1.對(duì)f(x)求導(dǎo),f'(x)=[(mx+t)'*(x²+1)-(mx+t)*(x²+1)']/(x²+1)²
=[m(x²+1)-2x(mx+t)]/(x²+1)²
在點(diǎn)M(0,f(0))處的切線(xiàn)斜率為1
即f'(0)=1
所以(m-0)/(0+1)²=1
m=1
2.f(x)=(x+t)/(x^2+1)≤2t =>t≥x/(2x^2+1)
s(x)=x/(2x^2+1)在區(qū)間[1,2]遞減,依題意只須t≥s(2)即可,則有t≥2/9
3.由韋達(dá)定理,a+b=-2t,ab=-1,b-a=2√(t^2+1)
若對(duì)于任意x∈[a,b],總存在x1,x2∈[a,b]使得f(x1)≤f(x)≦f(x2)恒成立,說(shuō)明x1,x2分別是區(qū)間[a,b]f(x)的最小最大值點(diǎn).
由問(wèn)1可得,f'(x)=-(x^2+2tx-1)/(x^2+1)^2,注意h(x)=x^2+2tx-1,不難發(fā)現(xiàn)函數(shù)f(x)在區(qū)間[a,b]
f'(x)≥0,f(x)遞增,則x1=a,x2=b
則g(t)=f(x2)-f(x1)=f(b)-f(a)=(b+t)/(b^2+1)-(a+t)/(a^2+1)
=(ba^2+b+ta^2-ab^2-a-tb^2)/(a^2b^2+a^2+b^2+1)
帶入數(shù)據(jù)得g(t)=√(t^2+1)=√5 =>t=±2