已知函数f(x)=ax²+bx+1(a,b为实数,a≠0,x∈R),

3个回答

  • 1.

    f(-1) = a - b + 1 = 0, b = a+1

    且函数f(x)的值域为[0,﹢无穷), 显然x = -1为该抛物线的对称轴,且抛物线开口向上, a >0

    对称轴x = -1 = -b/(2a) = -(a+1)/(2a)

    a = 1

    f(x)=x²+2x+1

    F(x) = x²+2x+1, x > 0

    = -(x²+2x+1), x < 0

    2.

    g(x) = x²+2x+1 - kx = x²+(2 - k)x+1

    g(x)的对称轴为x = -(2-k)/(2*1) = (k-2)/2

    当x∈[-2,2]时, g(x)是单调函数, 对称轴x = (k-2)/2 ≥2 或 x = (k-2)/2 ≤ -2

    即k ≥ 6或k ≤ -2

    3

    不影响结果,不妨设m > 0

    mn < 0, n < 0

    m + n > 0, m > |n|, m² > n²

    f(x)为偶函数, b = 0, f(x) = ax² + 1

    F(m) = f(m) = am²+ 1

    F(n) = -f(n) = -an² - 1

    F(m)+F(n) = a(m² - n²) > 0