Question

高中数学17题

Answer 1

17.f(0)=|b|<=1/2,-1/2<=b<=1/2,
f(4)=|4+4a+b|<=1/2,
∴-1/2<=4+4a+b<=1/2,
-9/2-b<=4a<=-7/2-b,
-5/4<=-9/8-b/4<=a<=-7/8-b/4<=-3/4.①
∴4/5<=-1/a<=4/3,
f(-1/a)=|2√(-1/a)-1+b|<=1/2,
-1/2<=2√(-1/a)-1+b<=1/2,
1/2-b<=2√(-1/a)<=3/2-b,
平方得1/4-b+b^2<=-4/a<=9/4-3b+b^2,
-4/(1/4-b+b^2)<=a<=-4/(9/4-3b+b^2),②
由①②，-9/8-b/4<=-4/(9/4-3b+b^2)，
两边都乘以-32（9/4-3b+b^2）,得(9+2b)(9-12b+4b^2)>=128,
g(b)=8b^3+12b^2-90b-47>=0,
g'(b)=24b^2+24b-90<0,g(b)是减函数，
g(b)<=g(-1/2)=-1+3+45-47=0,
∴b=-1/2,a=-1,
∴a+2b=-2.