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高一数学函数问题
f(x)=sin(2x+&/6)+sin(2x-&/6)+cos2x+a
(a为实数）
（1）求函数最小正周期
（2）求函数的单调递减区间
（3）若0&x&&/2时f(x)的最小值为-2， 求a的值。
(1)f(x)=sin(2x+π/6)+sin(2x-π/6)+cos2x+a (a为实数）
=2sin2x(√3/2)+cos2x+a
=√3sin2x+cos2x+a
=2sin(2x+π/6)+a
最小正周期T=2π/2=π
2kπ+π/2&=2x+π/6&=2kπ+3π/2
kπ+π/6&=x&=kπ+3π/2
所以函数的单调递减区间为[kπ+π/6,kπ+3π/2](k∈Z)
(3)0≤x≤π/2,则π/6≤2x+π/6≤7π/6
当2x+π/6=π/6,即x=0时，有fmin=f(0)=1+a=-2
=sin2xcos&/6+sin&/6cos2x+sin2xcos&/6-sin&/6cos2x+cos2x+a
=2sin2xcos6/&+cos2x+a
=*sin2x+cos2x+a
=2sin(2x+&/6)+a
最小周期：&
递减区间：&/6+k&&x&2&/3+k&
a的值：
*代表更号3
我想应该没错吧
您的举报已经提交成功，我们将尽快处理，谢谢！
已知函数f(x)=(√3*sin4x)/cos2x+a*sin²x在x=π/6时取到最大值.
(1)求函数f(x)的定义域.
函数f(x)=(...
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即f(x)=(m*2^x+m-2)&#47，-2&#47：f(-x)=(2^-x-1)&#47，2^x+1单调递增;(2)=0，即：2^x单调递增;3-t，那么;(2^x+1)为奇函数故，f(0)=(m+m-2)/(2^x+1)]又;(2^x+1)单调递减;(2^-x+1)=(1-2^x)/2：m=1即;(2^x+1)单调递增故;(2^x+1)检验;(2^x+1)在定义域R内单调递增那么;0，2&#47，2t&lt：t&-f(t-3)=f(3-t)故有;(2^x+1)=1-[2&#47：f(x)=(2^x+1-2)/(2^x+1)]=-f(x)……为奇函数f(x)=(2^x-1)/(2^x+1)=(2^x-1)&#47：f(t+1)&lt：f(t+1)+f(t-3)&(1+2^x)=-[(2^x-1)&#47：f(x)=1-2&#47，即，f(0)=0：t+1&lt
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出门在外也不愁高一数学，函数问题，在线等，急！_百度知道
高一数学，函数问题，在线等，急！
若函数h(x)在区间[1;x已知函数f(x)=x^2-ax+2a-1(a为实常数)（3）设h（x）=f(x)&#47，2]上为增函数
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a&0若函数h(x)在区间[1，所以
在区间[1，2]上为增函数，u(1)=1-2a+1&x，2]上为增函数，所以只需 当x=1时,2】上是增函数,若函数h(x)在区间[1，2]上的值为非负数令u=x^2-2a+1
u=x^2-2a+1
为二次函数 对称轴x=0所以
在【1;(x)=(x^2-2a+1)/x^2因为
x^2&gt，求实数a的取值范围h(x)=(x^2-ax+2a-1)&#47已知函数f(x)=x^2-ax+2a-1(a为实常数)（3）设h（x）=f(x)/=1实数a的取值范围
这个h'(x)是？
导数，哦，高一还没有学到
呃……能浅显一点么……高一的，着实看不懂
h(x)=(x^2-ax+2a-1)/x=x+(2a-1)/x-a令
x1,x2∈【1，2】且x1&x2h(x1)=x1+(2a-1)/x1-ah(x2)=x2+(2a-1)/x2-ah(x1)-h(x2)=x1+(2a-1)/x1-a-(x2+(2a-1)/x2-a)=(x1-x2)+(2a-1)(1/x1-1/x2)=(x1-x2)+(2a-1)((x2-x1)/x1*x2)=(x1-x2)(x1x2+1-2a)/x1x2
x1-x2&0所以
x1x2+1-2a&02a&x1x2+1
x1x2&1所以 2a&2a&1a=1时
h(x)=x+1/x-1
在【1,2】上也是增函数所以
实数a的取值范围
(-无穷，1】
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真是万分感谢捏~下次有问题咱就喊这位大神了哈~
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