已知函数f(x)=根号2asin(x-π/4)+a+b

显示全部楼层 · 2011-12-16 22:34:25

1.函数f(x)= √2sin(x-π/4)+1+b2kπ+π/2≤x-π/4≤2kπ+3π/2，k∈Z.解出x得减区间2。0≤x≤π-π/4≤ x-π/4≤3π/4-1/√2≤sin(x-π/4)≤1-1≤√2sin(x-π/4)≤√2a<0a√2≤a√2sin(x-π/4)≤-aa√2+a+b≤a√2sin(x-π/4)+a+b≤-a+a+bf(x)在[0,π]上的值域为[2,3],a√2+a+b=2, and b=3a=1-√2, b=3...

千问 · 2011-12-16 22:34:25

1.函数f(x)=√2sin(x-π/4)+1+b2kπ+π/2≤x-π/4≤2kπ+3π/2，k∈Z.解出x得减区间2。0≤x≤π-π/4≤x-π/4≤3π/4-1/√2≤sin(x-π/4)≤1-1≤√2sin(x-π/4)≤√2a<0a√2≤a√2sin(x-π/4)≤-aa√2+a+b≤a√2sin...