2014年数学竞赛预赛_数学专业）试题及答案.pdf

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1、6:Oy:3?:):;:KLd183I)mm(a,2014c10?)/:4m:150:100K?no8o151515152020100?5:1.kKL?3dm,?3?.2.K,?k69IP.3.XKx?,?3?,IK.!(?K 15)m?l1:x 41=y 32=z 81,l2:x+17=y+16=z+11.1y l1 l22 l1 l2R?IO3?l1?:l2?:?;,?(1)yyy l1k:r1=(4,3,8)v1=(1,2,1)l2k:r2=(1,1,1)v2=(7,6,1)q(r1 r2,v1,v2)=?549121761?6=0,?l1 l2(3)(2)l1?:P1=r1+t1v1

2、l2?:P2=r2+t2v2?P1P2=r2 r1+t2v2 t1v1=(5+7t2 t1,4 6t2+2t1,9+t2 t1).R?v=v1 v2=?ijk121761?=(4,6,8).1 1?5 d P1P2 v 1(5+7t2 t1):(4 6t2+2t1):(9+t2 t1)=4:6:8?t1=1,t2=0.?:r2+0v2=(1,1,1)3Rl?R?IOx+14=y+16=z+18.(9)(3)P1=r1+t1v1 P2=r2+t2v2?:12(3+t1+7t2,2 2t1 6t2,7+t1+t2).d:;,?v=v1 v2=(4,6,8).q:12(3,2,7)3?;,?4x+6

3、y+8z 40=0.(15)?!(?K 15)?f C0,1 K?NO1y?n N3?xn 0,1?f(xn)?n=Z10?f(x)?ndx.2ylimnxn=1.yyy:1f(0)nZ10?f(x)?ndx f(1)n,dY?05?xn?35(3)du f Nxn?(5)2?0,d f?K5N5?f(xn)?nZ11?f(1?)?n=?f(1?)?n,?f(xn)n?f(1?),l?liminfnf(xn)f(1?).1 2?5 6:Oy:3?:):;:KLdd f?N5liminfnxn 1?.d?5,limxn=1.(15)n!(?K 15)?V 4m 0,1?N?mX?f1,.,fn

4、V.ye?d1f1,.,fn52a1,.,an 0,1?det(fi(aj)6=0,p det L1?.yyy 2)1).1f1+nfn=0.a1,.,anO“?|1f1(a1)+nfn(a1)=0.1f1(an)+nfn(an)=05?|?X?(fi(aj)T,dd det(fi(aj)6=0?1=n=0.(6)1)2).8B.k n=1(w,g?n=k(Kn=k+1df1,.,fk+15 f1,.,fk5.da1,.,ak 0,1?det(fi(aj)kk6=0*?F(x)=detf1(a1)f1(ak)f1(x).fk(a1)fk(ak)fk(x)fk+1(a1)fk+1(ak)fk+1

5、(x).U?m?F(x)=1f1(x)+kfk(x)+k+1fk+1(x),1,.,k+15?k+16=0,ddf1,.,fk+15 F(x)0,l?ak+1 0,1?F(ak+1)6=0.=a1,.,ak+1 0,1,det(fi(aj)6=0.y.(15)o!(?K 15)?f(x)3 R k?,f(x),f0(x),f00(x)u,b?3?a,b?f00(x)6 af(x)+bf0(x)x R.(i)y:limxf0(x)=0;(ii)y:3 c?f0(x)6 cf(x).(iii)?c.yyy:d f 9 f0N4O?dlimxf(x)limxf0(x)3(2)1 3?5 n?x 3

6、x(0,1)?f(x+1)f(x)=f0(x+x)f0(x)0.?x 4 0?klimxf0(x)=0.(5)?c=b+b2+4a2.K c b 0,abc=c.ukf00(x)cf0(x)6(b c)f0(x)+af(x)=(b c)(f0(x)cf(x).e(bc)x(f0(x)cf(x)N4?5?T?x 4 0dk f0(x)cf(x)6 0.=f0(x)6 cf(x).(10)c Z?f(x)=ecxk f00(x)=af(x)+bf0(x).(15)!(?K 20)?m?.y?n,l,3 m?X?Xn+Xl=I+100002100032100.m 1m 2m 310mm 1m 221

7、.yyy 1-H=01.10,K?CXn+Xl=2I+2H+3H2+mHm1.(3)2)?/X10000a11000a2a1100.am1am2am310amam1am2a11?X,Kk X=I+a1H+a2H2+amHm1.(JXn=(I+a1H+a2H2+amHm1)n=I+(na1)H+(na2+f1(a1)H2+(nam+fm1(a1,am1)Hm1,f1(a1)d a1(.fm1(a1,am1)d a1,am1(.1 4?5 6:Oy:3?:):;:KLdaq/,kXl=I+(la1)H+(la2+g1(a1)H2+(lam+gm1(a1,am1)Hm1.(12)3)*?e?|(n+

8、l)a1=2,(n+l)a2+(f1(a1)+g1(a1)=3,(n+l)am+?fm1(a1,.,am1)+gm1(a1,am1)?=m.?wT|k)K?y(20)8!(?K 20)?(0,1),an?vliminfnn?anan+1 1?=(0,+).y limnnkan=0,k 0.yyy dl,m an N4dlimnan=a 0.e a 0,K?n an an+11/n=n?anan+1 1?an+1=a2 0.Xn+11nuXn=1(anan+1)u?d?w,?a1a.gAk a=0.(10)-bn=nkan.Kkn?bnbn+1 1?=?nn+1?k?n?anan+1 1?n?(1+1n)k 1?.(1+1n)k 1 kn(n ),d9?liminfnn?bnbn+1 1?=,(n ).ddmy?limnbn=0,=limnnkan=0.(20)1 5?5