若函数f(x)=(sinx+cosx)+2cos²x-m在[0,π/2]上有零点,求m的取值范围更正:f(x)=(sinx+cosx)²+2cos²x-m在[0,π/2]上有零点,求m的取值范围
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![若函数f(x)=(sinx+cosx)+2cos²x-m在[0,π/2]上有零点,求m的取值范围更正:f(x)=(sinx+cosx)²+2cos²x-m在[0,π/2]上有零点,求m的取值范围](/uploads/image/z/2535263-71-3.jpg?t=%E8%8B%A5%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%3D%EF%BC%88sinx%2Bcosx%EF%BC%89%2B2cos%26%23178%3Bx-m%E5%9C%A8%5B0%2C%CF%80%2F2%5D%E4%B8%8A%E6%9C%89%E9%9B%B6%E7%82%B9%2C%E6%B1%82m%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4%E6%9B%B4%E6%AD%A3%EF%BC%9Af%EF%BC%88x%EF%BC%89%3D%EF%BC%88sinx%2Bcosx%EF%BC%89%26%23178%3B%2B2cos%26%23178%3Bx-m%E5%9C%A8%5B0%EF%BC%8C%CF%80%2F2%5D%E4%B8%8A%E6%9C%89%E9%9B%B6%E7%82%B9%EF%BC%8C%E6%B1%82m%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
若函数f(x)=(sinx+cosx)+2cos²x-m在[0,π/2]上有零点,求m的取值范围更正:f(x)=(sinx+cosx)²+2cos²x-m在[0,π/2]上有零点,求m的取值范围
若函数f(x)=(sinx+cosx)+2cos²x-m在[0,π/2]上有零点,求m的取值范围
更正:f(x)=(sinx+cosx)²+2cos²x-m在[0,π/2]上有零点,求m的取值范围
若函数f(x)=(sinx+cosx)+2cos²x-m在[0,π/2]上有零点,求m的取值范围更正:f(x)=(sinx+cosx)²+2cos²x-m在[0,π/2]上有零点,求m的取值范围
题目有误吧,
f(x)=(sinx+cosx)²+2cos²x-m在[0,π/2]上有零点
即方程 (sinx+cosx)²+2cos²x-m=0在[0,π/2]上有解
即求 m=(sinx+cosx)²+2cos²x的值域
所以 m=(sinx+cosx)²+2cos²x
=1+2sinxcosx+(1+cos2x)
=sin2x+cos2x+2
=√2sin(2x+π/4)+2
2x+π/4∈【π/4,5π/4]
所以 sin(2x+π/4)∈【-√2/2,1]
所以 m∈【1,2+√2】
先求sinx+cosx的取值范围,用辅助角法,再求2cosx平方的范围,从而得出m的范围