已知函数f(x)=x3-ax2+1在区间(0,2)内单调递减,则实数a的取值范围是( ) A.a≥3 B.a=3 C.a≤3 D.0<a<3
已知函数f(x)=x3-ax2+1在区间(0,2)内单调递减,则实数a的取值范围是( )
A. a≥3
B. a=3
C. a≤3
D. 0<a<3
数学人气:773 ℃时间:2019-08-20 00:38:07
优质解答
∵函数f(x)=x
3-ax
2+1在(0,2)内单调递减,
∴f′(x)=3x
2-2ax≤0在(0,2)内恒成立,
即
a≥x在(0,2)内恒成立,
∵
x<3,
∴a≥3,
故选A
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