Question

已知a、b、c是實(shí)數(shù),函數(shù)f(x)=ax²+bx+c,g(x)=ax+b,當(dāng)-1≤x≤1時,│f(x)│≤1.

(Ⅰ)證明:│c│≤l;

(Ⅱ)證明:當(dāng)-1≤x≤1時,│g(x)│≤2;

(Ⅲ)設(shè)a>0,當(dāng)-1≤x≤1時,g(x)的最大值為2,求f(x).

Answer 1

 (Ⅰ)證明:由條件當(dāng)-1≤x≤1時,│f(x)│≤1,取x=0得

│c│=│f(0)│≤1,

即│c│≤1.                                                 

(Ⅱ)證法一:

當(dāng)a>0時,g(x)=ax+b在[-1,1]上是增函數(shù),

∴g(-1)≤g(x)≤g(1),

∵│f(x)│≤1(-1≤x≤1),│c│≤1,

∴g(1)=a+b=f(1)-c≤│f(1)│+│c│≤2,

g(-1)=-a+b=-f(-1)+c≥-(│f(-1)│+│c│≥－2,

由此得│g(x)│≤2;                                                       

當(dāng)a<0時,g(x)=ax+b在[-1,1]上是減函數(shù),

∴g(-1)≥g(x)≥g(1),

∵│f(x)│≤1(-1≤x≤1),│c│≤1,

∴g(-1)=-a+b=-f(-1)+c≤│f(-1)│+│c│≤2,

g(1)=a+b=f(1)-c≥-(│f(1)│+│c│)≥-2,

由此得│g(x)│≤2;                                                        

當(dāng)a=0時,g(x)=b,f(x)=bx+c.

∵-1≤x≤1,

∴│g(x)│=│f(1)-c│≤│f(1)│+│c│≤2.

綜上得│g(x)│≤2. 

證法二：

由可得

=

=

當(dāng)時，有

                                                                    

根據(jù)含絕對值的不等式的性質(zhì),得

即│g(x)│≤2.                   

(Ⅲ)因?yàn)閍>0,g(x)在[-1,1]上是增函數(shù),當(dāng)x=1時取得最大值2,

即g(1)=a+b=f(1)-f(0)=2.①

∵-1≤f(0)=f(1)-2≤1-2=-1,

∴c=f(0)=-1.                                

因?yàn)楫?dāng)-1≤x≤1時,f(x)≥-1,即f(x)≥f(0),

根據(jù)二次函數(shù)的性質(zhì),直線x=0為f(x)的圖象的對稱軸,由此得

即.

由①   得a=2.

所以   f(x)=2x²-1.