设正整数 n = a 0 ⋅ 2 0 + a 1 ⋅ 2 1 + a 2 ⋅ 2 2 ⋯ + a k - 1 ⋅ 2 k - 1 + a k ⋅ 2 k ,其中 a i ∈ 0 , 1 ,记 ω n = a 0 + a 1 + ⋯ + a k .则( )
ω 2 n = ω n
ω 2 n + 3 = ω n + 1
ω 8 n + 5 = ω 4 n + 3
ω 2 n - 1 = n
若函数y=x2+(2a-1)x+1在区间(-∞,2上是减函数,则实数a的取值范围是( )
已知f(x)=,则f(-1)+f(4)的值为()
已知函数f(x+1)=3x+2,则f(x)的解析式是()
已知集合()
若,则()