高数7道选择题，第一张是判断奇偶性 高数判断奇偶性
\u9ad8\u7b49\u6570\u5b66\u5224\u65ad\u9898\u3001\u5947\u5076\u6027\uff1f3\u30016\u30018\u9519
1.f(-x)=f(x)\uff0c\u662f\u5076\u51fd\u6570
2.f(-x)+f(x)=0\uff0c\u662f\u5947\u51fd\u6570
\u89e3\u6cd5\uff1a\u4f9d\u9898\u610f\u53ef\u5f97
1-e^-(-x)
,
-x\u22640
\uff08
-x\u22640
\uff0c\u5219x>0)
f(-x)={
e^\uff08-x\uff09-1,
-x\uff1e0
(-x>0,
\u5219x
\u6240\u4ee5\u5c31\u53ef\u5316\u4e3a
1-e^(x)
,
x\u22650
f(-x)={
e^\uff08-x\uff09-1,
x\uff1c0
\u5199\u6cd5\u4e0a\u4e0b\u6362\u4f4d\u7f6e\uff0c\u4e14\u63d0\u53d6\u4e00\u4e2a\u8d1f\u53f7\u3002\u5373
e^(x)-1
,
x\uff1e0
f(-x)=
-{
1-e^\uff08-x\uff09,
x\u22640
=-f(x\uff09
\u6240\u4ee5\u662f\u5947\u51fd\u6570\u3002
\uff08\u6ce8\u505a\u9898\u65f6\u53ea\u662f\u6ca1\u6709\u6ce8\u610fx=0\u7684\u7ec6\u8282\uff0c\u5e94\u8be5\u5355\u72ec\u5199\u4e00\u4e0b\u6216\u8005\u8ba8\u8bbax=0\u7684\u60c5\u51b5\uff09
=ln(1/(x+√(1+x^2)))
=-ln(x+√(1+x^2))
所以ln(x+√(1+x^2))是奇函数
.
-xarccos(-x/(1+(-x)^2))
=-xarccos(-x/(1+x^2))
=-x(π-arccos(x/(1+x^2)))
所以xarccos(x/(1+x^2))没有奇偶性
.
2.
lim[x→0+]exp(1/x)=+∞
lim[x→0-]exp(1/x)=0
所以选D
4.
lim[x→0+]xsin(1/x)=0
lim[x→0+]exp(1/x)=+∞
lim[x→0+]lnx=-∞
lim[x→0+]ln(1/x)=+∞
所以选A
.
2.
lim[]((x^2+1)/(1+x)-ax-b)
(x^2+1)/(1+x)-ax-b
=x-1+2/(1+x)-ax-b
=(1-a)x-1-b+2/(1+x)
所以由lim[x→∞]((x^2+1)/(1+x)-ax-b)=1
得：-1-b=1
即：b=-2
所以选B
.
3.
sinx无极限
lim[x→+∞]exp(-x)=0
lim[x→-∞]exp(-x)=+∞
所以lim[x→∞]exp(-x)无极限
lim[x→∞]((x+1)/(x^2+1))=lim[x→∞](1/(x-1))=0
lim[x→+∞]arctanx=π/2
lim[x→-∞]arctanx=-π/2
所以lim[x→∞]arctanx无极限
选择C
.
4.
x^3/(x^2+1)-x^2/(x-1)
=(x/(x^2+1)-1/(x-1))x^2
=-(x+1)x^2/(x^2+1)/(x-1)
=-(1+1/x)/(1+1/x^2)/(1-1/x)
所以lim[x→∞](x^3/(x^2+1)-x^2/(x-1))=-1
.
5.
lim[x→∞]((3x^2-1)/(x^3+2x+1))
=lim[x→∞]((3-1/x^2)/(1+2/x^2+1/x^3)/x)
=0
所以选A
绛旓細f(-x)=loga(鈭(x^2+1)-x)=loga((鈭(x^2+1)-x)(鈭(x^2+1)+x) / (鈭(x^2+1)+x))=loga(1 / (鈭(x^2+1)+x))=loga(鈭(x^2+1)+x)^(-1)=-loga(鈭(x^2+1)+x)=-f(x)鍥犳锛宖(x)涓哄鍑芥暟 5.C 涓ゅ嚱鏁颁簰涓哄弽鍑芥暟锛岃嚜鐒跺叧浜巠=x瀵圭О 7.A 涓婁笅鍚屾椂闄や互n^2...
绛旓細鎮ㄦ鏌ヤ竴涓嬶細
绛旓細瀵规暟瀛︿竴鏉ヨ:楂樼瓑鏁板閮ㄥ垎鍗82鍒 閫夋嫨棰 4 濉4 璁＄畻棰 5鍒 鏁板浜 楂樻暟鍗116鍒 閫夋嫨棰6鍒 濉5 澶ч 7鍒嗙嚎浠(鍖呮嫭鏁板浜)鍗34鍒 2 2 1姒傝鍗 34鍒 2 1 2瀵逛簬鏁板鐨勫涔,棣栧厛浠庣嚎鎬т唬鏁板紑濮,鍥犱负绾夸唬涓庢璁烘槸鏁板涓渶绠鍗曠殑鍒嗘敮,鍏堣涓嬬嚎浠ｄ竴鍏卞垎6绔,閲嶇偣绔犲湪绗簩绔犵殑鐭╅樀銆佺涓夌珷鐨勫悜閲,绗涓...
绛旓細绠鍗曡绠椾竴涓嬪氨琛岋紝绛旀濡傚浘鎵绀
绛旓細1=1 2.lim x瓒嬩簬0 sin2x/3x=2/3(鍒╃敤涓婇缁撹锛3.lim x瓒嬩簬0 x涔樹互cot3x =lim x瓒嬩簬0 锛1/3)3x涔樹互cos3x/sinsx=1/3 (鍒╃敤绗竴棰樼粨璁猴級4.lim x->鏃犵┓ sinx/2x =0 5.limx瓒嬩簬鏃犵┓(1+X/X)2x =e^2 6.lim x->0(1+X)1/x =e 7.lim x->鏃犵┓(1-1/X)x=1/e ...
绛旓細鍦ㄨ鐐逛袱渚э紝tanx鐨勫煎紓鍙凤紝鎺ヨ繎浜0锛屽掓暟涔嬪悗锛屽垎鍒槸卤鏃犵┓澶э紝涓嶈繛缁紝涓斾笉鍙幓銆傦紙3锛墄瓒嬭繎浜0,1/x瓒嬭繎浜幝辨棤绌峰ぇ锛宑osx鐨勫间笉纭畾锛屽洜姝わ紝涓嶅彲鍘汇傦紙4锛墄浠庡乏渚ц秼杩戜簬1锛y瓒嬭繎浜0锛寈浠庡彸渚ц秼杩戜簬1锛寉瓒嬭繎浜2锛屼笉鍚岋紝涓嶅彲鍘汇傜湅宸﹀彸鏋侀檺鏄惁鐩稿悓锛屾槸鍒ゆ柇鏄惁鍙幓鐨勫熀鏈柟娉曘
绛旓細瑙ｏ細鈭碘埆<0,1>f(x)dx=a 鈭粹埆<0,1>f(y)dy=a 鏁呪埆鈭<D>f(x)f(y)dxdy=鈭<0,1>f(x)dx鈭<0,1>f(y)dy=a*a=a^2銆
绛旓細鎴戝綋鏃跺績宸茬粡鍑変簡涓鍗婁簡,浣嗘槸鍋氬埌鍚庨潰鎱㈡參灏辨劅瑙夋潵浜,褰撶劧涔熸湁閬囧埌涓嶄細鐨棰,鍍2020骞磋瘉鏄庨涓婃潵灏辨病鏈変粈涔堟濊矾,涓ゅ垎閽熶箣鍐呴偅閬撳ぇ棰樻垜娌℃湁鍐欏嚭鏉ュ氨鐩存帴璺宠繃鍘,璺宠繃鍘讳箣鍚庢垜灏卞幓鍋氱嚎浠ｅ拰姒傜巼璁虹殑棰樸傛鐜囪鐨勯鍋氬緱杩樻槸姣旇緝濂界殑,鍋氱嚎浠ｇ殑棰樹篃鏄劅瑙夋潵浜,鎰熻涓鏉ヤ簡涔嬪悗鍐嶅洖鍘诲仛鍓嶉潰鐨閫夋嫨濉┖,鎴戝彂鐜板ソ鍍忕‘瀹炴槸...
绛旓細瑙ｏ細1銆棰橈紝瑕佸嚱鏁版湁鎰忎箟锛屽垯x-1鈮0銆亁-2鈮0銆4-x<0锛屸埓鍏跺畾涔夊煙涓簕x涓1鈮<2}鈭獅x涓2<x<4}銆2銆侀锛寁=鈭歺銆乽=lnv銆乼=鈭歶澶嶅悎鑰屾垚銆3棰橈紝鈭靛師寮=(3/k)lim(x鈫0)sin3x/(3x)=3/k=5锛屸埓k=3/5銆4棰橈紝宸﹁竟=e^(-5k)锛屸埓k=2銆5棰橈紝鍒嗗瓙鍒嗘瘝鍚岄櫎浠^7锛屸埓鍘熷紡=...
绛旓細=2*鈭玹^6-2t^4+t^2 dt =2/7*t^7-4/5*t^5+2/3*t^3+C =2/7*(x+1)^(7/2)-4/5*(x+1)^(5/2)+2/3*(x+1)^(3/2)+C 鈭玿/(x-1)^0.5 dx 浠=(x-1)^0.5 x=t^2+1 dx=2tdt =2鈭(t^2+1)dt =2/3*t^3+2t+C =2/3*(x-1)^(3/2)...
