高一上期末數(shù)學(xué)考試解題疑問：10,對(duì)于函數(shù)f（x）=cosx+sinx,給出下列四個(gè)命題：
1）存在a∈(0,π/2),使f(a)=4/3;
2)存在a∈(0,π/2),使f(x+a)=f(x+3a) 3) 存在φ∈R,使函數(shù)f(x+φ)的圖像關(guān)于y軸對(duì)稱
4）函數(shù)f（x）的圖像關(guān)于點(diǎn)(3π/4,0)對(duì)稱
其中正確命題的個(gè)數(shù)是
A 1 B 2 C 3 D 4
1.f（x）=cosx+sinx=√2sin(x+π/4),
當(dāng)x∈（0,π/2）時(shí),(x+π/4) ∈（π/4,3π/4）,
所以sin(x+π/4) ∈（√2/2,1）,f（x）∈（1,√2）,
而4/3∈（1,√2）,所以第一個(gè)命題成立.
2.f(x+α)=f(x+3α),說明函數(shù)的周期是2α,
而f（x）=cosx+sinx=√2sin(x+π/4)的最小正周期是2π,
則2α=2π,α=π.π不屬于（0,π/2）.
所以第二個(gè)命題不成立.
3.θ=π/4時(shí),f(x+θ)= √2sin(x+π/4+π/4)= √2sin(x+π/2) =√2cosx是偶函數(shù),偶函數(shù)的圖像關(guān)于y軸對(duì)稱.
所以存在θ=π/4,使函數(shù)f(x+θ)的圖象關(guān)于y軸對(duì)稱,該命題成立.
4.x=3π/4時(shí),f（3π/4）=√2sin(3π/4 +π/4)=0,所以圖象關(guān)于點(diǎn)（3π/4,0）對(duì)稱,該命題成立
4.x=3π/4時(shí),f（3π/4）=√2sin(3π/4 +π/4)=0,為什么x=3π/4時(shí),f（3π/4）=0就說明圖像關(guān)于點(diǎn)（3π/4,0）對(duì)稱,能說明點(diǎn)（3π/4,0）在圖像上呀?