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求a,b的值已知fx=x^2+(2+loga)x+logb,f(

2007-12-15 09:25:17雾***
已知fx=x^2+(2+loga)x+logb,f(-1)=-2且fx>=2x恒成立求a,b的值已知fx=x^2+(2+loga)x+logb,f(-1)=-2且fx=2x恒成立:f(x)=x²+(2+lga)x+lgb f?

最佳回答

  • f(x)=x²+(2+lga)x+lgb f(-1) =1-2-lga+lgb =lgb-lga -1 =-2 ====>lgb=lga-1........(1) f(x)≥2x ==>x²+(2+lga)x+lgb≥2x x²+xlga+lgb≥0恒成立 这个关于x的二次方程△≤0 ==>(lga)²-4lgb≤0 (1)==> (lga)² -4(lga-1)≤0 (lga -2)²≤0 =>只有lga=2满足 ==>a=100 代入(1)==>b=10
    2007-12-15 10:07:54
  • f(x) = x² + (2+lga)x + lgb f(-1) = -2  1 - 2 - lga + lgb = -2  lgb = lga -1     f(x) > = 2x 恒成立  x² + (lga)x + lgb > = 0 恒成立 (lga)² - 4*lab (lga)² - 4(lga - 1) (lga)² - 4lga + 4 (lga - 2)² lga = 2 ===> lgb = 1 所以 a = 100 , b = 10
    2007-12-15 09:54:29
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