Question

已知函数f（x）=x+t/x,和点P(1,0),过点P作f（x）的两条切线PM,PN.设|MN|=g(t),求g(t)。

已知函数f（x）=x+t/x,和点P(1,0),过点P作f（x）的两条切线PM,PN.设|MN|=g(t),求g(t)

Answer 1

(1)
f(x)=x+t/x
f'(x)=1-t/x^2
设过A点与f(x)相切直线切点为(a,b)
切线斜率k=b/(a-1)=f'(a)=1-t/a^2
又f(a)=a+t/a=b

(a+t/a)/(a-1)=1-t/a^2
a^2+2at-t=0
t>0方程有两根,即为分别M,N横坐标,设为x1,x1
则M,N纵坐标分别为y1=x1+t/x1,y2=x2+t/x2

y1+y2=x1+x2+t/x1+t/x2=(x1+x2)+t(x1+x2)/x1x2
=-2t+t(-2t)/(-t)=0
y1y2=(x1+t/x1)(x2+t/x2)=x1x2+tx1/x2+tx2/x1+t^2/(x1x2)
=-t+t[(x1+x2)^2-2x1x2]/(-t)+t^2/(-t)
=-4t^2-4t

|MN|^2=(x1-x2)^2+(y1-y2)^2
=(x1+x2)^2-4x1x2+(y1+y2)^2-4y1y2
=(-2t)^2+4t+0+4(4t^2+4t)
=20(t^2+t)

g(t)=|MN|=2√[5(t^2+t)]

(2)
沿用(1)的设定和结论,M,N为不同两点x1≠x2
假设M,N,A共线,则直线AM,AN斜率相等
f'(x1)=f'(x2)
1-t/x1^2=1-t/x2^2
x1^2=x2^2
x1=-x2
又:
(x1+t/x1-1)/x1=(x2+t/x2-1)/x2
x1+t/x1-1=x1+t/x1+1
无解
所以不存在t使M,N,A共线

(3)
应当是在(1)的条件下
易知g(t)=2√[5(t^2+t)]在区间(2,n+64/n]单调递增
使不等式g(a1)+g(a2)+......g(am)<g(am+1)成立,
m要最大则a(i)尽量小,a(m+1)尽量大
又n+64/n>=16,而对于任意正整数n都要成立
所以必须满足g(a1)+g(a2)+......g(am)<g(16)
又g(a1)+g(a2)+......g(am)>mg(2)
所以:mg(2)<g(16)
m2√30<2√1360
m<√(136/3),m<=6