设二次函数f(x)=ax^2+bx+c的图像与x轴交于点(-1,0),且满足[f(x)-x]*[f(x)-(x^2+1)/2]≤0恒成立 (1)求f(1)的值(2)求f(x)的解析式(3)求证∑(1/f(k))>2n/(n+2).
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![设二次函数f(x)=ax^2+bx+c的图像与x轴交于点(-1,0),且满足[f(x)-x]*[f(x)-(x^2+1)/2]≤0恒成立 (1)求f(1)的值(2)求f(x)的解析式(3)求证∑(1/f(k))>2n/(n+2).](/uploads/image/z/3140261-53-1.jpg?t=%E8%AE%BE%E4%BA%8C%E6%AC%A1%E5%87%BD%E6%95%B0f%28x%29%3Dax%5E2%2Bbx%2Bc%E7%9A%84%E5%9B%BE%E5%83%8F%E4%B8%8Ex%E8%BD%B4%E4%BA%A4%E4%BA%8E%E7%82%B9%28-1%2C0%29%2C%E4%B8%94%E6%BB%A1%E8%B6%B3%5Bf%28x%29-x%5D%2A%5Bf%28x%29-%28x%5E2%2B1%29%2F2%5D%E2%89%A40%E6%81%92%E6%88%90%E7%AB%8B+%281%29%E6%B1%82f%281%29%E7%9A%84%E5%80%BC%EF%BC%882%EF%BC%89%E6%B1%82f%EF%BC%88x%EF%BC%89%E7%9A%84%E8%A7%A3%E6%9E%90%E5%BC%8F%EF%BC%883%EF%BC%89%E6%B1%82%E8%AF%81%E2%88%91%281%2Ff%28k%29%29%3E2n%2F%28n%2B2%29.)
设二次函数f(x)=ax^2+bx+c的图像与x轴交于点(-1,0),且满足[f(x)-x]*[f(x)-(x^2+1)/2]≤0恒成立 (1)求f(1)的值(2)求f(x)的解析式(3)求证∑(1/f(k))>2n/(n+2).
设二次函数f(x)=ax^2+bx+c的图像与x轴交于点(-1,0),且满足[f(x)-x]*[f(x)-(x^2+1)/2]≤0恒成立
(1)求f(1)的值
(2)求f(x)的解析式
(3)求证∑(1/f(k))>2n/(n+2)
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设二次函数f(x)=ax^2+bx+c的图像与x轴交于点(-1,0),且满足[f(x)-x]*[f(x)-(x^2+1)/2]≤0恒成立 (1)求f(1)的值(2)求f(x)的解析式(3)求证∑(1/f(k))>2n/(n+2).
(1)由于[f(x)-x]*[f(x)-(x^2+1)/2]≤0恒成立,将x=1待入得出(a+b+c-1)^2≤0,因此(a+b+c-1)^2=0,得到f(1)=1.
(2)由f(-1)=0,f(1)=1,得到b=1/2,a+c=1/2.将这两个结果带出那个不等式得到[ax^2+(1/2-1)x+c]*[(a-1/2)x^2+1/2x+c-1/2]≤0.化简下
[ax^2-1/2x+c]*[(-c)x^2+1/2x+(-a)]≤0,
再得[ax^2-1/2x+c]*[cx^2+1/2x-a]》0,要使这个恒成立只要a=c=1/4.
f(x)=1/4x^2+1/2x+1/4.
(3)∑(1/f(k))=4∑1/(k^2+2k+1)=4∑1/(k+1)^2>4∑1/(k+1)(k+2)=4[(1/2-1/3)+(1/3-1/4)……+1/(n+1)-1/(n+2)]=4(1/2-1/(n+2))=2n/(n+2).
证完!