(2)设直线l是曲线y=f(x)切线,证明直线l与直线x=1和直线y=x所围三角形的面积为定值题干是,设函数f(x)=ax+1/(x+b) (a,b是整数),曲线y=f(x)在点(2,f(2))处的切线方程为y=3,(1)求f(x)的解析式;提问字
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![(2)设直线l是曲线y=f(x)切线,证明直线l与直线x=1和直线y=x所围三角形的面积为定值题干是,设函数f(x)=ax+1/(x+b) (a,b是整数),曲线y=f(x)在点(2,f(2))处的切线方程为y=3,(1)求f(x)的解析式;提问字](/uploads/image/z/1731082-58-2.jpg?t=%EF%BC%882%EF%BC%89%E8%AE%BE%E7%9B%B4%E7%BA%BFl%E6%98%AF%E6%9B%B2%E7%BA%BFy%3Df%28x%29%E5%88%87%E7%BA%BF%2C%E8%AF%81%E6%98%8E%E7%9B%B4%E7%BA%BFl%E4%B8%8E%E7%9B%B4%E7%BA%BFx%3D1%E5%92%8C%E7%9B%B4%E7%BA%BFy%3Dx%E6%89%80%E5%9B%B4%E4%B8%89%E8%A7%92%E5%BD%A2%E7%9A%84%E9%9D%A2%E7%A7%AF%E4%B8%BA%E5%AE%9A%E5%80%BC%E9%A2%98%E5%B9%B2%E6%98%AF%2C%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3Dax%2B1%2F%28x%2Bb%29+%28a%2Cb%E6%98%AF%E6%95%B4%E6%95%B0%29%2C%E6%9B%B2%E7%BA%BFy%3Df%28x%29%E5%9C%A8%E7%82%B9%282%2Cf%282%29%29%E5%A4%84%E7%9A%84%E5%88%87%E7%BA%BF%E6%96%B9%E7%A8%8B%E4%B8%BAy%3D3%2C%EF%BC%881%EF%BC%89%E6%B1%82f%28x%29%E7%9A%84%E8%A7%A3%E6%9E%90%E5%BC%8F%EF%BC%9B%E6%8F%90%E9%97%AE%E5%AD%97)
(2)设直线l是曲线y=f(x)切线,证明直线l与直线x=1和直线y=x所围三角形的面积为定值题干是,设函数f(x)=ax+1/(x+b) (a,b是整数),曲线y=f(x)在点(2,f(2))处的切线方程为y=3,(1)求f(x)的解析式;提问字
(2)设直线l是曲线y=f(x)切线,证明直线l与直线x=1和直线y=x所围三角形的面积为定值
题干是,设函数f(x)=ax+1/(x+b) (a,b是整数),曲线y=f(x)在点(2,f(2))处的切线方程为y=3,(1)求f(x)的解析式;提问字数不够,所以只把最主要的问题写上了!
(2)设直线l是曲线y=f(x)切线,证明直线l与直线x=1和直线y=x所围三角形的面积为定值题干是,设函数f(x)=ax+1/(x+b) (a,b是整数),曲线y=f(x)在点(2,f(2))处的切线方程为y=3,(1)求f(x)的解析式;提问字
1.
f(x)'=a-1/(x+b)^2
f(2)'=a-1/(2+b)^2=0
a、b是整数,所以1/(2+b)^2=1,否则不可能满足题意
所以b+2=+-1,b=-1或b=-3
a=1,又f(2)=3,所以3=2+1/(2+b),b=-1
所以f(x)=4x-1/(x-1)
2.设x=t点处的切线为y=f(t)'(x-t)+f(t)
即y=[1-1/(t-1)^2](x-t)+t+1/(t-1)=x-x/(t-1)^2+t/(t-1)^2+1/(t-1)
它与x=1的焦点即将x=1代入有,y=1-1/(t-1)^2+t/(t-1)^2+1/(t-1)=1+(-1+t+t-1)/(t-1)^2=1+(2t-2)/(t-1)^2=1+2/(t-1),交点在(1,1+2/(t-1))
它与y=x的交点即将y=x代入有,x=x-x/(t-1)^2+t/(t-1)^2+1/(t-1)有x=(t-1)^2*[t/(t-1)^2+1/(t-1)]=t+t-1=2t-1,交点在(2t-1,2t-1)
y=x与x=1的交点在(1,1)
所以该三角形的面积S=1/2×|[1+2/(t-1)-1]×[2t-1-1]|=1/2×|2/(t-1) ×2(t-1)|=2