高数选择题，判断函数的奇偶性，有界函数还是周期函数，如f(x)=cos(x+1)/(x^2+1)是 函数 f(x)=2+sinX /1+X^2 是 （ ） 有界...
\u8fd9\u4e24\u4e2a\u600e\u4e48\u5224\u65ad\u662f\u6709\u754c\u5468\u671f\u8fd8\u662f\u5947\u5076\u51fd\u6570\uff1f5\u3001\u662f\u5076\u51fd\u6570\uff01
f(x)=(2+sinx)/(1+x^2)
\u6709\u754c\u51fd\u6570\u51fd\u6570\u7684\u503c\u57df\u4e3a\uff08a,b),b>a,\u82e5b=a,(a,a),\u8be5\u533a\u95f4\u7b49\u4ef7\u4e8eaa\u4e14xa\u548c<a\u662f\u4e00\u5bf9\u77db\u76fe\u7684\u6761\u4ef6\uff0c\u4e8c\u8005\u4e0d\u53ef\u80fd\u540c\u65f6\u6210\u7acb\uff0c\u6216\u8005\u8bf4\u4f7f\u4e8c\u8005\u540c\u65f6\u6210\u7acb\u7684\u5b9e\u6570\u662f\u4e0d\u5b58\u5728\u7684\uff0c\u4e5f\u5c31\u662f\u8bf4\u6ee1\u8db3\u8fd9\u4e2a\uff08a,a)\u533a\u95f4\u7684\u5b9e\u6570x\u662f\u4e0d\u5b58\u5728\uff0c\u8fd9\u4e2a\u533a\u95f4\u662f\u7a7a\u96c6\uff0c\u662f\u65e0\u610f\u4e49\u7684\uff0c\u6240\u4ee5b/=a,b\u53d6\u4e0d\u5230a,\u8be5\u533a\u95f4\u7684\u4e24\u4e2a\u7aef\u70b9\u90fd\u4e3a\u5e38\u6570\u6570\u503c\uff0c
\u4e0d\u662f\uff08-\u65e0\u7a77\uff0ca)\u6216\u8005\uff08b,+\u65e0\u7a77\uff09\uff0c\u6216\u8005\uff08-\u65e0\u7a77\uff0c+\u65e0\u7a77\uff09\uff0c\u533a\u95f4\u4e2d\u6709\u4e00\u4e2a\u662f\u65e0\u7a77\uff0c\u90a3\u4e48\u5c31\u4e0d\u662f\u6709\u754c\u51fd\u6570\u3002
\u5b9a\u4e49\u57df\uff1a\u5206\u5b502+sinx,x:R,\u5206\u6bcd/=0,1+x^2/=0,
x^2/=-1
x^2=-1
\u4efb\u610f\u5b9e\u6570\u7684\u5e73\u65b9>=0,\u4e0d\u53ef\u80fd<0,x^2=-1:(-\u65e0\u7a77\uff0c0\uff09\u5728\u5b9e\u6570\u8303\u56f4\u5185\u65e0\u89e3\uff0c
x^2=-1\u7684\u89e3\u96c6\u662f\u7a7a\u96c6\uff0c
x^2/=-1\u7684\u89e3\u96c6\u662fx^2=-1\u7684\u89e3\u96c6\u7684\u8865\u96c6\uff0cCu(\u7a7a\u96c6\uff09=R,
\u5b9a\u4e49\u57df\u662f\u5206\u5b50\u7684\u5b9a\u4e49\u57df\u548c\u5206\u6bcd\u7684\u5b9a\u4e49\u57df\u53d6\u4ea4\u96c6\uff0cR\u4ea4R=R,\u4e24\u4e2a\u96c6\u5408\u76f8\u7b49\uff0c\u90a3\u4e48\u4e24\u4e2a\u96c6\u5408\u7684\u7684\u4ea4\u96c6\u7b49\u4e8e\u8fd9\u4e24\u4e2a\u96c6\u5408\u4efb\u610f\u4e00\u4e2a\u96c6\u5408\u672c\u8eab\uff0cA\u4ea4B=A\u4ea4A=A=R,\u56e0\u4e3a\u4e24\u4e2a\u96c6\u5408\u76f8\u540c\uff0c\u6240\u4ee5\u8fd9\u4e24\u4e2a\u96c6\u5408\u672c\u8eab\u90fd\u76f8\u7b49\uff0cA\u4ea4B=A=B=R
\u5b9a\u4e49\u57df\u5173\u4e8e\u539f\u70b9\u5bf9\u79f0\uff0c
\u5982\u679c\u662f\u5947\u51fd\u6570\uff0cf0:R,f(0)=\uff082+sin0)/(1+0^2)=(2+0)/(1+0)=2/1=2/=0
\u5947\u51fd\u6570\u5982\u679c\u5b9a\u4e49\u57df\u5305\u542b0\uff0c\u90a3\u4e48f(0)=0,\u73b0\u5728f\uff080\uff09/=0
\u6240\u4ee5\u8be5\u51fd\u6570\u4e0d\u662f\u5947\u51fd\u6570\u3002
C\u6392\u9664
2.\u5982\u679c\u662f\u5076\u51fd\u6570\uff0cf(-x)\u6052\u7b49\u4e8ef(x)
(2+sin(-x))/(1+(-x)^2)=(2+sinx)/(1+x^2)
(2-sinx)/(1+x^2)=(2+sinx)/(1+x^2)
\u5206\u6bcd1+x^2>=1>0,1+x^2>1>0\u63a8\u51fa1+x^2>0or1+x^2=1>0,1+x^2>0,\u4e24\u79cd\u60c5\u51b5\u7684\u7ed3\u679c\u76f8\u540c\uff0c\u6240\u4ee5\u7ed3\u679c\u80fd\u5408\u5e76\uff0c1+x^2>0,\u63a8\u51fa1+x^2/=0,
\u6240\u4ee5\u80fd\u6d88\u6389
2-sinx=2+sinx
2sinx=0
sinx=0
sinx\u6052\u7b49\u4e8e0\uff0csinx\u5728x:R\u4e0a\u7684\u503c\u57df\u662f[-1,1],0\u5c5e\u4e8e[-1,1],\u4e0d\u4e00\u5b9a\u6052\u7b49\u4e8e0\uff0c\u6052\u7b49\u4e8e0\u610f\u601d\u65e0\u8bbax\u5728R\u4e2d\u53d6\u4f55\u503c\uff0csinx=0\u5bf9x:R\u6052\u6210\u7acb\uff0csinx\u4e0d\u6052\u7b49\u4e8e0\uff0c\u5728R\u4e2d\u5b58\u5728\u81f3\u5c11\u4e00\u4e2ax0,\u4f7f\u5f97sinx0/=0,\u90a3\u4e48sinx\u5c31\u4e0d\u6052\u7b49\u4e8e0\uff0c\u53cd\u4f8b\uff0cx0=pai/2,sinx0=sinpai/2=1/=0,\u4e3e\u51fa\u4e861\u4e2a\u53cd\u4f8b\uff0c\u81f3\u5c11\u4e00\u4e2a\uff0c\u53cd\u4f8b\u4e2a\u6570>=1,\u4e2a\u6570\uff1aN,N:0,1,2,3,.......+\u65e0\u7a77\uff0c>=1,\u6700\u5c0f\u503c\u662f1\uff0c\u90a3\u4e48\u4ece1\u5f00\u59cb\u53d6\uff0c1,2\uff0c3......+\u65e0\u7a77\uff0c
1\uff1aN*
\u6240\u4ee5\u63a8\u7ffb\u4e86sinx\u6052\u7b49\u4e8e0\u7684\u7ed3\u8bba\uff0c\u6240\u4ee5sinx\u4e0d\u6052\u7b49\u4e8e0
\u6240\u4ee5f(x)\u4e0d\u662f\u5076\u51fd\u6570\u3002
D\u6392\u9664
\u5047\u8bbe\u662f\u6700\u5c0f\u6b63\u5468\u671f\u4e3aT(T>0)\u6700\u5c0f\u6b63\u5468\u671f\u662f\u6240\u6709\u6b63\u5468\u671f\u4e2d\u6700\u5c0f\u7684\u6b63\u5468\u671f\uff0c\u6bd4\u5982sinx\u7684\u6700\u5c0f\u6b63\u5468\u671f\u662f2pai,\u5b83\u5468\u671f\u7684\u901a\u9879\u662fk*2pai,k:Z,k/=0,k=0,T=0,\u5468\u671f\u662f\u4e0d\u80fd\u4e3a0\u7684\uff0c\u6240\u4ee5k/=0,k\u662f\u975e\u96f6\u6574\u6570\uff0c\u7136\u540e\u6700\u5c0f\u6b63\u5468\u671f\uff0ck*2pai>0,k>0,k>0\u7684\u6574\u6570\uff0c1,2.3....+\u65e0\u7a77\uff0c
k*2pai=2paik,2pai>0,\u6240\u4ee5\u662f\u6b63\u6bd4\u4f8b\u51fd\u6570\uff0c\u51fd\u6570\u7ecf\u8fc7\u4e00\u4e09\u8c61\u9650\uff0c\u5b9a\u4e49\u57df\u662f1,2,3....+\u65e0\u7a77\uff0c\u662f\u6b63\u6574\u6570\uff0cx>=1>0,x>0,\u7b2c\u4e09\u8c61\u9650x0,\u6240\u4ee5\u628a\u7b2c\u4e09\u8c61\u9650\u53bb\u9664\u6389\uff0c\u53ea\u4fdd\u7559\u7b2c\u4e00\u8c61\u9650\u7684\uff0c\u800c\u4e14x>0,\u6240\u4ee5\uff080,0\uff09\u8fd9\u4e2a\u70b9\u53d6\u4e0d\u5230\uff0c\u662f\u4ece\uff081,2pai)\uff0c\uff082,4pai),\uff083,6pai),......(k,2paik),........\u4e00\u76f4\u53d6\u4e0b\u53bb\u7684\u79bb\u6563\u7684\u70b9\uff0ck:N*,
\u90a3\u4e48\u70b9\u7684\u7eb5\u5750\u6807\u662f\u51fd\u6570\u7684\u6b63\u5468\u671f\uff0c\u90a3\u4e482pai,4pai,6pai,.......2paik,k:N*
\u662f\u5355\u8c03\u9012\u589e\u7684\u6570\u5217\uff0cTmin=T1=2pai,
\u5468\u671f\u51fd\u6570\u6ee1\u8db3f\uff08x+T\uff09=f(x)\u5bf9R\u5185\u7684\u4efb\u610fx\u90fd\u6210\u7acb
(2+sin\uff08X +T))/(1+(X+T)^2 )=(2+sinx)/(1+x^2)
(2+sin(x+T))(1+x^2)=(2+sinx)(1+(x+T)^2)
2+2x^2+sin(x+T)+sin(x+T)x^2=2+2(x+T)^2+sinx+sinx(x+T)^2
2x^2+sinxcosT+cosxsinT+x^2(sinxcosT+cosxsinT)=2(x^2+2Tx+T^2)+sinx+sinx(x^2+2Tx+T^2)
2x^2+sinxcosT+cosxsinT+x^2sinxcosT+x^2cosxsinT=2x^2+4Tx+2T^2+sinx+x^2sinx+2Txsinx+sinxT^2
sinxcosT+cosxsinT+x^2sinxcosT+x^2cosxsinT=4Tx+2T^2+sinx+x^2sinx+2Txsinx+sinxT^2
\u9000\u4e0d\u51fa\u6765\uff0c\u4e0d\u5b58\u5728T>0,\u4f7f\u5f97f(x+T)=f(x)\u5bf9\u4e8ex:R\u6052\u6210\u7acb\u3002
\u4e0d\u662f\u5468\u671f\u51fd\u6570\uff0cB\u6392\u9664\uff0c
\u6392\u9664\u6cd5\uff0c\u9009A,B\uff0cC,D\u90fd\u6392\u9664
f(x)=(2+sinx)/(1+x^2)
x:R,\u5206\u5b50\u7684\u503c\u57df[1,3],\u5206\u6bcd\u7684\u503c\u57dfy=1+x^2,a=1>0,\u6709\u6700\u5c0f\u503c\uff0c\u5bf9\u79f0\u8f74x=0,fmin=f(0)=1,f(x)>=fmin=f(0\uff09=1\uff0cf(x)>=1,[1,+\u65e0\u7a77\uff09
\u4ece\u6781\u9650\u7684\u89d2\u5ea6\u8003\u8651\uff0cx\u8d8b\u5411\u4e8e-\u65e0\u7a77\uff0cx^2\u8d8b\u5411\u4e8e\u6b63\u65e0\u7a77\uff0c1+x^2=1+\u6b63\u65e0\u7a77\u8d8b\u5411\u4e8e\u6b63\u65e0\u7a77\uff0cx\u8d8b\u5411\u4e8e\u8d1f\u65e0\u7a77\uff0csinx\u662f\u9707\u8361\u7684\uff0c\u5f53x\u8d8b\u5411\u4e8e\u8d1f\u65e0\u7a77\u65f6\u5019\uff0csinx\u5728[-1,1]\u4e2d\u4e0d\u65ad\u5730\u53d8\u5316\uff0c2+sinx\u5728[1,3]\u8fd9\u4e2a\u8303\u56f4\u4e0d\u65ad\u5730\u53d8\u5316\uff0c
\u8bbea=2+sinx,a:[1,3],a\u662f\u5728[1,3]\u4e2d\u53d8\u5316\u7684\u5e38\u6570\uff0ca>=1>0,a>0,a\u662f\u6b63\u5e38\u6570\uff0c
limx\u8d8b\u5411\u4e8e\u8d1f\u65e0\u7a77a/\u8d1f\u65e0\u7a77=alimx\u8d8b\u5411\u4e8e\u8d1f\u65e0\u7a771/\u8d1f\u65e0\u7a77=ax0-,
0-\u662f0,0-\u662f<0\u8d8b\u5411\u4e8e0\uff0c\u8be5\u503c\u662f\u4ece0\u7684\u5de6\u8fb9\u65e0\u9650\u5730\u63a5\u8fd1\u4e8e0,\u8be5\u70b9\u57280\u7684\u5de6\u8fb9\uff0c\u90a3\u4e48\u8be5\u6570<0,<0\u8d8b\u5411\u4e8e0\uff0c\u8d8b\u5411\u4e8e0-,
\u90a3\u4e48\u8be5\u51fd\u6570\u6709\u4e2a\u4e0a\u754c0\uff0cf(x)<0
\u5f53x\u8d8b\u5411\u4e8e+\u65e0\u7a77\u65f6\uff0cx^2\u8d8b\u5411\u4e8e+\u65e0\u7a77\uff0c1+x^2=1+\u6b63\u65e0\u7a77\u8d8b\u5411\u4e8e+\u65e0\u7a77\uff0c\u5206\u5b50\uff0csinx\u662f\u5728[-1,1]\u4e2d\u9707\u8361\u7684\uff0cx\u8d8b\u5411\u4e8e+\u65e0\u7a77\uff0csinx:[-1,1],2+sinx:[1,3],\u4ee4a=2+sinx,a:[1,3]
a>=1>0,a>0,a\u662f\u6b63\u5e38\u6570\uff0c\u56e0\u4e3aa\u662f\u5728[1,3]\u8fd9\u4e2a\u8303\u56f4\u53d8\u5316\u7684\u6b63\u5e38\u6570\uff0c
limx\u8d8b\u5411\u4e8e+\u65e0\u7a77a/\u6b63\u65e0\u7a77=alimx\u8d8b\u5411\u4e8e\u6b63\u65e0\u7a771/+\u65e0\u7a77=ax0+,0+,\u8d8b\u5411\u4e8e0\uff0c>0,\u4ece0\u7684\u53f3\u8fb9\u65e0\u9650\u5730\u63a5\u8fd1\u4e8e0,\u8be5\u503c\u662f\u6bd40\u5927\uff0c\u4f46\u662f\u65e0\u9650\u5730\u63a5\u8fd1\u4e8e0,
ax0+,0+\u8d8b\u5411\u4e8e0\uff0cax\u8d8b\u5411\u4e8e0\u8d8b\u5411\u4e8eax0=0,a,0+\u53d6\u5411\u4e8e0\uff0ca>0,0+>0,\u6b63\u6b63\u5f97\u6b63\uff0c\u6240\u4ee5ax0+>0,>0\u8d8b\u5411\u4e8e0\u90a3\u4e48\u662f\u8d8b\u5411\u4e8e0+,\u90a3\u4e48f(x)>0
f(x)\u662f\u6709\u754c\u51fd\u6570
A
解：
f(x)=cos(x+1)/(x^2+1)
(1)奇偶性：非奇非偶
f(x)-f(-x)
=[cos(x+1)-cos(-x+1)]/(x²+1)
=-2sinxsin1/(x²+1)
No≡0
f(x)+f(-x)
=[cos(x+1)+cos(-x+1)]/(x²+1)
=2cosxcos1/(x²+1)
No≡0
∴f(x)非奇非偶
(2)有界性
|f(x)|
=|cos(x+1)|*|1/(x^2+1)|
≤1*1
=1
∴f(x)是有界函数
绛旓細闈炲闈炲伓锛屾湁鐣 瑙ｏ細f(x)=cos(x+1)/(x^2+1)(1)濂囧伓鎬锛氶潪濂囬潪鍋 f(x)-f(-x)=[cos(x+1)-cos(-x+1)]/(x²+1)=-2sinxsin1/(x²+1)No鈮0 f(x)+f(-x)=[cos(x+1)+cos(-x+1)]/(x²+1)=2cosxcos1/(x²+1)No鈮0 鈭磃(x)闈炲闈炲伓 ...
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绛旓細1 鍏堝垎瑙ｅ嚱鏁颁负甯歌鐨勪竴鑸嚱鏁帮紝姣斿澶氶」寮弜^n锛屼笁瑙鍑芥暟锛屽垽鏂鍋舵 2 鏍规嵁鍒嗚В鐨勫嚱鏁颁箣闂寸殑杩愮畻娉曞垯鍒ゆ柇锛屼竴鑸彧鏈変笁绉嶇f(x)g(x)銆乫(x)+g(x锛夛紝f(g(x))锛堥櫎娉曟垨鍑忔硶鍙互鍙樻垚鐩稿簲鐨勪箻娉曞拰鍔犳硶锛3 鑻锛坸锛夈乬(x锛夊叾涓竴涓负濂囧嚱鏁帮紝鍙︿竴涓负鍋跺嚱鏁帮紝鍒檉(x)g(x)濂囥乫(x)+g(x...
绛旓細渚嬪y=x^a锛寉=sinx,y=C锛圕鏄父鏁帮級绛夌瓑锛2.璁颁綇濂囧伓鍑芥暟鐨鈥滆繍绠楁硶鍒欌濓紝渚嬪锛1锛夊悓濂囧伓鎬х殑鍑芥暟鐩稿姞鍑锛屽鍋舵涓嶅彉锛2锛夊紓濂囧伓鎬х殑鍑芥暟鐩稿姞鍑忥紝闈炲闈炲伓锛2锛夊悓濂囧伓鎬х殑鍑芥暟鐩镐箻闄ゅ緱鍋跺嚱鏁帮紱3锛夊紓濂囧伓鎬х殑鍑芥暟鐩镐箻闄ゅ緱濂囧嚱鏁般傝繖鏍蜂綘灏卞彲浠ュ簲浠樺ぇ澶氭暟鐨勬绫閫夋嫨棰浜嗐
绛旓細鈭<0,y>dx鈭<0,x>xf(u)du 鏄潪濂囬潪鍋鍑芥暟銆傚綋 a鈮0 鏃讹紝鈭玠x鈭玿f(u)du 涓鑸槸闈炲闈炲伓鍑芥暟銆備緥濡 鈭<1,y>dx鈭<1,x>2xudu = 鈭<1,y>xdx鈭<1,x>2udu = 鈭<1,y>x(x^2-1)dx = [x^4/4-x^2/2]<1,y> = y^4/4-y^2/2+1/4....
绛旓細璇佹槑棰/璁＄畻棰橈紝鍙兘鏍规嵁濂囧伓鎬鐨勫畾涔夋潵璇佹槑濂囧伓鎬 閫夋嫨棰锛氫笉鑰冭檻鐗规畩鍑芥暟f(x)=0鐨勮瘽锛屽+濂=濂囷紝濂-濂=濂 濂嚸楀=鍋讹紝濂/濂=鍋 鍋+鍋=鍋讹紝鍋-鍋=鍋 鍋睹楀伓=鍋讹紝鍋/鍋=鍋 濂+鍋讹紝鏃犳硶鐩存帴鍒ゆ柇 濂-鍋讹紝鏃犳硶鐩存帴鍒ゆ柇 濂嚸楀伓=濂 濂嚸峰伓=濂 ...
绛旓細棣栧厛f(0)=1 鑰寈澶т簬0鏃讹紝f(x)=1+x 閭ｄ箞f(-x)=1-(-x)=1+x 浜庢槸寰楀埌f(x)=f(-x)鑰屼笖瀹氫箟鍩熸槸瀵圭О鐨 鎵浠ヨ繖灏辨槸鍋鍑芥暟
绛旓細鍥炵瓟锛氱敤瀹氫箟鍗冲彲
绛旓細a锛0锛屾槸濂鍑芥暟銆俛鈮0锛屼笉纭畾銆
