已知函数f(x)=ax 2 +bx+c ( 1 3 ≤a≤1)的图象过点A(0,1)且直线2x+y-1=0与y

1个回答

  • (1)f(0)=1c=1(2分)

    y=-2x+1

    y=a x 2 +bx+1 ax 2+(b+2)x=0有等根(5分)

    b=-2(7分)

    (2)f(x)=ax 2-2x+1=a(x-

    1

    a ) 2+1-

    1

    a (8分)

    1

    3 ≤a≤1∴1≤

    1

    a ≤3恒有N(a)=1-

    1

    a (10分)

    当1≤

    1

    a ≤2即

    1

    2 ≤a≤1时M(a)=9a-5

    M(a)-N(a)=2 a=

    7

    9 a=

    4-

    7

    9 <

    1

    2 (舍去)(12分)

    当2<

    1

    a ≤3即

    1

    3 ≤a<

    1

    2 时M(a)=a-1

    M(a)-N(a)=2a=2±

    3 >

    1

    2 都舍去

    综上 a=

    4+

    7

    9 (15分)