已知a>0,函数f(x)=ln(2-x)+ax^2(x<2),1.设曲线y=f(x)在点(1,f(1))处的切线为l,若l与园(x+1)^2+y^2=1相离,求a的范围.2.求函数f(x)在[0,1]的最大值求详解
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![已知a>0,函数f(x)=ln(2-x)+ax^2(x<2),1.设曲线y=f(x)在点(1,f(1))处的切线为l,若l与园(x+1)^2+y^2=1相离,求a的范围.2.求函数f(x)在[0,1]的最大值求详解](/uploads/image/z/3766376-56-6.jpg?t=%E5%B7%B2%E7%9F%A5a%EF%BC%9E0%2C%E5%87%BD%E6%95%B0f%28x%29%3Dln%282-x%29%2Bax%5E2%28x%EF%BC%9C2%29%2C1.%E8%AE%BE%E6%9B%B2%E7%BA%BFy%3Df%28x%29%E5%9C%A8%E7%82%B9%281%2Cf%281%29%EF%BC%89%E5%A4%84%E7%9A%84%E5%88%87%E7%BA%BF%E4%B8%BAl%2C%E8%8B%A5l%E4%B8%8E%E5%9B%AD%EF%BC%88x%2B1%EF%BC%89%5E2%2By%5E2%3D1%E7%9B%B8%E7%A6%BB%2C%E6%B1%82a%E7%9A%84%E8%8C%83%E5%9B%B4.2.%E6%B1%82%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%E5%9C%A8%5B0%2C1%5D%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E6%B1%82%E8%AF%A6%E8%A7%A3)
已知a>0,函数f(x)=ln(2-x)+ax^2(x<2),1.设曲线y=f(x)在点(1,f(1))处的切线为l,若l与园(x+1)^2+y^2=1相离,求a的范围.2.求函数f(x)在[0,1]的最大值求详解
已知a>0,函数f(x)=ln(2-x)+ax^2(x<2),
1.设曲线y=f(x)在点(1,f(1))处的切线为l,若l与园(x+1)^2+y^2=1相离,求a的范围.
2.求函数f(x)在[0,1]的最大值
求详解
已知a>0,函数f(x)=ln(2-x)+ax^2(x<2),1.设曲线y=f(x)在点(1,f(1))处的切线为l,若l与园(x+1)^2+y^2=1相离,求a的范围.2.求函数f(x)在[0,1]的最大值求详解
f'(x)=-1/(2-x)+2ax
在点(1,f(1))处的切线斜率
f'(1)=-1/(2-1)+2a=2a-1
而f(1)=a
则直线方程为:
y-a=(2a-1)(x-1)
l与园(x+1)^2+y^2=1相离,则
圆心到直线距离大于半径
|2(2a-1)-a|/√[(2a-1)^2+1]>1
则|3a-2|/√[(2a-1)^2+1]>1
5a^2-8a+2>0
则(4-√6)/5
f'(x)=-1/(2-x)+2ax
在点(1,f(1))处的切线斜率
f'(1)=-1/(2-1)+2a=2a-1
而f(1)=a
则直线方程为:
y-a=(2a-1)(x-1)
l与园(x+1)^2+y^2=1相离,则
圆心到直线距离大于半径
|2(2a-1)-a|/√[(2a-1)^2+1]>1
则|3a-2|/√[(2...
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f'(x)=-1/(2-x)+2ax
在点(1,f(1))处的切线斜率
f'(1)=-1/(2-1)+2a=2a-1
而f(1)=a
则直线方程为:
y-a=(2a-1)(x-1)
l与园(x+1)^2+y^2=1相离,则
圆心到直线距离大于半径
|2(2a-1)-a|/√[(2a-1)^2+1]>1
则|3a-2|/√[(2a-1)^2+1]>1
5a^2-8a+2>0
则(4-√6)/5又a>0
则:0f'(x)=-1/(2-x)+2ax=0
则:2ax^2-4ax+1=0
x=a+√(a^2-a/2),
x=a-√(a^2-a/2),
f''(x)=-1/(2-x)^2+2a
带入驻点值,则
f''(a-√(a^2-a/2))<0,取得最大值
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