(shù)C．f[?(x)]，?[f(x)]都是偶函數(shù)D．f[?(x)]是偶函數(shù)，?[f(x)]是奇函數(shù)f[?(?x)]=f[?(x)]；?[f(?x)]=?[?f(x)]=?[f(x)]6．已知等式f(x+y)=f(x)+f(y)對于一切實數(shù)成立，則f(x)=。AA．奇函數(shù)B．偶函數(shù)C．非奇非偶D．既奇又偶y=?2x；f(?x)=f(x)+f(?2x)=f(x)+2f(?x)?f(x)+f(?x)=0 a2+b2 a2+b2A．(?3,?7)B．(?3,7)C．3(,?7)D．不存在系式：f(x+2)?f(x)=f2()成立，則當f(x)是以2期的周期函數(shù)時，必有f1()=。BT=2，f(x+2)=f(x)，f(x+2)?f(x)=f2()?f2()=0x=?1，f1()?f(?1)=0；f(x)奇函數(shù)?2f1()=010．設f(x)=?g(x)=?，則f[g(x)]=。Bx<0x≥0x<0x≥0x<0x≥0x<0x≥0x≥0，g(x)=x≥0?f[g(x)]=x+1x<0，g(x)=x2≥0?f[g(x)]=x2+1二、填空題(每題2分，共20分)x2+6x+11f(x+2)=f[(x+4)?2]=(x+4)2?2(x+4)+3fxexfgx)]=1?x2，則g(x)=。ln(1?x2)f[g(x)]=eln(1?x2)3．已知f(x)=?的反函數(shù)為?(x)，則?3()=。23=x2?1x=2(x>1)4．設f(x)=?和g(x)=?，則當xxx<0，g(x)=?x>0?f[g(x)]=(?x)2=x2xx?5≥x2≥0??≤x≤56．已知2f(x)+f1(?x)=ex，f(x)=。2(ex?e1?x)2(ex?e1?x)8．已知f(x)為偶函數(shù)，且x>0時，f(x)=x1(?x)，則x<0時，f(x)的表達式為。?x1(+x)f(?x)=f(x)；x<0??x>0，f(x)=f(?x)=(?x)[1?(?x)]=?x1(+x)9．設f(x)=ax5+bx3+cx+1，若f(?k)=2，則10．設函數(shù)f(x)=(ax+a?x)，g(x)=(ax?a?x)，其中a>1，已知f(x)g(y)+f(y)g(x)=g(u)，則u=。x+yf(x)g(y)+f(y)g(x+y=(ax+y+ay?x?ax?y?a?(x+y))++(ax+y?ay?x+ax?y?a?(x+y))=(ax+y?a?(x+y))三、計算題(每題7分，共56分)1．已知z=f2(x+y)+2y?4x，且當y=1時，z=4x2，fxxxfxxx?2，令u=2x+1，x=(u?1)，代入得，f(u)=u2?3故，f(x)=x2?3，z=2(x+y)2?3+2y?4x=4x2+4xy+y2?4x+2y?32．己知f(x)=ln(x+x2+1)，g(x)與f(x)的圖形對圖形對稱于直線y=x，所以g(x)是f(x)的反函數(shù)。nxey=+x，e?y=?x，則x=(ey?e?y)，故g(x)=(ex?e?x)?1?2x2x<?13．設f(x)=??x3?1≤x≤2，寫出f(x)反函數(shù)x>2yyxx，y=x3的反函2反函數(shù)為y=(x+16)(x>8)f(x)=g(x2)+g(x+3)的定義域。解：由條件得Dg(x):0<x≤4?x∈0(,4]Dg(x2):0<x2≤4?0<|x|≤2?x∈[?2,0)U0(,2]；Dg(x+3):0<x+3≤4??3<x≤1?x∈(?3,1] (1)f(?x)；(2)f[f(x)]。x≤0，f(x)=x2≥0?f[f(x)]=(x2)2+x2=x4+x2x>0，f(x)=x2+x≥0?f[f(x)]=(x2+x)2+(x2+x)f[f(x)]=?(1)f(x)的定義域；(2)f(lnx)的定義域。e7．討論f(x)=?的奇偶性。f(?x)=?=?f(x)奇函數(shù)或f(x)=(?x2+1)+4xf(?x)=[?(?x)2+1]+4(?x)=?[x(?x2+1)+4x]=?f(x)奇函數(shù)x8．設f(x)的定義域為0[,1]，求f(x+a)+f(x?a)∵a>0?1+a>1?a∴a≤x≤1?a欲使Df(x+a)nDf(x?a)≠Φ令a≤1?a?a≤0.5所以f(