设函数f(x)=ax+b/x(a,b∈R),若f(x)在点(1,f(1))处的切线的斜率为1,.用a表示b.设函数f(x)=ax+b/x(a,b∈R),若f(x)在点(1,f(1))处的切线的斜率为1,.用a表示b.设g(x)=lnx-f(x),若g
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![设函数f(x)=ax+b/x(a,b∈R),若f(x)在点(1,f(1))处的切线的斜率为1,.用a表示b.设函数f(x)=ax+b/x(a,b∈R),若f(x)在点(1,f(1))处的切线的斜率为1,.用a表示b.设g(x)=lnx-f(x),若g](/uploads/image/z/1980773-53-3.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%3Dax%2Bb%2Fx%EF%BC%88a%2Cb%E2%88%88R%EF%BC%89%2C%E8%8B%A5f%EF%BC%88x%EF%BC%89%E5%9C%A8%E7%82%B9%EF%BC%881%2Cf%EF%BC%881%EF%BC%89%EF%BC%89%E5%A4%84%E7%9A%84%E5%88%87%E7%BA%BF%E7%9A%84%E6%96%9C%E7%8E%87%E4%B8%BA1%2C.%E7%94%A8a%E8%A1%A8%E7%A4%BAb.%E8%AE%BE%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%3Dax%2Bb%2Fx%EF%BC%88a%2Cb%E2%88%88R%EF%BC%89%2C%E8%8B%A5f%EF%BC%88x%EF%BC%89%E5%9C%A8%E7%82%B9%EF%BC%881%2Cf%EF%BC%881%EF%BC%89%EF%BC%89%E5%A4%84%E7%9A%84%E5%88%87%E7%BA%BF%E7%9A%84%E6%96%9C%E7%8E%87%E4%B8%BA1%2C.%E7%94%A8a%E8%A1%A8%E7%A4%BAb.%E8%AE%BEg%EF%BC%88x%EF%BC%89%3Dlnx-f%EF%BC%88x%EF%BC%89%2C%E8%8B%A5g)
设函数f(x)=ax+b/x(a,b∈R),若f(x)在点(1,f(1))处的切线的斜率为1,.用a表示b.设函数f(x)=ax+b/x(a,b∈R),若f(x)在点(1,f(1))处的切线的斜率为1,.用a表示b.设g(x)=lnx-f(x),若g
设函数f(x)=ax+b/x(a,b∈R),若f(x)在点(1,f(1))处的切线的斜率为1,.用a表示b.
设函数f(x)=ax+b/x(a,b∈R),若f(x)在点(1,f(1))处的切线的斜率为1,.用a表示b.设g(x)=lnx-f(x),若g(x)≤-1对定义域内的x恒成立.(1)求实数a的取值范围.(2)对任意的θ∈[0,π/2),证明:g(1-sinθ)≤g(1+sinθ)
设函数f(x)=ax+b/x(a,b∈R),若f(x)在点(1,f(1))处的切线的斜率为1,.用a表示b.设函数f(x)=ax+b/x(a,b∈R),若f(x)在点(1,f(1))处的切线的斜率为1,.用a表示b.设g(x)=lnx-f(x),若g
f'(x)=a-b/x^2
由f'(1)=a-b=1 ==>b=a-1
2
g(x)=lnx- ax-(a-1)/x (x>0)
若g(x)≤-1对定义域内的x恒成立
需g(x)max≤-1
g'(x)=1/x-a+(a-1)/x^2=[-ax^2+x+(a-1)]/x^2
= (x-1)[-ax-(a-1)]/x^2
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f'(x)=a-b/x^2
由f'(1)=a-b=1 ==>b=a-1
2
g(x)=lnx- ax-(a-1)/x (x>0)
若g(x)≤-1对定义域内的x恒成立
需g(x)max≤-1
g'(x)=1/x-a+(a-1)/x^2=[-ax^2+x+(a-1)]/x^2
= (x-1)[-ax-(a-1)]/x^2
a=0时,g'(x)=(x-1)/x^2,
x∈(0,1),g'(x)<0,g(x)递减,x∈(1,+∞),g'(x)>0,g(x)递增
g(x)无最大值
a≠0时,g'(x)=-a(x-1)[x-(1/a-1)]/x^2
当a<0时,1/a-1<-1 ,x>1时,g'(x)>0, g(x)递增,g(x)无最大值
当01,
x∈(0,1),g'(x)<0,g(x)递减,x∈(1,1/a -1),g'(x)> 0,g(x)递增
x∈(1/a-1,+∞),g'(x)<0 ,g(x)递减
g(x)max=g(1/a-1)=ln(1/a-1)-a(1/a-1)-(a-1)/(1/a-1)
=ln(1/a-1)-1+2a>0 不合题意
当a=1/2时,g'(x)=-(x-1)^2/(2x^2)≤0 ,g(x)递减,不合题意
当1>a>1/2时, 0<1/a-1<1
0<1/a-1<1, (0,1/a-1) 减,(1/a-1,1)增,(1,+∞)减
g(x)max=g(1)=-2a+1≤-1 ==>a≥1 与1/2 a≥1时,1/a-1≤0
(0,1)增,(1,+∞)减
g(x)max=g(1)=-2a+1≤-1 ==> a≥1 符合题意
∴符合条件的 实数a的取值范围是a≥1 。
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