高等数学!急急急,求答案
2个回答
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(1)
lim(x->∞) (1 +1/x)^10 =1
lim(x->∞) (1 +1/10)^x ->+∞
ans: A
(2)
f(x) = ∫(1->x) sint/t dt
f'(x) = sinx/x
∫(0->1) xf(x) dx
=(1/2)∫(0->1) f(x) dx^2
=(1/2)x^2.f(x)]|(0->1) -(1/2)∫(0->1) x^2.f'(x) dx
=(1/2)x^2.f(x)]|(0->1) -(1/2)∫(0->1) xsinx dx
=(1/2)f(1) +(1/2)∫(0->1) x dcosx
=0 + (1/2)[xcosx]|(0->1) -(1/2)∫(0->1) cosx dx
=(1/2)cos1 - (1/2)[sinx]|(0->1)
=(1/2)cos1 - (1/2)sin1
ans :B
(3)
f(x) = x^p +(1-x)^p
f'(x) =px^(p-1) -p(1-x)^(p-1)
f'(x)=0
px^(p-1) -p(1-x)^(p-1)=0
x= 1-x
x=1/2
f'(x)| x=1/2 + <0 , f'(x)|x=1/2- >0
x=1/2 (min)
min f(x)
=f(1/2)
= (1/2)^p +(1/2)^p
=(1/2)^(p-1)
ans: C
lim(x->∞) (1 +1/x)^10 =1
lim(x->∞) (1 +1/10)^x ->+∞
ans: A
(2)
f(x) = ∫(1->x) sint/t dt
f'(x) = sinx/x
∫(0->1) xf(x) dx
=(1/2)∫(0->1) f(x) dx^2
=(1/2)x^2.f(x)]|(0->1) -(1/2)∫(0->1) x^2.f'(x) dx
=(1/2)x^2.f(x)]|(0->1) -(1/2)∫(0->1) xsinx dx
=(1/2)f(1) +(1/2)∫(0->1) x dcosx
=0 + (1/2)[xcosx]|(0->1) -(1/2)∫(0->1) cosx dx
=(1/2)cos1 - (1/2)[sinx]|(0->1)
=(1/2)cos1 - (1/2)sin1
ans :B
(3)
f(x) = x^p +(1-x)^p
f'(x) =px^(p-1) -p(1-x)^(p-1)
f'(x)=0
px^(p-1) -p(1-x)^(p-1)=0
x= 1-x
x=1/2
f'(x)| x=1/2 + <0 , f'(x)|x=1/2- >0
x=1/2 (min)
min f(x)
=f(1/2)
= (1/2)^p +(1/2)^p
=(1/2)^(p-1)
ans: C
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