已知奇函数f(x)的定义域为R,且f(x)在[0,+∞)上为增函数,当0≤θ≤π/2时,是否存在实数m奇函数f(x)的定义域为R,且在[0,+∞)上是增函数,当0≤θ≤π/2时,是否存在实数m,使f(4m-2mcosθ)-f(2sin²θ
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![已知奇函数f(x)的定义域为R,且f(x)在[0,+∞)上为增函数,当0≤θ≤π/2时,是否存在实数m奇函数f(x)的定义域为R,且在[0,+∞)上是增函数,当0≤θ≤π/2时,是否存在实数m,使f(4m-2mcosθ)-f(2sin²θ](/uploads/image/z/1993578-42-8.jpg?t=%E5%B7%B2%E7%9F%A5%E5%A5%87%E5%87%BD%E6%95%B0f%28x%29%E7%9A%84%E5%AE%9A%E4%B9%89%E5%9F%9F%E4%B8%BAR%2C%E4%B8%94f%28x%29%E5%9C%A8%5B0%2C%EF%BC%8B%E2%88%9E%EF%BC%89%E4%B8%8A%E4%B8%BA%E5%A2%9E%E5%87%BD%E6%95%B0%2C%E5%BD%930%E2%89%A4%CE%B8%E2%89%A4%CF%80%2F2%E6%97%B6%2C%E6%98%AF%E5%90%A6%E5%AD%98%E5%9C%A8%E5%AE%9E%E6%95%B0m%E5%A5%87%E5%87%BD%E6%95%B0f%28x%29%E7%9A%84%E5%AE%9A%E4%B9%89%E5%9F%9F%E4%B8%BAR%2C%E4%B8%94%E5%9C%A8%5B0%2C%2B%E2%88%9E%29%E4%B8%8A%E6%98%AF%E5%A2%9E%E5%87%BD%E6%95%B0%2C%E5%BD%930%E2%89%A4%CE%B8%E2%89%A4%CF%80%2F2%E6%97%B6%2C%E6%98%AF%E5%90%A6%E5%AD%98%E5%9C%A8%E5%AE%9E%E6%95%B0m%2C%E4%BD%BFf%EF%BC%884m-2mcos%CE%B8%EF%BC%89-f%EF%BC%882sin%26%23178%3B%CE%B8)
已知奇函数f(x)的定义域为R,且f(x)在[0,+∞)上为增函数,当0≤θ≤π/2时,是否存在实数m奇函数f(x)的定义域为R,且在[0,+∞)上是增函数,当0≤θ≤π/2时,是否存在实数m,使f(4m-2mcosθ)-f(2sin²θ
已知奇函数f(x)的定义域为R,且f(x)在[0,+∞)上为增函数,当0≤θ≤π/2时,是否存在实数m
奇函数f(x)的定义域为R,且在[0,+∞)上是增函数,当0≤θ≤π/2时,是否存在实数m,使f(4m-2mcosθ)-f(2sin²θ+2)>f(0)对所有θ∈[0,π/2]求出所有适合条件的实数m.
m>4-2倍根号2.
已知奇函数f(x)的定义域为R,且f(x)在[0,+∞)上为增函数,当0≤θ≤π/2时,是否存在实数m奇函数f(x)的定义域为R,且在[0,+∞)上是增函数,当0≤θ≤π/2时,是否存在实数m,使f(4m-2mcosθ)-f(2sin²θ
由题意,f(x)在x=0处有定义且在[0,+∞)上是增函数,
故f(x)在(-∞,+∞)上连续且为增函数
由f(0)=-f(-0),得f(0)=0
f(cos2θ-3)+f(4m-2mcosθ)>f(0)=0
移向变形得
f(cos2θ-3)>-f(4m-2mcosθ)=f(2mcosθ-4m)
∴由f(x)(-∞,+∞)上连续且为增函数,得
cos2θ-3>2mcosθ-4m
2cos²θ-4-2mcosθ+4m>0
cos²θ-mcosθ+(2m-2)>0
根据题意,θ∈[0,π/2]时,cosθ∈[0,1]
令t=cosθ∈[0,1]
则,题目变成t∈[0,1]时,t²-mt+(2m-2)>0恒成立,求m的取值范围
令f(t)=t²-mt+(2m-2),此函数对应的抛物线开口向上,对称轴t=m/2,
分类讨论:
①当此抛物线对称轴t=m/2在区间[0,1]内时,m∈[0,2],
函数最小值(2m-2)-m²/4>0即可,此时m²-8m+81,与m2,
只要f(1)>0即可,此时1-m+2m-2=m-1>0,推出m>1,
∴m>2
综上所述,m的取值范围是(4-2√2,+∞)