已知f(x)=三分之2kx a

2x看作3分之6x,相加后是3分之7x,3分之7x等于3分之56.即是7x等于56,x=8

1)令f（x）的导数为0,将X=三分之四代入,即可求a2）用数形结合的方法,已知f（x）的表达式,画出在[-1,1]上f（x）的图像,用X=m在图像上上下移动,有两个交点即为两个不同实数根,根据图像求

f(x)=2sin(2x-π/3)+1当2x-π/3∈[-π/2+2kπ,π/2+2kπ]单调递增,解得x∈[-π/12+kπ,5π/12+kπ]当2x-π/3∈[π/2+2kπ,3π/2+2kπ]单

y=x/3-2x=-5x/3

f(x)=sin(2x+三分之派)+sin(2x-三分之派)+2cos方x-1=sin2x*cos(3分之π)+cos2x*sin(3分之π)+sin2x*cos(3分之π)-cos2x*sin(3分

1f(x)=2sinX平方+2跟3sinxcosx+a把x=π/3代入f(x),sinπ/3=跟3/2cosπ/3=1/2f(x)=2*3/4+3/2+a=4a=1f(x)=跟3sin2x-cos2x

函数f(x)=cos(2x-π/3)+sin^2x-cos^2x=cos2xcos(π/3)+sin2xsin(π/3)-cos2x=√3/2sin2x-1/2cos2x=sin(2x-π/6)函数f

因为f(π/3)=f(π/6)且之间能取到最小值且无最大值所以f（π/4）=-1w*π/4+π/3=2kπ+3π/2得到w=14/3+8kk=整数因为之间无最大值所以周期大于π/3即w小于6所以k=0

1.π,52.-π/12+kπ,5π/12+kπ再问：能给我详细的过程吗，我可以加分再答：1.2π/2=π，3+2=52.令2x-π/3=-π/2+2kπ，x=.-π/12+kπ,令2x-π/3=π/

x=三分之2y+4,代入y=三分之3x-4,得y=3分之2y+4-4,则y=3分之2y,则y=0则x=3分之4

(1)f（X）+f(Y)=f(X+Y),当x=y=0,f（0）=0当y=-x,f（X）+f(-x)=f(0)=0,∴f（-X）=-f（X）（2）设x1>x2,f(x1)-f(x2)=f(x1)+f(-

cos(x-π/6)=-√3/3cosx+cos(x-π/3)=cos(x-π/6+π/6)+cos(x-π/6-π/6)=cos(x-π/6)cosπ/6-sin(x-π/6)sinπ/6+cos(

因为cos(a+b)=cosacosb-sinasinbcos(a-b)=cosacosb+sinasinb相加得cos(a+b)+cos(a-b)=2cosacosb即cosacosb=[cos(a

f(x)=2cos(x+π/3)[sin(x+π/3)-√3cos(x+π/3)]=2cos(x+π/3)sin(x+π/3)-2√3cos²(x+π/3)=sin(2x+2π/3)-√3[

f(x)=acos²x+bsinxcosx=cosx(acosx+bsinx)∵f(0)=cos0(acos0+bsin0)=2解得a=2同理,∵f(π/3)=1/2+（√3/2）且a=2解

（1）f=（2x+π/3）+3根号3/2正周期T=2π/2=π对称轴2x+π/3=π/2+2kπ∴2x=π/6+2kπ∴x=π/12+kπ,k∈z（2）∵-π/2+2kπ≤2x+π/3≤π/2+2kπ

f的导数为-cos（三分之派-x）,转换为cos（π+三分之派-x）,f=cos（六分之π+x）即向右平移二分之π第一道选择C

1f(x)=2sinX平方+2跟3sinxcosx+a把x=π/3代入f(x),sinπ/3=跟3/2cosπ/3=1/2f(x)=2*3/4+3/2+a=4a=1f(x)=跟3sin2x-cos2x

f(x)=2COS²x+2√3sinxcosx=(1+cos2x)+√3sin2x=√3sin2x+cos2x+1=2sin(2x+π/6)+1,X∈[-π/6,π/3],2x+π/6∈[-

再答：再问：cosα的值取正还是负再答：不用管再答：没有用cos