最终版第九届数学竞赛预赛答案(数学类).pdf

6:Oy:3?:) :;: KLd 13I)mmY (a, 2017c10?) /: 4m:150:100 K?no8o 151515201520100 ? 5: 1. kKL?3dm, ?3?. 2. K, ?k69IP. 3. XKx?, ?3?, IK. ! (?K 15 ) 3m?IX, ?“V ? ? 1. y: ? ? m+p X n=m bn+1 bn bn+1 m+p X n=m bn+1 bn bm+p+1 = bm+p+1 bm bm+p+1 = 1 bm bm+p+1 . dd bn k.,p?. ? an k(13) K?1? anU an+1. X: an= en 2 ., ? P n=1 an+1an an+1lnan+1 . (15) 3 n!yK15? = W1,W2,.,Wrr ? ? ?_n?E?8. eT8u? 4(=, M,N ,kMN ), yr i=1Wi = 0 ?=?r i=1tr(Wi) = 0,tr(Wi)LWi?,. yyy75d,?5?.(2) 5ku_?W kWW1,.,WWr . ?k W WW1,WW2,.,WWr = W1,W2,.,Wr, =W = ,W .(7 PS = r i=1Wi, K WS = S,W . ? S 2 = rS, = S2 rS = 0. eS? A?K2 r = 0, = = 0r. (r i=1tr(Wi) = 0S?A?U0. dkS rI_X?S? ?)?w 2g5? S(S rI) = S2 rS = 0dm (S rI)1=?S = 0.y (15) 4 6:Oy:3?:) :;: KLd o!?K20“a9n? ? ? ?A(=, A?=AT?uA), P?kS8T T = (X,Y )|X Rnn,Y Rnn,XY = aI + A, In? ?Rnnkn?8“y?T?(X,Y ) (M,N), 7kXN + Y TMT 6= 0. yyyy. e XN + Y TMT = 0,Kk NTXT+ MY = 0. (2) ,?d(X,Y ) T? XY + (XY )T= 2aI, = XY + Y TXT = 2aI. aqk MN + NTMT= 2aI. d XY T MNT ! YN XTMT ! = 2a I0 0I ! , (10) ? 1 2a YN XTMT ! XY T MNT ! = I0 0I ! , ? Y Y T + NNT= 0 Y = 0,N = 0 ?XY = 0, XY = aI + A 6= 0g.y(20) 5 !?K15? f(x) = arctanx, A . e ? ? B = lim n ? n X k=1 f ?k n ? An ? 3, A,B. ): I. k A=lim n 1 n n X k=1 f ?k n ? = Z 1 0 f(x)dx =xarctanx? ?1 0 Z 1 0 x 1 + x2 dx = 4 ln2 2 . (6 ) u x ?k1 n , k n ?, (1 6 k 6 n), dn, 3 n,k ?k1 n , k n ? ? f(x) = f ?k n ? + f0? k n ?x k n ? + f00(n,k) 2 ?x k n ?2. . (9 ) u, ? ? ? n X k=1 f ?k n ? nA + n X k=1 n Zk n k1 n f0? k n ?x k n ? dx ? ? ? = ? ? ? n X k=1 n Zk n k1 n h f ?k n ? + f0? k n ?x k n ? f(x) i dx ? ? ? 6M n X k=1 n Zk n k1 n ?x k n ?2 dx = M 3n, M = 1 2 maxx0,1|f00(x)|. (12 ) d, lim n ? n X k=1 f ?k n ? An ? = lim n n X k=1 n Zk n k1 n f0? k n ?x k n ? dx =lim n 1 2n n X k=1 f0? k n ? = 1 2 Z 1 0 f0(x)dx = 8 . (15 ) 6 6:Oy:3?:) :;: KLd II. k A=lim n 1 n n X k=1 f ?k n ? = Z 1 0 f(x)dx =xarctanx? ?1 0 Z 1 0 x 1 + x2 dx = 4 ln2 2 . (6 ) u x ?k1 n , k n ?, (1 6 k 6 n), dn, 3 n,k ?k1 n , k n ? ? f ?k n ? = f(x) + f0(x)? k n x?+ f00(n,k) 2 ?k n x?2. . (9 ) u, ? ? ? n X k=1 f ?k n ? nA n X k=1 n Zk n k1 n f0(x)? k n x?dx ? ? ? = ? ? ? n X k=1 n Zk n k1 n h f ?k n ? f(x) f0(x)? k n x? i dx ? ? ? 6M n X k=1 n Zk n k1 n ?k n x?2dx = M 3n, M = 1 2 maxx0,1|f00(x)|. (12 ) d, lim n ? n X k=1 f ?k n ? An ? = lim n n X k=1 n Zk n k1 n f0(x)? k n x?dx =lim n n X k=1 nf0(n,k) Zk n k1 n ?k n x?dx =lim n 1 2n n X k=1 f0(n,k) = 1 2 Z 1 0 f0(x)dx = 8 , n,k ?k1 n , k n ?. (15 ) III. k A=lim n 1 n n X k=1 f ?k n ? = Z 1 0 f(x)dx =xarctanx? ?1 0 Z 1 0 x 1 + x2 dx = 4 ln2 2 . 7 (6 ) u x ?k1 2 n , k+1 2 n ?, (1 6 k 6 n), dn, 3 n,k ?k1 2 n , k+1 2 n ? ? f(x) = f ?k n ? + f0? k n ?x k n ? + f00(n,k) 2 ?x k n ?2. . (9 ) u, ? ? ? n X k=1 f ?k n ? nA n Z 1+ 1 2n 1 f(x)dx + n Z1 2n 0 f(x)dx ? ? ? = ? ? ? n X k=1 n Z k+1 2 n k1 2 n h f ?k n ? f(x) + f0? k n ?k n x? i dx ? ? ? 6M n X k=1 n Zk n k1 n ?k n x?2dx = M 3n, M = 1 2 maxx0,1|f00(x)|. (12 ) d, lim n ? n X k=1 f ?k n ? An ? = lim n n Z 1+ 1 2n 1 f(x)dx lim n n Z1 2n 0 f(x)dx = f(1) 2 f(0) 2 = 8 . (15 ) 8 6:Oy:3?:) :;: KLd 8!?K20?f(x) = 1x2+x3(x 0,1), ? ? Oe4nd lim n R 1 0 fn(x)ln(x + 2)dx R 1 0 fn(x)dx . ): f(x) Y. 5? f(x) = 1 x2(1 x), k 0 1 2 , 06 R 1 fn(x)dx R 0 fn(x)dx = R 1 1 fn(x)dx R 0 fn(x)dx + R 1 fn(x)dx R 0 fn(x)dx = R 0 ?1 x(1 x)2?n dx R 0 ?1 x2(1 x)?n dx + R 1 fn(x)dx R 0 fn(x)dx 6 R 0 ?1 x 4 ?n dx R 0 ?1 x2?n dx + R 1 fn(x)dx R 0 fn(x)dx 6 R 0 ?1 x 4 ?n dx R 1 n 0 ?1 x n?ndx + (1 2)Mn mn = 4 n+1 ? 1 ?1 4 ?n+1? n n+1 ? 1 ?1 1 n ?n+1 ? + (1 ) ?M mn ?n . l? lim n R 1 fn(x)dx R 0 fn(x)dx = 0. (14 ) = 9 (555: ?4. ?K, lim n R 1 fn(x)dx R 0 fn(x)dx = 0 4 , lim n R 1 1 fn(x)dx R 0 fn(x)dx = 0 4 . ) = u (0,ln 5 4), ? = 2(e 1), K (0, 1 2), ln 2+ 2 = . , dc(, 3 N 1 ? n N k R 1 fn(x)dx R 0 fn(x)dx 6 . l?qk ? ? ? R 1 0 fn(x)ln(x + 2)dx R 1 0 fn(x)dx ln2 ? ? ? = R 1 0 fn(x)ln x+2 2