已知函数f(x)=sinwx(w>0),若在[0,1]上单调递增,则w取值范围解析:∵函数f(x)=sinwx(w>0),单调递增区:2kπ-π/2<=wx<=2kπ+π/2==>2kπ/w-π/(2w)<=x<=2kπ/w+π/(2w)令-π/(2w)<=0==>w>0π/(2w)>=1==>w<=π/2∴w取值范围为0...
已知函数f(x)=sin wx(0)在区间[0,3上单调递增,在区间2元33上单调递减;如图,四边形OACB中,a,b,C为△ABC的内角A,B,C的对边,且满足4@-c
函数f(x)=sinwx的最小正周期是4,因此有2*π / w = 4 得w = π/2动直线x = t与f(x),g(x)的交点分别为A( t,sin(πt/2) ),B( t,cos(πt/2) )因此| 向量AB | = | sin(πt/2) - cos(πt/2) | = | 根号2 * sin(πt/2 -... APP内打开 结果2 举报 答案没错的话应该是...
解答一 举报 当x∈(π/2,π)时,wx+π/4∈(πw/2+π/4,πw+π/4)而函数y=sinx的单调递减区间为[π/2,3π/2]那么πw/2+π/4≥π/2,πw+π/4≤3π/2所以1/2≤w≤5/4,即w的取值范围是[1/2,5/4] 解析看不懂?免费查看同类题视频解析查看解答 ...
因为-1<=sin(wx+φ)<=1 所以必有f(x1)=-1,f(x2)=1 则x1和x2最近的差半个周期 T=2π/|w| 所以|x1-x2|最小值=T/2=π/|w|
解析：∵函数f(x)=sinwx（w>0）f(x)的图像经过(2π/3，0)点 f(2π/3)=sinw2π/3=0==> w2π/3=2π==>w=3或w2π/3=π==>w=3/2 ∵f(x)在区间（0，π/3）上是增函数 f(x)单调增区间：wx∈[2kπ-π/2, 2kπ+π/2]==>x∈[2kπ/w-π/(2w),2kπ/w+π/(2w...
f(x)=sinwxsin(wx+π/6)-√3/4(w>0)=[cos(π/6)-cos(2wx+π/6)]/2-√3/4 =-1/2cos(2wx+π/6)T=π/4*2=π/2 2π/(2w)=π/2 w=2 f(x)=-1/2cos(4x+π/6)1) 4x+π/6=2kπ x=kπ/2-π/24 ∵23π/24∈[11π/12,9π/8]∴f(23π/24)=-1/2 ...
已知函数f(x)=2sinwx,其中常数w>0.(1)若y=f(x)在[,]上单调递增,求w的取值范围;(2)令w=2,将函数y=f(x)的图象向左平移个单位,再向上平移1个单位,得到函数y=g(x)的图象,区间[a,b](a,b∈R且a0,根据题意有(z)/umε/(uj) (z)/u-(a)i/u ,解得0<w≤.(2)f(x)=2sin2x,...
解: (1)f(x)=sin(ωx)sin(ωx+π/3)+cos^2(ωx)=sin(ωx)*[ sin(ωx)1/2+cos(ωx)√3/2]+cos^2(ωx) =1/2sin^2(ωx)+cos^2(ωx)+√3/2sin(ωx)cos(ωx) =1/2[sin^2(ωx)+cos^2(ωx)]+ √3/4sin(2ωx)+1/4(1+cos(2ωx)) =1/2sin...
满足|f(x1)-g(x2)|＝2的x1,x2 其|x1-x2|min=T/4 得(2π/ω)/4=π/4,ω=2 sin2x＜cos2x (√2)sin(2x-π/4)<0 sin(2x-π/4)<0 -π+2kπ<2x-π/4<2kπ,k∈Z -3π/8+kπ<x<π/8+kπ,k∈Z 所以sin2x＜cos2x的解集是{x|-3π/8+kπ<x<π/8+kπ,k∈Z...