考研数学三(一元函数微分学与一元函数积分学)模拟试卷2附答案解析

{“msg”:”返回成功”,”muban”:[{“ename”:”ksn3choose”,”cunt”:7,”grade”:”4.6″,”cname”:”选择题”,”shiti”:[{“groupCodePrimQuestion”:””,”primQuestion”:”设f(x)二阶连续可导,且[*]=-1,则( ).”,”discription”:”因为f(x)二阶连续可导,且[*]=-1,所以[*]f”(x)=0,即f”(0)=0.又\n[*]=-1<0,由极限的保号性,存在δ>0,当0<|x|<δ时,有[*]<0,即当x∈(-δ,0)时,f”(x)>0,当x∈(0,δ)时,f”(x)<0,所以(0,f(0))为曲线y=f(x)的拐点,选C.”,”firstPic”:””,”questionCode”:”id_ve03005j_ksn3choose_111_191119″,”subPrimPic”:””,”refAnswer”:”C”,”fourthPic”:””,”discPic”:”2019m11x/ct_ve03005j_ks3choose_111_191119a,2019m11x/ct_ve03005j_ks3choose_111_191119aa,2019m11x/ct_ve03005j_ks3choose_111_191119aaa,2019m11x/ct_ve03005j_ks3choose_111_191119aaaa”,”second”:”f(0)是f(x)的极大值”,”subjectTypeEname”:”ksn3choose”,”originalText”:””,”thirdPic”:””,”secondPic”:””,”third”:”(0,f(0))是曲线y=f(x)的拐点”,”fifthPic”:””,”audioFiles”:””,”children”:[],”primPic”:”2019m11x/ct_ve03005j_ks3choose_111_191119″,”fifth”:””,”fourth”:”x=0是f(x)的驻点但不是极值点”,”answerPic”:””,”first”:”f(0)是f(x)的极小值”},{“groupCodePrimQuestion”:””,”primQuestion”:”设函数f(x)满足关系f”(x)+f’2(x)=z,且f’(0)=0,则( ).”,”discription”:”由f’(0)=0得f”(0)=0,f’”(x)=1-2f’(x)f”(x),f’”(0)=1>0,由极限保号性,存在δ>0,当0<|x|<δ时,f’”(x)>0,再由f”(0)=0,得\n[*],故(0,f(0))是曲线y=f(x),选C.”,”firstPic”:””,”questionCode”:”id_ve03005j_ksn3choose_112_191119″,”subPrimPic”:””,”refAnswer”:”C”,”fourthPic”:””,”discPic”:”2019m11x/ct_ve03005j_ks3choose_112_191119″,”second”:”f(0)是f(x)的极大值”,”subjectTypeEname”:”ksn3choose”,”originalText”:””,”thirdPic”:””,”secondPic”:””,”third”:”(0,f(0))是y=f(x)的拐点”,”fifthPic”:””,”audioFiles”:””,”children”:[],”primPic”:””,”fifth”:””,”fourth”:”(0,f(0))不是y=f(x)的拐点”,”answerPic”:””,”first”:”f(0)是f(x)的极小值”},{“groupCodePrimQuestion”:””,”primQuestion”:”下列说法正确的是( ).”,”discription”:”令f’(x)=[*]f”(0)=0,但[*]f’(x)不存在,所以A不对;\n若最大值在端点取到则不是极大值,所以B不对;C显然不对,选D.”,”firstPic”:””,”questionCode”:”id_ve03005j_ksn3choose_113_191119″,”subPrimPic”:””,”refAnswer”:”D”,”fourthPic”:””,”discPic”:”2019m11x/ct_ve03005j_ks3choose_113_191119,2019m11x/ct_ve03005j_ks3choose_113_191119a”,”second”:”f(x)在[a,b]上的最大值一定是其极大值”,”subjectTypeEname”:”ksn3choose”,”originalText”:””,”thirdPic”:””,”secondPic”:””,”third”:”f(x)在(a,b)内的极大值一定是其最大值”,”fifthPic”:””,”audioFiles”:””,”children”:[],”primPic”:””,”fifth”:””,”fourth”:”若f(x)在[a,b]上连续,在(a,b)内可导,且f(x)在(a,b)内有唯一的极值点,则该极值点一定为最值点”,”answerPic”:””,”first”:”设f(x)在x0二阶可导,则f”(x)在x=x0处连续”},{“groupCodePrimQuestion”:””,”primQuestion”:”设f(x)在[a,+∞)上二阶可导,f(a)<0,f’(a)=0,且f”(x)≥k(k>0),则f(x)在(a,+∞)内的零点个数为( ).”,”discription”:”因为f’(a)=0,且f”(x)≥k(k>0),所以f(x)=f(a)+f’(a)(x-a)+\n[*](x-a)2≥f(a)+[*](x-a)2,其中ξ介于a与x之间.而[*]f(a)+[*](x-a)2=+∞,\n故[*]f(x)=+∞,再由f(a)<0得f(x)在(a,+∞)内至少有一个零点.又因为\nf’(a)=0,且f”(x)≥k(k>0),所以f’(x)>0(x>a),即f(x)在[a,+∞)单调增加,所以零点是唯一的,选B.”,”firstPic”:””,”questionCode”:”id_ve03005j_ksn3choose_114_191119″,”subPrimPic”:””,”refAnswer”:”B”,”fourthPic”:””,”discPic”:”2019m11x/ct_ve03005j_ks3choose_114_191119,2019m11x/ct_ve03005j_ks3choose_114_191119a,2019m11x/ct_ve03005j_ks3choose_114_191119aa,2019m11x/ct_ve03005j_ks3choose_114_191119b,2019m11x/ct_ve03005j_ks3choose_114_191119ba”,”second”:”1个”,”subjectTypeEname”:”ksn3choose”,”originalText”:””,”thirdPic”:””,”secondPic”:””,”third”:”2个”,”fifthPic”:””,”audioFiles”:””,”children”:[],”primPic”:””,”fifth”:””,”fourth”:”3个”,”answerPic”:””,”first”:”0个”},{“groupCodePrimQuestion”:””,”primQuestion”:”设k>0,则函数f(x)=lnx-[*]+k的零点个数为( ).”,”discription”:”函数f(x)的定义域为(0,+∞),由f’(x)=[*]=0得x=e,当0<x<e时,\nf’(x)>0;当x>e时,f’(x)<0,由驻点的唯一性知x=e为函数f(x)的最大值点,\n最大值为f(e)=k>0,又[*]f(x)=-∞,[*]f(x)=-∞,于是f(x)在(0,+∞)内有且仅有两个零点,选C.”,”firstPic”:””,”questionCode”:”id_ve03005j_ksn3choose_115_191119″,”subPrimPic”:””,”refAnswer”:”C”,”fourthPic”:””,”discPic”:”2019m11x/ct_ve03005j_ks3choose_115_191119a,2019m11x/ct_ve03005j_ks3choose_115_191119aa,2019m11x/ct_ve03005j_ks3choose_115_191119aaa”,”second”:”1个”,”subjectTypeEname”:”ksn3choose”,”originalText”:””,”thirdPic”:””,”secondPic”:””,”third”:”2个”,”fifthPic”:””,”audioFiles”:””,”children”:[],”primPic”:”2019m11x/ct_ve03005j_ks3choose_115_191119″,”fifth”:””,”fourth”:”3个”,”answerPic”:””,”first”:”0个”},{“groupCodePrimQuestion”:””,”primQuestion”:”曲线y=[*]的渐近线的条数为( ).”,”discription”:”因为[*]y=∞,所以曲线y=[*]无水平渐近线;\n由[*]y=∞,[*]y=+∞,得曲线y=[*]有两条铅直渐近线;\n由[*]=1,[*](y-x)=0,得曲线y=[*]有一条斜渐近线y=x,选D.”,”firstPic”:””,”questionCode”:”id_ve03005j_ksn3choose_116_191119″,”subPrimPic”:””,”refAnswer”:”D”,”fourthPic”:””,”discPic”:”2019m11x/ct_ve03005j_ks3choose_116_191119a,2019m11x/ct_ve03005j_ks3choose_116_191119b,2019m11x/ct_ve03005j_ks3choose_116_191119c,2019m11x/ct_ve03005j_ks3choose_116_191119d,2019m11x/ct_ve03005j_ks3choose_116_191119e,2019m11x/ct_ve03005j_ks3choose_116_191119f,2019m11x/ct_ve03005j_ks3choose_116_191119j,2019m11x/ct_ve03005j_ks3choose_116_191119h”,”second”:”1条”,”subjectTypeEname”:”ksn3choose”,”originalText”:””,”thirdPic”:””,”secondPic”:””,”third”:”2条”,”fifthPic”:””,”audioFiles”:””,”children”:[],”primPic”:”2019m11x/ct_ve03005j_ks3choose_116_191119″,”fifth”:””,”fourth”:”3条”,”answerPic”:””,”first”:”0条”},{“groupCodePrimQuestion”:””,”primQuestion”:”设函数f(x)在(-∞,+∞)内连续,其导数的图形如右图,则f(x)有( ).\n[*]”,”discription”:”设当x<0时,f’(x)与x轴的两个交点为(x1,0),(x2,0),其中x1<x2;\n当x>0时,f’(x)与x轴的两个交点为(x3,0),(x4,0),其中x3<x4.\n当x<x1时,f’(x)>0,当x∈(x1,x2)时,f’(x)<0,则x=x1为f(x)的极大值点;\n当x∈(x2,0)时,f’(x)>0,则x=x2为f(x)的极小值点;当x∈(0,x3)时,f’(x)<0,则x=0为f(x)的极大值点;\n当x∈(x3,x4)时,f’(x)>0,则x=x3为f(x)的极小值点;\n当x>x4时,f’(x)<0,则x=x4为f(x)的极大值点,即f(x)有三个极大值点,两个极小值点,又f”(x)有两个零点,根据一阶导数在两个零点两侧的增减性可得,y=f(x)有两个拐点,选C.”,”firstPic”:””,”questionCode”:”id_ve03005j_ksn3choose_117_191119″,”subPrimPic”:””,”refAnswer”:”C”,”fourthPic”:””,”discPic”:””,”second”:”两个极大值点,两个极小值点,两个拐点”,”subjectTypeEname”:”ksn3choose”,”originalText”:””,”thirdPic”:””,”secondPic”:””,”third”:”三个极大值点,两个极小值点,两个拐点”,”fifthPic”:””,”audioFiles”:””,”children”:[],”primPic”:”2019m11x/ct_ve03005j_ks3choose_117_191119″,”fifth”:””,”fourth”:”两个极大值点,三个极小值点,两个拐点”,”answerPic”:””,”first”:”两个极大值点,两个极小值点,一个拐点”}],”basic”:”1″},{“ename”:”ksn3cloze”,”cunt”:8,”grade”:”3.0″,”cname”:”填空题”,”shiti”:[{“groupCodePrimQuestion”:””,”primQuestion”:”设f(sin2x)=[*],则[*]=_____________.”,”discription”:”由f(sin2x)=[*],得f(x)=[*],\n于是[*]dx[*]”,”firstPic”:””,”questionCode”:”id_ve03005j_ksn3cloze_190_191119″,”subPrimPic”:””,”refAnswer”:”arcsin2[*]+C”,”fourthPic”:””,”discPic”:”2019m11x/ct_ve03005j_ks3cloze_190_191119aa,2019m11x/ct_ve03005j_ks3cloze_190_191119b,2019m11x/ct_ve03005j_ks3cloze_190_191119c,2019m11x/ct_ve03005j_ks3cloze_190_191119d”,”second”:””,”subjectTypeEname”:”ksn3cloze”,”originalText”:””,”thirdPic”:””,”secondPic”:””,”third”:””,”fifthPic”:””,”audioFiles”:””,”children”:[],”primPic”:”2019m11x/ct_ve03005j_ks3cloze_190_191119,2019m11x/ct_ve03005j_ks3cloze_190_191119a”,”fifth”:””,”fourth”:””,”answerPic”:”2019m11x/ct_ve03005j_ks3cloze_190_191119da”,”first”:””},{“groupCodePrimQuestion”:””,”primQuestion”:”设f(lnx)=[*],则[*]f(x)dx=_____________.”,”discription”:”由f(lnx)=[*]得f(x)=[*],\n则[*]f(x)dx=[*]dx=-[*]ln(1+ex)d(e-x)\n=-e-xln(1+ex)+[*]dx=-[*]dx\n=-[*]d(1+e-x)=-[*]-ln(1+e-x)+C.”,”firstPic”:””,”questionCode”:”id_ve03005j_ksn3cloze_191_191119″,”subPrimPic”:””,”refAnswer”:”﹣[*]-ln(1+e-x)+C”,”fourthPic”:””,”discPic”:”2019m11x/ct_ve03005j_ks3cloze_191_191119aa,2019m11x/ct_ve03005j_ks3cloze_191_191119b,2019m11x/ct_ve03005j_ks3cloze_191_191119c,2019m11x/ct_ve03005j_ks3cloze_191_191119d,2019m11x/ct_ve03005j_ks3cloze_191_191119e,2019m11x/ct_ve03005j_ks3cloze_191_191119f,2019m11x/ct_ve03005j_ks3cloze_191_191119g,2019m11x/ct_ve03005j_ks3cloze_191_191119h,2019m11x/ct_ve03005j_ks3cloze_191_191119i”,”second”:””,”subjectTypeEname”:”ksn3cloze”,”originalText”:””,”thirdPic”:””,”secondPic”:””,”third”:””,”fifthPic”:””,”audioFiles”:””,”children”:[],”primPic”:”2019m11x/ct_ve03005j_ks3cloze_191_191119,2019m11x/ct_ve03005j_ks3cloze_191_191119a”,”fifth”:””,”fourth”:””,”answerPic”:”2019m11x/ct_ve03005j_ks3cloze_191_191119ia”,”first”:””},{“groupCodePrimQuestion”:””,”primQuestion”:”设[*]xf(x)dx=arcsinx+C,则[*]=_____________.”,”discription”:”由[*]xf(x)dx=arcsinx+C得xf(x)=[*],所以f(x)=[*],\n[*]”,”firstPic”:””,”questionCode”:”id_ve03005j_ksn3cloze_192_191119″,”subPrimPic”:””,”refAnswer”:”[*]”,”fourthPic”:””,”discPic”:”2019m11x/ct_ve03005j_ks3cloze_192_191119aa,2019m11x/ct_ve03005j_ks3cloze_192_191119b,2019m11x/ct_ve03005j_ks3cloze_192_191119c,2019m11x/ct_ve03005j_ks3cloze_192_191119d”,”second”:””,”subjectTypeEname”:”ksn3cloze”,”originalText”:””,”thirdPic”:””,”secondPic”:””,”third”:””,”fifthPic”:””,”audioFiles”:””,”children”:[],”primPic”:”2019m11x/ct_ve03005j_ks3cloze_192_191119,2019m11x/ct_ve03005j_ks3cloze_192_191119a”,”fifth”:””,”fourth”:””,”answerPic”:”2019m11x/ct_ve03005j_ks3cloze_192_191119da”,”first”:””},{“groupCodePrimQuestion”:””,”primQuestion”:”设f(x)为连续函数,且满足[*]f(xt)dt=f(x)+xsinx,则f(x)=_____________.”,”discription”:”由[*]f(xt)dt=f(x)+xsinx,得[*]f(xt)d(xt)=xf(x)+x2sinx,即\n[*]f(t)dt=xf(x)+x2sinx,两边求导得f’(x)=-2sinx-xcosx,积分得\nf(x)=cosx-xsinx+C.”,”firstPic”:””,”questionCode”:”id_ve03005j_ksn3cloze_193_191119″,”subPrimPic”:””,”refAnswer”:”cosx-xsinx+C”,”fourthPic”:””,”discPic”:”2019m11x/ct_ve03005j_ks3cloze_193_191119a,2019m11x/ct_ve03005j_ks3cloze_193_191119b,2019m11x/ct_ve03005j_ks3cloze_193_191119ba”,”second”:””,”subjectTypeEname”:”ksn3cloze”,”originalText”:””,”thirdPic”:””,”secondPic”:””,”third”:””,”fifthPic”:””,”audioFiles”:””,”children”:[],”primPic”:”2019m11x/ct_ve03005j_ks3cloze_193_191119″,”fifth”:””,”fourth”:””,”answerPic”:””,”first”:””},{“groupCodePrimQuestion”:””,”primQuestion”:”[*]dx=_____________.”,”discription”:”[*]”,”firstPic”:””,”questionCode”:”id_ve03005j_ksn3cloze_194_191119″,”subPrimPic”:””,”refAnswer”:”[*]”,”fourthPic”:””,”discPic”:”2019m11x/ct_ve03005j_ks3cloze_194_191119a”,”second”:””,”subjectTypeEname”:”ksn3cloze”,”originalText”:””,”thirdPic”:””,”secondPic”:””,”third”:””,”fifthPic”:””,”audioFiles”:””,”children”:[],”primPic”:”2019m11x/ct_ve03005j_ks3cloze_194_191119″,”fifth”:””,”fourth”:””,”answerPic”:”2019m11x/ct_ve03005j_ks3cloze_194_191119aa”,”first”:””},{“groupCodePrimQuestion”:””,”primQuestion”:”[*]arctan[*]dx=_____________.”,”discription”:”[*]xarctan[*]dx[*]arctantdt=[*]arctantd(t4)\n=[*]t4arctant-[*]dt=[*]t4arctant-[*]dt\n=[*]t4arctant-[*](t2-1+[*])dt\n=[*]t4arctant-[*]([*]t3-t+arctant)+C\n=[*](x2-1)arctan[*]+C”,”firstPic”:””,”questionCode”:”id_ve03005j_ksn3cloze_195_191119″,”subPrimPic”:””,”refAnswer”:”[*](x2-1)arctan[*]+C”,”fourthPic”:””,”discPic”:”2019m11x/ct_ve03005ks195_v,2019m11x/ct_ve03005ks195_b,2019m11x/ct_ve03005ks395_c,2019m11x/ct_ve03005ks195_d,2019m11x/ct_ve03005ks195_e,2019m11x/ct_ve03005ks195_f,2019m11x/ct_ve03005ks195_g,2019m11x/ct_ve03005ks195_h,2019m11x/ct_ve03005ks195_i,2019m11x/ct_ve03005ks195_j,2019m11x/ct_ve03005ks195_k,2019m11x/ct_ve03005ks195_l,2019m11x/ct_ve03005ks195_m,2019m11x/ct_ve03005ks195_n,2019m11x/ct_ve03005ks195_o,2019m11x/ct_ve03005ks195_p”,”second”:””,”subjectTypeEname”:”ksn3cloze”,”originalText”:””,”thirdPic”:””,”secondPic”:””,”third”:””,”fifthPic”:””,”audioFiles”:””,”children”:[],”primPic”:”2019m11x/ct_ve03005j_ks3cloze_195_191119,2019m11x/ct_ve03005j_ks3cloze_195_191119a”,”fifth”:””,”fourth”:””,”answerPic”:”2019m11x/ct_ve03005j_ks3cloze_195_191119pa,2019m11x/ct_ve03005j_ks3cloze_195_191119paa”,”first”:””},{“groupCodePrimQuestion”:””,”primQuestion”:”[*]arcsinxdx=_____________.”,”discription”:”[*]arcsindx[*]tcos2tdt=[*]t(1+cos2t)dt\n=[*]td(sin2t)=[*](tsin2t-[*]sin2tdt)\n=[*]t2+[*]tsin2t+[*]cos2t+C\n=[*]arcsin2x+[*]\nx[*]arcsinx-[*]x2+C.”,”firstPic”:””,”questionCode”:”id_ve03005j_ksn3cloze_196_191119″,”subPrimPic”:””,”refAnswer”:”[*]arcsin2x+[*]arcsinx-[*]x2+C”,”fourthPic”:””,”discPic”:”2019m11x/ct_ve03005j_ks3clo196_191119a,2019m11x/ct_ve03005j_ks3clo196_191119b,2019m11x/ct_ve03005j_ks3clo196_191119c,2019m11x/ct_ve03005j_ks3clo196_191119d,2019m11x/ct_ve03005j_ks3cl196_191119e,2019m11x/ct_ve03005j_ks3cl196_191119f,2019m11x/ct_ve03005j_ks3cl196_191119j,2019m11x/ct_ve03005j_ks3cl196_191119h,2019m11x/ct_ve03005j_ks3cl196_191119i,2019m11x/ct_ve03005j_ks3cl196_191119g,2019m11x/ct_ve03005j_ks3cl196_191119k,2019m11x/ct_ve03005j_ks3cl196_191119l,2019m11x/ct_ve03005j_ks3cl196_191119m”,”second”:””,”subjectTypeEname”:”ksn3cloze”,”originalText”:””,”thirdPic”:””,”secondPic”:””,”third”:””,”fifthPic”:””,”audioFiles”:””,”children”:[],”primPic”:”2019m11x/ct_ve03005j_ks3cloze_196_191119″,”fifth”:””,”fourth”:””,”answerPic”:”2019m11x/ct_ve03005j_ks3cloze_196_191119ma,2019m11x/ct_ve03005j_ks3cloze_196_191119ba,2019m11x/ct_ve03005j_ks3cloze_196_191119cc”,”first”:””},{“groupCodePrimQuestion”:””,”primQuestion”:”[*]dx=_____________.”,”discription”:”[*]dx=2[*]lnxd([*])=2[*]lnx-2[*]dx,\n因为[*],\n所以[*].”,”firstPic”:””,”questionCode”:”id_ve03005j_ksn3cloze_197_191119″,”subPrimPic”:””,”refAnswer”:”2[*]lnx-4[*]+C”,”fourthPic”:””,”discPic”:”2019m11x/ct_ve03005j_ksn3cloze_197_191119a,2019m11x/ct_ve03005j_ksn3cloze_197_191119b,2019m11x/ct_ve03005j_ksn3cloze_197_191119c,2019m11x/ct_ve03005j_ksn3cloze_197_191119d,2019m11x/ct_ve03005j_ksn3cloze_197_191119e,2019m11x/ct_ve03005j_ksn3cloze_197_191119f,2019m11x/ct_ve03005j_ksn3cloze_197_191119g”,”second”:””,”subjectTypeEname”:”ksn3cloze”,”originalText”:””,”thirdPic”:””,”secondPic”:””,”third”:””,”fifthPic”:””,”audioFiles”:””,”children”:[],”primPic”:”2019m11x/ct_ve03005j_ks3cloze_197_191119″,”fifth”:””,”fourth”:””,”answerPic”:”2019m11x/ct_ve03005j_ks3cloze_197_191119da,2019m11x/ct_ve03005j_ks3cloze_197_191119db”,”first”:””}],”basic”:”3″},{“ename”:”ksn3shot”,”cunt”:9,”grade”:”10.4″,”cname”:”解答题”,”shiti”:[{“groupCodePrimQuestion”:””,”primQuestion”:”设f(x)在[0,2]上连续,在(0,2)内二阶可导,且[*]=0,又f(2)=2[*]f(x)dx,证明:存在ξ∈(0,2),使得f’(ξ)+f”(ξ)=0.”,”discription”:””,”firstPic”:””,”questionCode”:”id_ve03005j_ksn3shot_128_191119″,”subPrimPic”:””,”refAnswer”:”由[*]=0,得f(1)=-1,\n又[*],所以f’(1)=0.\n由积分中值定理得f(2)=2[*]f(x)dx=f(c),其中c∈[1,[*]],\n由罗尔定理,存在X0∈(c,2)[*](1,2),使得f’(x0)=0.\n令φ(x)=exf’(x),则φ(1)=φ(x0)=0,\n由罗尔定理,存在ξ∈(1,x0)[*](0,2),使得φ’(ξ)=0,\n而φ’(x)=ex[f’(x)+f”(x)]且ex≠0,所以f’(ξ)+f”(ξ)=0.”,”fourthPic”:””,”discPic”:””,”second”:””,”subjectTypeEname”:”ksn3shot”,”originalText”:””,”thirdPic”:””,”secondPic”:””,”third”:””,”fifthPic”:””,”audioFiles”:””,”children”:[],”primPic”:”2019m11x/ct_ve03005j_ks3shot_128_191119,2019m11x/ct_ve03005j_ks3shot_128_191119a”,”fifth”:””,”fourth”:””,”answerPic”:”2019m11x/ct_ve03005j_ks3shot_128_191119b,2019m11x/ct_ve03005j_ks3shot_128_191119c,2019m11x/ct_ve03005j_ks3shot_128_191119d,2019m11x/ct_ve03005j_ks3shot_128_191119e,2019m11x/ct_ve03005j_ks3shot_128_191119f,2019m11x/ct_ve03005j_ks3shot_128_191119g”,”first”:””},{“groupCodePrimQuestion”:””,”primQuestion”:”设f(x)在[0,1]上可导,f(0)=0,|f’(x)|≤[*]|f(x)|.证明:f(x)=0,x∈[0,1].”,”discription”:””,”firstPic”:””,”questionCode”:”id_ve03005j_ksn3shot_129_191119″,”subPrimPic”:””,”refAnswer”:”因为f(x)在[0,1]上可导,所以f(x)在[0,1]上连续,从而|f(x)|在[0,1]上\n连续,故|f(x)|在[0,1]上取到最大值M,即存在x0∈[0,1],使得|f(x)0|=M.\n当x0=0时,则M=0,所以f(x)[*]0,x∈[0,1];\n当x0≠0时,M=|f(x0)|=|f(x0)-f(0)|=|f’(ξ)|x0≤|f’(ξ)|≤[*]|f(ξ)|≤[*],其中ξ∈(0,x0),故M=0,于是f(x)[*]0,x∈[0,1].”,”fourthPic”:””,”discPic”:””,”second”:””,”subjectTypeEname”:”ksn3shot”,”originalText”:””,”thirdPic”:””,”secondPic”:””,”third”:””,”fifthPic”:””,”audioFiles”:””,”children”:[],”primPic”:”2019m11x/ct_ve03005j_ks3shot_129_191119″,”fifth”:””,”fourth”:””,”answerPic”:”2019m11x/ct_ve03005j_ksn3shot_129_191119a,2019m11x/ct_ve03005j_ksn3shot_129_191119aa,2019m11x/ct_ve03005j_ksn3shot_129_191119aaa,2019m11x/ct_ve03005j_ksn3shot_129_191119b”,”first”:””},{“groupCodePrimQuestion”:””,”primQuestion”:”设f(x)∈C[a,b],在(a,b)内可导,f(a)=f(b)=1.证明:存在ξ,η∈(a,b),使得[*]=(ea+eb)[f’(η)+f(η)].”,”discription”:””,”firstPic”:””,”questionCode”:”id_ve03005j_ksn3shot_130_191119″,”subPrimPic”:””,”refAnswer”:”令φ(x)=exf(x),由微分中值定理,存在η∈(a,b),使得\n[*]=eη[f’(η)+f(η)],再由f(a)=f(b)=1,得[*]=eη[f’(η)+f(η)], 从而[*]=(ea+eb)eη[f’(η)+f(η)],令φ(x)=e2x,由微分中值定理,存在ξ∈(a,b),使[*]=2e, 即2e=(ea+eb)eη[f’(η)+f(η)],或2e2ξ-η=(ea+eb)[f’(η)+f(η)].”,”fourthPic”:””,”discPic”:””,”second”:””,”subjectTypeEname”:”ksn3shot”,”originalText”:””,”thirdPic”:””,”secondPic”:””,”third”:””,”fifthPic”:””,”audioFiles”:””,”children”:[],”primPic”:”2019m11x/ct_ve03005j_ks3shot_130_191119″,”fifth”:””,”fourth”:””,”answerPic”:”2019m11x/ct_ve03005j_ks3shot_130_191119a,2019m11x/ct_ve03005j_ks3shot_130_191119aa,2019m11x/ct_ve03005j_ks3shot_130_191119aaa,2019m11x/ct_ve03005j_ks3shot_130_191119b”,”first”:””},{“groupCodePrimQuestion”:””,”primQuestion”:”设f(x)二阶可导,f(0)=f(1)=0且[*]f(x)=-1.证明:存在ξ∈(0,1),使得f”(ξ)≥8.”,”discription”:””,”firstPic”:””,”questionCode”:”id_ve03005j_ksn3shot_131_191119″,”subPrimPic”:””,”refAnswer”:”因为f(x)在[0,1]上二阶可导,所以f(x)在[0,1]上连续且f(0)=f(1)=0,\n[*]f(x)=-1,由闭区间上连续函数最值定理知,f(x)在[0,1]取到最小值且最小值在(0,1)内达到,即存在c∈(0,1),使得f(c)=-1,再由费马定理知f’(c)=0,根据泰勒公式f(0)=f(c)+f’(c)(0-c)+[*](0-c)2,ξ1∈(0,c)f(1)=f(c)+f’(c)(1-c)+[*](1-c)2,ξ2∈(c,1) 整理得 f”(ξ1)=[*],f”(ξ2)=[*].当c∈(0,[*]]时,f”(ξ1)=[*]≥8,取ξ=ξ1;当c∈([*],1)时,f”(ξ2)=[*]≥8,取ξ=ξ2.所以存在ξ∈(0,1),使得f”(ξ)≥8.”,”fourthPic”:””,”discPic”:””,”second”:””,”subjectTypeEname”:”ksn3shot”,”originalText”:””,”thirdPic”:””,”secondPic”:””,”third”:””,”fifthPic”:””,”audioFiles”:””,”children”:[],”primPic”:”2019m11x/ct_ve03005j_ks3shot_131_191119″,”fifth”:””,”fourth”:””,”answerPic”:”2019m11x/ct_ve03005j_ks3shot_131_191119a,2019m11x/ct_ve03005j_ks3shot_131_191119v,2019m11x/ct_ve03005j_ks3shot_131_191119c,2019m11x/ct_ve03005j_ks3shot_131_191119d,2019m11x/ct_ve03005j_ks3shot_131_191119e,2019m11x/ct_ve03005j_ks3shot_131_191119f,2019m11x/ct_ve03005j_ks3shot_131_191119g,2019m11x/ct_ve03005j_ks3shot_131_191119h,2019m11x/ct_ve03005j_ks3shot_131_191119i”,”first”:””},{“groupCodePrimQuestion”:””,”primQuestion”:”一质点从时间t=0开始直线运动,移动了单位距离使用了单位时间,且初速度和末速度都为零.证明:在运动过程中存在某个时刻点,其加速度绝对值不小于4.”,”discription”:””,”firstPic”:””,”questionCode”:”id_ve03005j_ksn3shot_132_191119″,”subPrimPic”:””,”refAnswer”:”设运动规律为S=S(t),显然S(0)=0,S’(0)=0,S(1)=1,S’(1)=0.由泰勒公式\nS([*])=S(0)+[*],ξ1∈(0,[*]),\nS([*])=S(1)+[*],ξ2∈([*],1),\n两式相减,得S”(ξ2)-S”(ξ1)=-8[*]|S”(ξ1)|+|S”(ξ2)|≥8.\n当|S”(ξ1)|≥|S”(ξ2)|时,|S”(ξ1)|≥4;当|S”(ξ1)|<|S”(ξ2)|时,|S”(ξ2)|≥4.”,”fourthPic”:””,”discPic”:””,”second”:””,”subjectTypeEname”:”ksn3shot”,”originalText”:””,”thirdPic”:””,”secondPic”:””,”third”:””,”fifthPic”:””,”audioFiles”:””,”children”:[],”primPic”:””,”fifth”:””,”fourth”:””,”answerPic”:”2019m11x/ct_ve03005j_ks3shot_132_191119,2019m11x/ct_ve03005j_ks3shot_132_191119a,2019m11x/ct_ve03005j_ks3shot_132_191119b,2019m11x/ct_ve03005j_ks3shot_132_191119c,2019m11x/ct_ve03005j_ks3shot_132_191119d,2019m11x/ct_ve03005j_ks3shot_132_191119e,2019m11x/ct_ve03005j_ks3shot_132_191119f”,”first”:””},{“groupCodePrimQuestion”:””,”primQuestion”:”设f(x)在[0,1]_k–阶可导,且|f”(x)|≤1(x∈[0,1]),又f(0)=f(1),证明:\n|f’(x)|≤[*](x∈[0,1]).”,”discription”:””,”firstPic”:””,”questionCode”:”id_ve03005j_ksn3shot_133_191119″,”subPrimPic”:””,”refAnswer”:”由泰勒公式得\nf(0)=f(x)-f’(x)x+[*]f”(ξ1)x2,ξ1∈(0,x),\nf(1)=f(x)+f’(x)(1-x)+[*]f”(ξ2)(1-x)2,ξ2∈(x,1),\n两式相减,得f’(x)=[*]f”(ξ1)x2-[*]f”(ξ2)(1-x)2.\n两边取绝对值,再由|f”(x)|≤1,得|f’(x)|≤[*][x2+(1-x)2]=(x-[*])2+[*].”,”fourthPic”:””,”discPic”:””,”second”:””,”subjectTypeEname”:”ksn3shot”,”originalText”:””,”thirdPic”:””,”secondPic”:””,”third”:””,”fifthPic”:””,”audioFiles”:””,”children”:[],”primPic”:”2019m11x/ct_ve03005j_ks3shot_133_191119″,”fifth”:””,”fourth”:””,”answerPic”:”2019m11x/ct_ve03005j_ks3shot_133_191119a,2019m11x/ct_ve03005j_ks3shot_133_191119b,2019m11x/ct_ve03005j_ks3shot_133_191119c,2019m11x/ct_ve03005j_ks3shot_133_191119d,2019m11x/ct_ve03005j_ks3shot_133_191119e,2019m11x/ct_ve03005j_ks3shot_133_191119f,2019m11x/ct_ve03005j_ks3shot_133_191119g”,”first”:””},{“groupCodePrimQuestion”:”设f(x)在(-1,1)内二阶连续可导,且f”(x)≠0.证明:”,”primQuestion”:”设f(x)在(-1,1)内二阶连续可导,且f”(x)≠0.证明:”,”discription”:””,”firstPic”:””,”questionCode”:”id_ve03005j_ksn3shot_134_191119″,”subPrimPic”:””,”refAnswer”:””,”fourthPic”:””,”discPic”:””,”secondQuestion”:””,”second”:””,”subjectTypeEname”:”ksn3shot”,”originalText”:””,”thirdPic”:””,”secondPic”:””,”third”:””,”fifthPic”:””,”audioFiles”:””,”children”:[{“primQuestion”:””,”discription”:””,”firstPic”:””,”questionCode”:”id_ve03005j_ksn3shot_134_191119_*001″,”subPrimPic”:””,”refAnswer”:”对任意x∈(-1,1),根据微分中值定理,得\nf(x)=f(0)+xf’[Θ(x)x],其中0<Θ(x)<1.\n因为f”(x)∈C(-1,1)且f”(x)≠0,所以f”(x)在(-1,1)内保号,不妨设f”(x)>0,\n则f’(x)在(-1,1)内单调增加,又由于x≠0,所以Θ(x)是唯一的.”,”fourthPic”:””,”discPic”:””,”secondQuestion”:”对(-1,1)内任一点x≠0,存在唯一的θ(x)∈(0,1),使得f(x)=f(0)+xf’[θ(x)x];”,”second”:””,”subjectTypeEname”:”ksn3shot”,”originalText”:””,”thirdPic”:””,”secondPic”:””,”third”:””,”fifthPic”:””,”audioFiles”:””,”primPic”:””,”fifth”:””,”fourth”:””,”answerPic”:””,”first”:””},{“primQuestion”:””,”discription”:””,”firstPic”:””,”questionCode”:”id_ve03005j_ksn3shot_134_191119_*002″,”subPrimPic”:”2019m11x/ct_ve03005j_ks3shot_134_191119,2019m11x/ct_ve03005j_ks3shot_134_191119a”,”refAnswer”:”由泰勒公式,得\nf(x)=f(0)+f’(0)x+[*]x2,其中ξ介于0与x之间,\n而f(x)=f(0)+xf’[θ(x)x],所以有\nf’[Θ(x)x]=f’(0)+[*]·Θ(x)=[*],\n令x→0,再由二阶导数的连续性及非零性,得[*](x)=[*].”,”fourthPic”:””,”discPic”:””,”secondQuestion”:”[*]θ(x)=[*]”,”second”:””,”subjectTypeEname”:”ksn3shot”,”originalText”:””,”thirdPic”:””,”secondPic”:””,”third”:””,”fifthPic”:””,”audioFiles”:””,”primPic”:””,”fifth”:””,”fourth”:””,”answerPic”:”2019m11x/ct_ve03005j_ks3shot_134_191119aa,2019m11x/ct_ve03005j_ks3shot_134_191119b,2019m11x/ct_ve03005j_ks3shot_134_191119c,2019m11x/ct_ve03005j_ks3shot_134_191119d,2019m11x/ct_ve03005j_ks3shot_134_191119e”,”first”:””}],”primPic”:””,”fifth”:””,”fourth”:””,”answerPic”:””,”first”:””},{“groupCodePrimQuestion”:””,”primQuestion”:”设f(x)在[a,b]_上二阶可导,且f’(a)=f’(b)=0.证明:存在ξ∈(a,b),使得|f”(ξ)|≥[*]|f(b)-f(a)|·”,”discription”:””,”firstPic”:””,”questionCode”:”id_ve03005j_ksn3shot_137_191119″,”subPrimPic”:””,”refAnswer”:”由泰勒公式得\nf([*])=f(a)+f’(a)([*]-a)+[*]([*]-a)2,ξ1∈(a,[*]),\nf([*])=f(b)+f’(b)([*]-b)+[*]([*]-b)2,ξ2∈([*],b),\n即f([*])=f(a)+[*]f”(ξ1),f([*])=f(b)+[*]f”(ξ2),\n两式相减得f(b)-f(a)=[*][f”(ξ1)-f”(ξ2)],\n取绝对值得|f(b)-f(a)|≤[*][ |f”(ξ1)|+|f”(ξ2)|].\n(1)当f”(ξ1)|≥|f”(ξ2)|时,取ξ=ξ1,则有|f”(ξ)|≥[*]|f(b)-f(a)|;\n(2)当|f”(ξ1)|<|f”(ξ2)|时,取ξ=ξ2,则有|f”(ξ)|≥[*]|f(b)-f(a)|.”,”fourthPic”:””,”discPic”:””,”second”:””,”subjectTypeEname”:”ksn3shot”,”originalText”:””,”thirdPic”:””,”secondPic”:””,”third”:””,”fifthPic”:””,”audioFiles”:””,”children”:[],”primPic”:”2019m11x/ct_ve03005j_ks3shot_137_191119″,”fifth”:””,”fourth”:””,”answerPic”:”2019m11x/ct_ve03005j_137a,2019m11x/ct_ve03005j_137b,2019m11x/ct_ve03005j_137c,2019m11x/ct_ve03005j_k_137d,2019m11x/ct_ve03005j_k_137e,2019m11x/ct_ve03005j_k_137f,2019m11x/ct_ve03005j_k_137g,2019m11x/ct_ve03005j_k_137h,2019m11x/ct_ve03005j_k_137i,2019m11x/ct_ve03005j_k_137j,2019m11x/ct_ve03005j_k_137k,2019m11x/ct_ve03005j_k_137l,2019m11x/ct_ve03005j_k_137m,2019m11x/ct_ve03005j_k_137n,2019m11x/ct_ve03005j_k_137o,2019m11x/ct_ve03005j_k_137p,2019m11x/ct_ve03005j_k_137q,2019m11x/ct_ve03005j_k_137r”,”first”:””}],”basic”:”4″}],”examTypeCode”:”ve03005″,”code”:1,”examstyle”:”3″,”examName”:”考研数学三(一元函数微分学与一元函数积分学)模拟试卷2″,”count”:24,”examID”:”8b289758-b898-4838-abc0-c6cd73a007e8″,”planID”:”jia”,”timelimit”:180}

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