已知向量a=(cos(3x/4),sin(3x/4)),向量b=(cos(x/4+π/3),-sin(x/4+π/3))
1:令f(x)=(向量a+向量b)的平方,求f(x)的解析式和单调区间
2:若x属于-π/6,5π/6都是闭区间,求f(x)的最大最小值
人气:358 ℃ 时间:2020-03-24 22:28:53
解答
1.f(x)=[cos(3x/4)+cos(x/4+π/3)]^2+[sin(3x/4)-sin(x/4+π/3)]^2
=cos^2(3x/4)+sin^2(3x/4)+cos^2(x/4+π/3)+sin^2(x/4+π/3)
+2cos(3x/4)cos(x/4+π/3)-2sin(3x/4)sin(x/4+π/3)
=1+1+2cos(3x/4+x/4+π/3)=2cos(x+π/3)+2
单增区间(2kπ-π/3,2kπ+2π/3),单减区间(2kπ+2π/3,2kπ+5π/3),极值点2kπ-π/3,2kπ+2π/3,k∈Z
2.[-π/6,5π/6]包含一个极小值点2π/3,两个端点中f(-π/6)较大.所以最大值是f(-π/6)=根3+2,
最小值是f(2π/3)=0
推荐
- 已知向量a=(cos(3x/2),sin(3x/2)),b=(cos(x/2),-sin(x/2)),且x∈[0,π/2]
- 已知向量a=[cos(3x/2),sin(3x/2)],已知向量b=[cos(x/2),-sin(x/2)],x属于[0,兀/3]
- 已知向量a=[cos(3x/2),sin(3x/2)],向量b=[cos(x/2),-sin(x/2)],且x[0,π/2]
- 设向量a=(cos(x/2),sin(x/2)),向量b=(sin(3x/2),cos(3x/2)),x∈[0,π/2].
- 已知向量a=(cos 3x/2,sin 3x/2),b=(cosx/2,-sinx/2),x∈[-π/3,π/2],若a+b的绝对值为1/3,求cosx的值
- 问点化学反应离子式
- 1.Would you mind closeing the music? OK.I'll do it right __________.
- “莫问桑田事,但看桑落州.数家新住处,昔日大江流.古岸崩欲尽,平沙长未休.想应百年后,人世更悠悠.”唐朝诗人胡玢的诗描述河流沿岸的变化,结合下图和所学知识,回答13-14题.13.关于图中信息说法正确的是:( )
猜你喜欢