作业帮若f(x)=(1-λ)sinx-(1/2+1/2λ)cosx+1在[-π/2,π/2]是增函数,求λ取值范围!
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若f(x)=(1-λ)sinx-(1/2+1/2λ)cosx+1在[-π/2,π/2]是增函数,求λ取值范围!
若f(x)=(1-λ)sinx-(1/2+1/2λ)cosx+1在[-π/2,π/2]是增函数,求λ取值范围!
若f(x)=(1-λ)sinx-(1/2+1/2λ)cosx+1在[-π/2,π/2]是增函数,求λ取值范围!
若f(x)=(1-λ)sinx-(1/2+1/2λ)cosx+1在[-π/2,π/2]是增函数,求λ取值范围!
解析：设cosθ=(1-λ)/√[5/4+5λ^2/4-3λ/2],sinθ=(1/2+1/2λ)/√[5/4+5λ^2/4-3λ/2]
∴f(x)=√[5/4+5λ^2/4-3λ/2]sin(x-θ)+1
要使f(x)在[-π/2,π/2]是增函数,须使θ=0
令1/2+1/2λ=0==>λ=-1
∴f(x)在[-π/2,π/2]是增函数,λ=-1
f'(x)=(1-λ)cosx+(1/2+λ/2)sinx>=0在[-π/2,π/2]成立，
∴λ=-1.
首先要知道asinx+bcosx=√(a^2+b^2)sin(x+arctan(b/a)
则λ=1时，f(x)=-cosx+1,满足在[-兀/2，兀/2]上单调递增；
λ不为1，则f(x)=√[(1-λ)²+1/4*(1+λ)²]*sin(x+arctan [(1/2+λ/2)/(λ-1)])
又∵y=sinx在[-兀/2，兀/2]上单调递增
∴...
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首先要知道asinx+bcosx=√(a^2+b^2)sin(x+arctan(b/a)
则λ=1时，f(x)=-cosx+1,满足在[-兀/2，兀/2]上单调递增；
λ不为1，则f(x)=√[(1-λ)²+1/4*(1+λ)²]*sin(x+arctan [(1/2+λ/2)/(λ-1)])
又∵y=sinx在[-兀/2，兀/2]上单调递增
∴arctan(1/2+λ/2)/(λ-1)=0，即(1/2+λ/2)/(λ-1)=0
解得λ=-1
∴λ为1或者-1
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