已知函数y=sinx^2+sin2x+3cox^2x(1)最小值是此时的x的集合 (2)求函数单调减区间

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/15 09:18:43
已知函数y=sinx^2+sin2x+3cox^2x(1)最小值是此时的x的集合 (2)求函数单调减区间
xUNA~l;Ф!] hjh`#\,$P&`%h0Q[E0X|CeW̜)l^l;{~3Sll0^o.悫Ԇ?7nK5,lqcxaOc4N=i^sA[~PHj:U; >{9|Kt*guHnI؝Lxr^tB;3ӍY8: /ʐۭ[g䇦ԋdzmnJ^/rt-#4#AVp2Q@Rq>2PtZBa/IHʔq};9 8 2ϧ VԩhR-%ȾnNyY;O'*="~Oa8 ֶ%4_q+2NhF$nͲJ 1ʈKpBu|ܼ덍^~2&%i4[=fo>hI{_@$LbUbs:]ҽre C%|<$(8)83\J#o>A >ߘg>g=~•Ua' Vp+@mDG.5 S8d`.A2i- a!WRIrV6w!ɁeMP \PQtCGL3(h aTU9a{

已知函数y=sinx^2+sin2x+3cox^2x(1)最小值是此时的x的集合 (2)求函数单调减区间
已知函数y=sinx^2+sin2x+3cox^2x
(1)最小值是此时的x的集合 (2)求函数单调减区间

已知函数y=sinx^2+sin2x+3cox^2x(1)最小值是此时的x的集合 (2)求函数单调减区间
y=sinx^2+sin2x+3cos^2x
=1-cos^2x+sin2x+3cos^2x
=1+2cos^2x+sin2x
=1+1+cos2x+sin2x
=2+√2sin(2x+π/4)
y最小值时x=Kπ+5π/8,K∈Z
x∈[Kπ-3π/8,Kπ+π/8]时,函数单调递增
x∈[Kπ+π/8,Kπ+5π/8]时,函数单调递减

(1)
y=sinx^2+sin2x+3cox^2x
=1/2(1-cos2x)+sin2x+3/2(1+cos2x)
=sin2x+cos2x+2
=√2(√2/2sin2x+√2/2cos2x)+2
=√2sin(2x+π/4)+2

函数最小值为2-√2
此时2x+π/4=2kπ-π/2,k∈Z
即x的集合为{...

全部展开

(1)
y=sinx^2+sin2x+3cox^2x
=1/2(1-cos2x)+sin2x+3/2(1+cos2x)
=sin2x+cos2x+2
=√2(√2/2sin2x+√2/2cos2x)+2
=√2sin(2x+π/4)+2

函数最小值为2-√2
此时2x+π/4=2kπ-π/2,k∈Z
即x的集合为{x|x=kπ-3π/8,k∈Z}
(2)
2kπ+π/2≤2x+π/4≤2kπ+3π/2
kπ+π/8≤x≤kπ+5π/8π,k∈Z
∴函数单调减区间:
[kπ+π/8,kπ+5π/8π],k∈Z.
参考http://58.130.5.100//

收起

y=sinx^2+sin2x+3cos^2x
=1-cos^2x+sin2x+3cos^2x
=sin2x+2cos^2x+1
=sin2x+cos2x+2
=√2sin(2x+π/4)+2
2x+π/4=2kπ+3π/2时,取最小值
2x=2kπ+5π/4
{x|x=kπ+5π/8 kEz}
当2x+π/4E[2kπ+π/2,2kπ+3π/2]时,是减的
2xE[2kπ+π/4,2kπ+5π/4]
xE[kπ+π/8,kπ+5π/8]时,是减的。

解y=sinx^2+sin2x+3cox^2x
=sinx^2+cox^2x+sin2x+2cox^2x
=1+sin2x+2cox^2x
=1+sin2x+2cox^2x-1+1
=sin2x+cos2x+2
=√2sin(2x+π/4)+2
即当2x+π/4=2kπ+3/2π,k属于Z,y有最小值√2*(-1)+2=2-√2
此时x=kπ...

全部展开

解y=sinx^2+sin2x+3cox^2x
=sinx^2+cox^2x+sin2x+2cox^2x
=1+sin2x+2cox^2x
=1+sin2x+2cox^2x-1+1
=sin2x+cos2x+2
=√2sin(2x+π/4)+2
即当2x+π/4=2kπ+3/2π,k属于Z,y有最小值√2*(-1)+2=2-√2
此时x=kπ+5π/4,k属于Z
当2kπ-1/2π≤2x+π/4≤2kπ+3/2π,k属于Z,y是减函数
即kπ-3/8π≤x≤kπ+5/8π,k属于Z,y是减函数

单调减区间[kπ-3/8π,kπ+5/8π],k属于Z

收起

(1)y=(1-cos2x)/2+sin2x+3(1+cos2x)/2=sin2x+cos2x+2
=√2sin(2x+π/4)+2
最小值为-√2+2
此时2x+π/4=2kπ+3π/2(k∈Z)
x=kπ+5π/8(k∈Z)
(2)令2kπ+π/2≤2x+π/4≤2kπ+3π/2
得kπ+π/8≤x≤kπ+5π/8
单调减区间[kπ+π/8,kπ+5π/8](k∈Z)