已知曲线C:y=4ln(x-1)-(x+1)^2,直线l:2x+y+2k-1=0,当x属于(1,3]时,l恒在C上方,求k的范围.
人气:141 ℃ 时间:2020-06-18 20:57:40
解答
答:
设f(x)=y1-y2
=-(2x+2k-1)-[4ln(x-1)-(x+1)²]
=-2x-2k+1-4ln(x-1)+x²+2x+1
=x²-4ln(x-1)-2k+2>0,x>1
求导得:
f'(x)=2x-4/(x-1)=2(x-2)(x+1)/(x-1),1
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