Question

求解这道数学题如图，求详细过程

Answer 1

（1）f(x)=2(cosx)^2+2√3*sinxcosx+m
=√3*sin2x+cos2x+m+1
=2sin(πππm+1;
（2）由已知，0<=x<=π/2, π/6<=2x+π/6<=7π/6,
-1/2<=sin(2x+π/6)<=1,
2sin(2x+π/6)+m+1的最大值是m+3, m+3=4, m=1;
（3）最小正周期 π，
2kπ+π/2<=2x+π/6<=2kπ+3π/2, k是整数（下同），
kπ+π/6<=x<=kπ+2π/3 ，
减区间 [kπ+π/6 , kπ+2π/3].

Answer 2

设p(m,n)为m选n 应该是求近似解由题可知0<x<1 可得 1/(1+x) =1- x + x^2 - x^3 + x^4 - 1/(1+x)^2=1- 2x + 3x^2 -

Answer 3

看不清楚！