高中數(shù)學(xué)第三章函數(shù)的應(yīng)用3.1函數(shù)與方程3.1.1方程的根與函數(shù)的零點(diǎn)練習(xí)新人教版.docx

3.1.1方程的根與函數(shù)的零點(diǎn)1.函數(shù)y=-x的零點(diǎn)是(D)(A)1 (B)-1(C)(1,0),(-1,0)(D)1,-1解析:由y=0,即-x=0,解得x=1或x=-1.所以函數(shù)的零點(diǎn)為1,-1.故 選D.2.二次函數(shù)f(x)=ax2+bx+c中,ac0,則該函數(shù)的零點(diǎn)個(gè)數(shù)是(B) (A)1(B)2(C)0(D)無(wú)法確定解析:因?yàn)閍c0,所以該函數(shù)有兩個(gè)零點(diǎn),故選B.3.已知定義在R上的函數(shù)f(x)的圖象是連續(xù)不斷的,且有對(duì)應(yīng)值 如表:x123f(x)3.42.6-3.7則函數(shù)f(x)一定存在零點(diǎn)的區(qū)間是(C)(A)(-,1)(B)(1,2)(C)(2,3) (D)(3,+)解析:若f(x)在a,b上連續(xù),且f(a)f(b)0,f(3)0所以f(x)在2,3上一定存在零點(diǎn). 選C.4.函數(shù)f(x)=ln x-x2+4x+5的零點(diǎn)個(gè)數(shù)為(C)(A)0(B)1(C)2(D)3解析:由數(shù)形結(jié)合可知函數(shù)y=ln x圖象與函數(shù)y=x2-4x-5圖象有2個(gè)交點(diǎn).所以函數(shù)f(x)有2個(gè)零點(diǎn).故C正確.5.函數(shù)f(x)=x2-ex的零點(diǎn)個(gè)數(shù)為(B)(A)0(B)1(C)2(D)3解析:函數(shù)f(x)=x2-ex的零點(diǎn)個(gè)數(shù),可轉(zhuǎn)化為函數(shù)y=x2和y=ex圖象的交點(diǎn)的個(gè)數(shù),在同一坐標(biāo)系中作出函數(shù)y=x2和y=ex的圖象,如圖所示,由圖象可知,當(dāng)x0時(shí),函數(shù)y=x2和y=ex的圖象只有一個(gè)交點(diǎn),當(dāng)x0時(shí),函數(shù)y=x2的圖象始終在函數(shù)y=ex的圖象的下方,沒(méi)有交點(diǎn),所以函數(shù)f(x)=x2-ex有且只有一個(gè)零點(diǎn),故選B.6.設(shè)函數(shù)f(x)=log2x+2x-3,則函數(shù)f(x)的零點(diǎn)所在的區(qū)間為(B)(A)(0,1)(B)(1,2)(C)(2,3)(D)(3,4)解析:因?yàn)楹瘮?shù)f(x)=log2x+2x-3,所以f(1)=log21+21-3=-10,所以在區(qū)間(1,2)內(nèi)函數(shù)存在零點(diǎn).故選B.7.函數(shù)f(x)=()x-x+2的零點(diǎn)所在的一個(gè)區(qū)間是(C)(A)(-1,0)(B)(0,1)(C)(2,3)(D)(1,2)解析:函數(shù)f(x)=()x-x+2,可得,f(-1)=50,f(0)=30,f(1)=0,f(2)=0,f(3)=-0,由零點(diǎn)存在定理可知,函數(shù)的零點(diǎn)在(2,3)內(nèi).故選C.8.已知x0是函數(shù)f(x)=2x+的一個(gè)零點(diǎn).若x1(1,x0),x2(x0,+),則(B)(A)f(x1)0,f(x2)0(B)f(x1)0(C)f(x1)0,f(x2)0,f(x2)0解析:因?yàn)閤0是函數(shù)f(x)=2x+的一個(gè)零點(diǎn),所以f(x0)=0,因?yàn)閒(x)=2x+是單調(diào)遞增函數(shù),且x1(1,x0),x2(x0,+),所以f(x1)f(x0)=0f(x2),故選B.9.已知函數(shù)f(x)=若方程f(x)-a=0有三個(gè)不同的實(shí)數(shù)根,則實(shí)數(shù)a的取值范圍為(A)(A)(0,1)(B)(0,2)(C)(0,3)(D)(1,3)解析:因?yàn)楹瘮?shù)f(x)=所以作出函數(shù)f(x)圖象,如圖所示,因?yàn)榉匠蘤(x)-a=0有三個(gè)不同的實(shí)數(shù)根,等價(jià)于函數(shù)y=f(x)的圖象與y=a有三個(gè)不同的交點(diǎn),根據(jù)圖象可知,當(dāng)0a1時(shí),函數(shù)y=f(x)的圖象與y=a有三個(gè)不同的交點(diǎn),方程f(x)-a=0有三個(gè)不同的實(shí)數(shù)根,a的取值范圍是(0,1),故選A.10.定義:區(qū)間x1,x2(x1x2)的長(zhǎng)度為x2-x1,已知函數(shù) y=2|x|的定義域?yàn)閍,b,值域?yàn)?,2,記區(qū)間a,b的最大長(zhǎng)度為m,最小長(zhǎng)度為n,則函數(shù) g(x)=mx-(x+2n)的零點(diǎn)個(gè)數(shù)是(B)(A)1(B)2(C)0(D)3解析:函數(shù)y=2|x|的值域?yàn)?,2,且函數(shù)在(-,0)單調(diào)遞減,在0,+)單調(diào)遞增,則0|x|1,即最大長(zhǎng)度m=2,最小長(zhǎng)度n=1.則函數(shù)g(x)=mx-(x+2n)=2x-(x+2),求零點(diǎn)個(gè)數(shù),可令y1=2x,y2=x+2,即兩函數(shù)圖象交點(diǎn)個(gè)數(shù).由圖象可看出共有兩個(gè)零點(diǎn).故選B.11.直線y=3與函數(shù)y=|x2-6x|圖象的交點(diǎn)個(gè)數(shù)為.解析:由y=|x2-6x|圖象如圖所示,則兩函數(shù)圖象交點(diǎn)個(gè)數(shù)為4.答案:412.已知函數(shù)f(x)=lg x+x-10的零點(diǎn)在區(qū)間(k,k+1)上,kZ,則k=.解析:由題意知函數(shù)f(x)為(0,+)上的增函數(shù).且f(9)=lg 9+9-10=lg 9-10,即f(9)f(10)0,且a1,則函數(shù)f(x)=ax+(x-1)2-2a的零點(diǎn)個(gè)數(shù)為.解析:設(shè)g(x)=2a-ax,h(x)=(x-1)2,注意到g(x)的圖象恒過(guò)定點(diǎn)(1,a),畫(huà)出它們的圖象,無(wú)論a1還是0a0時(shí),此函數(shù)圖象開(kāi)口向上,又f(0)=-10,結(jié)合二次函數(shù)圖象知符合題意;當(dāng)a0時(shí),此函數(shù)圖象開(kāi)口向下,又f(0)=-10,從而有即a=-.綜上可知,實(shí)數(shù)a的取值范圍為-0,+).16.已知函數(shù)f(x)=-log2x,若實(shí)數(shù)x0是方程f(x)=0的解,且0x1x0,則f(x1)(C)(A)恒為負(fù)值(B)等于0(C)恒為正值(D)不大于0解析:由f(x)=0,得()=log2x0,分別作出函數(shù)y=,y=log2x的圖象,由圖象可知,當(dāng)0x1log2x1,所以f(x1)=()-log2x10.故選C.17.已知函數(shù)f(x)=若函數(shù)g