答案第二问为什么可以直接带派

设函数f(x)=ax+cosx,x∈[0,π]. (Ⅰ)讨论f(x)单调性； (Ⅱ)设f(x)≤1+sinx,求a取值范围.
(1)解析：∵函数f(x)=ax+cosx,x∈[0,π].
F’(x)=a-sinx
当a=0时f(x)=cosx
∴f(x)[0,π]上单调减；

当a>0时令f’(x)=a-sinx=0
A∈(0,1)
x1=arcsina,x2=π-arcsina
f’’(x)=-cosx==>f(x1)<0f(x2)>0
∴f(x)x1处取极大值x2处取极小值
∴x∈[0,x1)或x∈[x2, π]时单调增；x∈[x1,x2)时单纤念调减；
A∈[1,+∞), f’(x)>=0f(x)单调增；

当a<0时f’(x)=a-sinx<0
∴f(x)[0,π]上单调减；

(2)解析：∵f(x)≤1+sinx
设h(x)=ax+cosx-sinx-1<=0 x∈[0,π]
h(x)=ax-√2sin(x-π/4)-1
令h’(x)=a-sinx-cosx=a-√2sin(x+π/4)=0
X1=√2/2a-π/4x2=π-√2/2a-π/4
h’’(x)=-cosx-sinx=-√2sin(x+π/4)
当a>=√2π/4时
∴h(x)x1处取极大值x2处取极小值
要满足f(x)≤1+sinx只须满散竖州足h(x1)<=0
h(x1)=√2/2a^2-π/4a+√2cos(√2/2a)-1<=0
h(π)=aπ-√2sin(π-π/4)-1=aπ-2<=0==>a<=2/冲蔽π
∵2/π<√2π/4∴当a<=2/π时x1<0
∴只要验证h(0)=-√2sin(-π/4)-1=0
当a<√2π/4时x1<0(舍)x2处取极小值
∴h(0)=-√2sin(-π/4)-1=0

综上：要满足f(x)≤1+sinx只须满足a<=2/π