已知函数f(x)=x^2+a,(x ∈R).(1)对任意的x1,x2∈R,比较1/2[f(x1)+f(x2)]与f[(x1+x2)/2]的大小(2)若x∈[-1,1]时,有f(x)的绝对值小于等于1,试求实数a的取值范围
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![已知函数f(x)=x^2+a,(x ∈R).(1)对任意的x1,x2∈R,比较1/2[f(x1)+f(x2)]与f[(x1+x2)/2]的大小(2)若x∈[-1,1]时,有f(x)的绝对值小于等于1,试求实数a的取值范围](/uploads/image/z/5347133-53-3.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dx%5E2%2Ba%2C%28x+%E2%88%88R%29.%281%29%E5%AF%B9%E4%BB%BB%E6%84%8F%E7%9A%84x1%2Cx2%E2%88%88R%2C%E6%AF%94%E8%BE%831%2F2%5Bf%28x1%29%2Bf%28x2%29%5D%E4%B8%8Ef%5B%28x1%2Bx2%29%2F2%5D%E7%9A%84%E5%A4%A7%E5%B0%8F%EF%BC%882%EF%BC%89%E8%8B%A5x%E2%88%88%5B-1%2C1%5D%E6%97%B6%2C%E6%9C%89f%28x%29%E7%9A%84%E7%BB%9D%E5%AF%B9%E5%80%BC%E5%B0%8F%E4%BA%8E%E7%AD%89%E4%BA%8E1%2C%E8%AF%95%E6%B1%82%E5%AE%9E%E6%95%B0a%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
已知函数f(x)=x^2+a,(x ∈R).(1)对任意的x1,x2∈R,比较1/2[f(x1)+f(x2)]与f[(x1+x2)/2]的大小(2)若x∈[-1,1]时,有f(x)的绝对值小于等于1,试求实数a的取值范围
已知函数f(x)=x^2+a,(x ∈R).(1)对任意的x1,x2∈R,比较1/2[f(x1)+f(x2)]与f[(x1+x2)/2]的大小
(2)若x∈[-1,1]时,有f(x)的绝对值小于等于1,试求实数a的取值范围
已知函数f(x)=x^2+a,(x ∈R).(1)对任意的x1,x2∈R,比较1/2[f(x1)+f(x2)]与f[(x1+x2)/2]的大小(2)若x∈[-1,1]时,有f(x)的绝对值小于等于1,试求实数a的取值范围
(1)f(x)=x^2+a
所以1/2[f(x1)+f(x2)]=1/2[(x1)²+a+(x2)²+a]
=[(x1)²+(x2)²]/2+a
f[(x1+x2)/2]=[(x1+x2)/2]²+a
=[(x1)²+2x1x2+(x2)²]/4+a
所以1/2[f(x1)+f(x2)]-f[(x1+x2)/2]
=[(x1)²+(x2)²]/2+a-{[(x1)²+2x1x2+(x2)²]/4+a}
=[(x1)²+(x2)²]/2--[(x1)²+2x1x2+(x2)²]/4
=[2(x1)²+2(x2)²-(x1)²-2x1x2-(x2)²]/4
=[(x1)+(x2)-2x1x2]/4
=(x1-x2)²/4≥0
所以1/2[f(x1)+f(x2)]-f[(x1+x2)/2]≥0
(2)x∈[-1,1] 所以0≤x²≤1
f(x)的绝对值小于等于1
所以-1≤f(x)≤1 即-1≤x²+a≤1
所以-1-x²≤a≤1-x²
所以-2≤a≤1