函数f(x)=1/ln(x-1)的连续区间是( ).A.[1,2)U(2,+∞)B.(1,2)U(2,+∞)C.(1,+∞)D.[1,+∞)
函数f(x)=1/ln(x-1)的连续区间是( ).
A.[1,2)U(2,+∞)
B.(1,2)U(2,+∞)
C.(1,+∞)
D.[1,+∞)
B.(1,2)U(2,+∞)
C.(1,+∞)
D.[1,+∞)
参考解析
解析:f(x)=1/ln(x-1)的定义域为:x-1>0,x-1≠1,即(1,2)∪(2,+∞).(1,2)及(2,+∞)均为f(x)的定义区间,又f(x)为初等函数,故应选B.
相关考题:
若f″(x)存在,则函数y=ln[f(x)]的二阶导数为:()A、(f″(x)f(x)-[f′(x)]2)/[f(x)]2B、f″(x)/f′(x)C、(f″(x)f(x)+[f′(x)]2)/[f(x)]2D、ln″[f(x)]·f″(x)
单选题设y=f[(2x-1)/(x+1)],f′(x)=ln(x1/3),则dy/dx=( )。Aln[(2x-1)/(x-1)]/(x+1)2Bln[(2x+1)/(x+1)]/(x+1)2Cln[(2x+1)/(x+1)]/(x-1)2Dln[(2x-1)/(x+1)]/(x+1)2
单选题若函数z=f(x,y)满足∂2z/∂y2=2,且f(x,1)=x+2,fy′(x,1)=x+1,则f(x,y)=( )。Ay2+(x-1)y+2By2+(x+1)y+2Cy2+(x-1)y-2Dy2+(x+1)y-2
单选题设函数f(x)=x2(x-1)(x-2),则f′(x)的零点个数为( )。A0B1C2D3