Known A =(1, x), B=(x2+x,-x) m is constant and m≤-2, find the inequality A• B+2> m (2 A B+1) the range of x. Known A =(1, x), B=(x2+x,-x) m is a constant and m≤-2, the inequality A• B+2> m (2 A B+1). Known A =(1, x), B=(x2+x,-x) m is constant and m≤-2, find the inequality A• B+2> m (2 A B +1) The range of x.

Known A =(1, x), B=(x2+x,-x) m is constant and m≤-2, find the inequality A• B+2> m (2 A B+1) the range of x. Known A =(1, x), B=(x2+x,-x) m is a constant and m≤-2, the inequality A• B+2> m (2 A B+1). Known A =(1, x), B=(x2+x,-x) m is constant and m≤-2, find the inequality A• B+2> m (2 A B +1) The range of x.

0

Given vector a=(sinwx,sinwx), b=(sinwx,-coswx),(w >0), the minimum positive period of function f (x)=a*b is π/2. I simplify to f (x)=1/2-√2/2sin (2wx +π/4) and then T =2π/2w to make it equal to π/2. So do you want to add the absolute value of 2w and then discuss it by category?

On the condition w >0, T=2π/|2w|=π/w=π/2, w=2.
So f (x)=1/2-√2/2sin (4x /4).
When 4x /4=2k /2, sin (4x /4)=1, f (x) has a minimum value of 1/2-√2/2,
In this case, the set of x is {x|x=kπ/2/16, k∈Z},
When 4x /4=2kπ-π/2, sin (4x /4)=-1, f (x) has a maximum value of 1/2 2/2,
In this case, the set of x is {x|x=kπ/2-3π/16, k∈Z}.

Given vector m=(2cos wx,-1), n=(sinwx-coswx,2), where w >0, function e (x)=m times the period of n+3 as beat, find the value of w... Given the vector m=(2cos wx,-1), n=(sinwx-coswx,2), where w >0, the function e (x)=m is multiplied by the period of n+3 to obtain the value of w.

M*n+3=2coswx (sinwx-coswx)+3
= Sin 2wx -2 cos^2wx +3
= Sin2wx -1- cos2wx +3
= Sin2wx - cos2wx +2
= Sin (2wx-pi/4)+2
W =1

M*n+3=2coswx (sinwx-coswx)+3
= Sin 2wx -2 cos^2wx +3
= Sin2wx -1- cos2wx +3
= Sin2wx - cos2wx +2
= Sin (2wx - pi/4)+2
W =1

Given vector a=(2coswx,1), b=(sinwx+coswx,-1), w∈R, w >0, let f (x)=a*b (x∈R), if the minimum positive period of f (x) is π/2 1. Find the value of w 2. Find the monotone interval of f (x)

It is known that:(1) f (x)=(2coswx,1)(sinwx+coswx,-1)=2coswx (sinwx+coswx)-1=2coswxcoswx -1+sin2wx=cos2w+sin2wx=√2/2*sin (2wx /4) then 2π/2w=π/2 and therefore w=2(2) because w=2, f (x)=√2/2*sin (4x /4) when 4x

Given vector a=(root number 3, coswx), vector b=(sinwx,1), function f (x)=vector a*vector b, and the minimum positive period is 4π. ... (2) Let a, B be [π/2,π], f (2a-π/3)=6/5, f (2B+2π/3)=-24/13, find the value of sin (a+B)? (3) If x belongs to [-π,π], find the range of function f (x)?

F (x)=(root 3, coswx)*(sinwxx,1)= root number 3*sinwx+coswx*1=2[(( root number 3)/2)*sinwx+(1/2) coswx ]=2sin (wx+pi/6)(1)2*pi/w=4*piw=1/2 f (x)=2 s in (1/2x+pi/6)(2) Strip x=2a-π/3 into f (x) with sina=3/5; x=2B+2π/3, bring in simplification with cos...

Given vector m=(1, coswx), vector n=(sinwx, root number 3),(w >0), function f (x)=m*n And the coordinates of the highest point on the image of f (x) are (π/12,2), and the coordinates of the lowest point adjacent to it are (7π/12,-2) the analytic expression of f (x)

F (x)= sin wx + root 3*cos wx=2 s in (wx /3)
The coordinate of one highest point is (π/12,2), and the coordinate of one lowest point adjacent to it is (7π/12,-2).
Period=2(7π/12-π/12)=π, so 2π/w=π, then w=2, and
F (x)=2s in (wx /3)=2s in (2x /3)