2ãå½æ°f(x)=(1/2)x²-lnxçåè°å¢åºé´
解ï¼å®ä¹åï¼x>0ï¼
f'(x)=x-(1/x)=(x²-1)/x=(x+1)(x-1)/xï¼å½1â¦x<+âæ¶f'(x)â§0ï¼æ
åå¢åºé´ä¸ºï¼[1ï¼+â)ï¼æ
åºéD.
5ã设F(x)=ãaï¼xãâ«xf(t)dtï¼åF'(x)=
解ï¼F'(x)=ãaï¼xãâ«{∂[xf(t)]/∂x}dt+xf(t)(dx/dx)=ãaï¼xãâ«f(t)dt+xf(t)ï¼æ
éC.
9ãy=(x-5)³+2
解ï¼ä»¤y'=3(x-5)²=0ï¼å¾é©»ç¹x=5ï¼æy''=6(x-5)ï¼å½x=5æ¶y''=0ï¼æ
x=5æ¯æç¹ï¼ä¸æ¯æå¼ç¹ï¼
å³è¯¥å½æ°ææç¹(5ï¼2)ï¼ä½æ æå¼ç¹ï¼æ
éC.
12ãæ±æéx➔âlim[xsin(2/x)+(2/x)sinx]
解ï¼å½x➔âæ¶2/x➔0ï¼å æ¤sin(2/x)∾2/xï¼
æ
åå¼=x➔âlim[x(2/x)+(2/x)sinx]=x➔âlim[2+(2/x)sinx]=2ã
22ãå¤æf(x)=e^(1/x)(x<0)ï¼f(x)=1(x=0)ï¼f(x)=(1/x)[ln(x²+x)-lnx](x>0)ï¼å¨x=0å¤çè¿ç»æ§ï¼è¥f(x)
å¨x=0å¤é´æï¼æåºé´æç¹çå称ã
解ï¼å·¦æéx➔-0limf(x)=x➔-0lime^(1/x)=0â f(0)=1ï¼
å³æéx➔+0limf(x)=x➔+0lim(1/x)[ln(x²+x)-lnx]=x➔+0lim(1/x)ln[(x²+x)/x]=x➔+0lim(1/x)ln(x+1)
=x➔+0lim1/(x+1)=1=f(0)ï¼æ
该å½æ°å¨x=0å¤å·¦ä¸è¿ç»ï¼å±ç¬¬ä¸ç±»é´æç¹(è·³è·åé´æç¹)
24ãå·²ç¥æ¹ç¨lnâ(x²+y²)=arctan(y/x)ç¡®å®äºy=y(x)ï¼æ±dy
解ï¼ä»¤F(xï¼y)=lnâ(x²+y²)-arctan(y/x)=0
ådy/dx=-(∂F/∂x)/(∂F/∂y)=-[x/(x²+y²)+(y/x²)/(1+y²/x²)]/[y/(x²+y²)-(1/x)/(1+y²/x²)]=(x+y)/(x-y)
æ
dy=[(x+y)/(x-y)]dx
27ãä¸æ²çº¿è¿åç¹ï¼ä¸å
¶ä¸ä»»æä¸ç¹(xï¼y)å¤çå线çæç为2x+yï¼æ±æ²çº¿çæ¹ç¨
解ï¼y'=2x+y...(1)ï¼y'-y=2xï¼å
æ±é½æ¬¡æ¹ç¨y'-y=0çé解ï¼å离åédy/y=dxï¼
积åä¹å¾lny=x+lnC₁ï¼æ
y=C₁e^xï¼å°C₁æ¢æxçå½æ°uï¼å¾y=ue^x.....(2)ï¼
对(2)å导æ°å¾dy/dx=ue^x+(e^x)(du/dx)......(3)ï¼
å°(2)å(3)代å
¥(1)å¼å¾ï¼ue^x+(e^x)(du/dx)=2x+ue^xï¼æ
å¾(e^x)(du/dx)=2xï¼
å离åéå¾du=2xe^(-x)dxï¼ç§¯åä¹å¾u=2â«xe^(-x)dx=-2â«xd[e^(-x)]=-2[xe^(-x)-â«e^(-x)dx]
=-2[xe^(-x)+e^(-x)]+C=-2(x+1)e^(-x)+Cï¼
代å
¥(2)å¼å¾é解为ï¼y=[-2(x+1)e^(-x)+C]e^x=-2(x+1)+Ce^xï¼
æ²çº¿è¿åç¹ï¼å³æy(0)=0ï¼ä»£å
¥å¾0=-2+Cï¼æ
C=2
æ以æ²çº¿æ¹ç¨ä¸ºy=-2(x+1)+2e^x.
28ãå·²ç¥ç´çº¿Lï¼(x-7)/5=(y-4)/1=(z-5)/4ä¸å¹³é¢Ïï¼3x-y+2z-5=0ç交ç¹ä¸ºMoï¼ä¸åç´çº¿Låç´
çç´çº¿æ¹ç¨ã
解ï¼å°ç´çº¿Lçæ¹åç¢éa={5ï¼1ï¼4}ï¼å°å
¶æ¹ç¨æ¹åæåæ°å½¢å¼ï¼x=5t+7ï¼y=t+4ï¼z=4t+5ï¼
代å
¥å¹³é¢Ïçæ¹ç¨å¾ï¼
3(5t+7)-(t+4)+2(4t+5)-5=22t+22=0ï¼æ
å¾t=-1ï¼äºæ¯å¾äº¤ç¹Mo(2ï¼3ï¼1)ï¼
设ææ±ç´çº¿L₁çæ¹ç¨ä¸º(x-2)/m=(y-3)/n=(z-1)/Lï¼å
¶æ¹åç¢é为b={mï¼nï¼L}ï¼å 为L₁â¥Lï¼æ
å
¶
ç¹ç§¯aâb=5m+n+4L=0ï¼æ
å¯åm=6ï¼n=2ï¼L=-8ãå½ç¶ä¹å¯å·å«çæ°ã
äºæ¯å¾ç´çº¿æ¹ç¨ä¸º(x-2)/6=(y-3)/2=(z-1)/(-8)
16ãæ±å®ç§¯åã-1ï¼1ãâ«[xcosx+â(1-x²)]dx
解ï¼åå¼=ã-1ï¼1ã[â«xcosxdx+â«â(1-x²)]dx]ã第ä¸ä¸ªç§¯åç被积å½æ°xcosxæ¯å¥å½æ°ï¼ä¸ç§¯ååºé´
å
³äºåç¹å¯¹ç§°ï¼æ
å
¶ç§¯å为0ï¼ç¬¬äºä¸ªç§¯åç被积å½æ°æ¯å¶å½æ°ãã
=ã0ï¼1ã2â«â(1-x²)dxï¼ã令x=sinuï¼ådx=cosuduï¼x=0æ¶u=0ï¼x=1æ¶u=Ï/2ï¼ä»£å
¥ã
=ã0ï¼Ï/2ãâ«2cos²udu=ã0ï¼Ï/2ã(1/2)â«(1+cos2u)d(2u)=(1/2)[2u+sin2u]ã0ï¼Ï/2ã=Ï/2.
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