相关x1x21的文学知识

已知函数f(x)=ax2+2ax+4(0<a<3),若x1<x2,x1+x2=1﹣a,则(  )A.f(x1)...

已知函数f(x)=ax2+2ax+4(0<a<3),若x1<x2,x1+x2=1﹣a,则(  )A.f(x1)...

问题详情:已知函数f(x)=ax2+2ax+4(0<a<3),若x1<x2,x1+x2=1﹣a,则()A.f(x1)<f(x2)      B.f(x1)>f(x2)C.f(x1)=f(x2)D.f(x1)<f(x2)和f(x1)=f(x2)都有可能【回答】A【解答】解:∵0<a<3,由函数表达式f(x)=ax2+2ax+4=a(x+1)2+4﹣a知,其对称轴为x=﹣1,又x1+x2=1﹣a,所以(x1+x2)=(1﹣a),...

设函数f(x)=ex﹣|ln(﹣x)|的两个零点为x1,x2,则(  )A.x1x2<0 B.x1x2=1  ...

设函数f(x)=ex﹣|ln(﹣x)|的两个零点为x1,x2,则(  )A.x1x2<0 B.x1x2=1  ...

问题详情:设函数f(x)=ex﹣|ln(﹣x)|的两个零点为x1,x2,则()A.x1x2<0 B.x1x2=1  C.x1x2>1D.0<x1x2<1【回答】D【解答】解:令f(x)=0,则|ln(﹣x)|=ex,作出y=|ln(﹣x)|和y=ex在R上的图象,可知恰有两个交点,设零点为x1,x2且|ln(﹣x1)|<|ln(﹣x2)|,x1<﹣1,x2>﹣1,故有>x2,即x1x2<1.又...