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1 The equation for one axis of symmetry for the image of the function y=sin(2x+5 2 ) is (
a.x=-π/2 b.x=-π/4 c. x=π/8 d. x=5π/4
Parse: y=sin(2x+5 2)=sin(2x+2).
The equation for the axis of symmetry is 2x+ 2 2k - 2(k z).
2x=2kπ-π==>x= kπ-π/2
or 2x+2 2k+2==>x=k2
x= 2 is the equation for one of its axes of symmetry.
If the inclination angle of the A2 line 3x+4y-1=0 is , then the value of cos is (
a.-4/5 d.-3/4
Analysis: The slope of the straight line 3x+4y-1=0 is k=tan=-a b=-3 4
The angle of inclination is the angle of the second quadrant, cos = -4 5
Select A3The known function f(x)=-sin x+2asinx+a-1,x r
1) Write the analytic expression of the maximum value of the function f(x) g(a);
2) If f(x) 1 is constant for all x r, find the range of the value of a.
Analysis: Formula the function to obtain f(x)=-(sinx-a) 2+a 2+a-1
When the function f(x) is sinx=a, i.e., -1<=a<=1, the maximum value is a 2+a-1
then, g(a) = a 2 + a-1 (-1< = a < = 1).
When a<-1, g(a)=-(-1-a) 2+a 2+a-1=-a-2
If a>1, g(a)=-(1-a) 2+a 2+a-1=3a-2
In summary: when a<-1, g(a)=-a-2
When -1<=a<=1, g(a)=a2+a-1
When a>1, g(a) = 3a-2
If f(x) 1 is constant for all x r, then:
a-2<=1==>a>=-3
a^2+a-1<=1==>-2<=a<=1
3a-2<=1==>a<=1
In summary: if f(x) 1 is constant for all x r, then -3<=a<=1
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1 a1.The equation for the axis of symmetry is.
x=2kπ+2/π(k∈z)
When k = -1, so x = 2
The slope of 2 a straight line 3x + 4y = 0 k = tan = -3 4
cos α =-4/5.
1.f(x)=-(sinx-a) 2+a 2+a-1 if a<-1, g(a)=-(-1-a) 2+a 2+a-1=-a-2 if a>1, g(a)=-(1-a) 2+a 2+a-1=3a-21<=x<=1, g(a)=a 2+a-1
1 is equivalent to g(a)<1
Solution-3
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1.The equation for the axis of symmetry is.
x=2kπ+2/π(k∈z)
When k = -1, so x = 2
2.Draw a coordinate axis and intersect x with (1 3,0).
Intersect with y with (0,1 4).
So cos = -4 5
Question 3: Wait.
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(1) The domain is defined as r, that is, for any real number x, x +ax+a≠0 is x +ax+a=0, so δ=a -4a<0 solution gives 00 g(x)=x +ax+a=(x+a 2) +a-a 4g(x) decreasing at (- a 2), at (-a 2.).+ Incremental.
Thus f (x) is increasing at (-a 2) and at (-a 2.).+ decrement.
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y'=e^x-e=0
e^x=ex=1x<1,e^x1,y'>0, increment function.
So x=1 is the minimum point.
So x=1y, good luck placement=e-e=0
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This is a special value method that is common in functions.
Because "for any x, y r, there is f(xy+1)=f(x)f(y)-f(y)-x+2".
You might as well make x=0, y=0, and substitute it with: f(1)=f(0)f(0)-f(0)-0+2
f(0)=1
So f(1)=2
In order to eliminate y, let y=0, and substitute it to f(1)=f(x)f(0)-f(0)-x+2
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Assuming y=0, then f(y)=1 is brought into the original formula.
Therefore, f(1)=f(x)-x+1 (1) and then y=1 x=0
then there is f(1) = f(1)-f(1)-0+2
i.e. f(1)=2 (2) synthesis(1)(2).
The solution yields f(x)=x+1
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You should be a freshman in high school!
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Swap xy, subtract it, and pull it out.
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Let x=0, y=0, and substitute f(0+1)=f(0)f(0)-f(0)+2, that is, f(1)=1-1+2=2
Let y=0 then f(x*0+1)=f(x)f(0)-f(0)-x+2, i.e. f(1)=f(x)-1-x+2=2
So f(x)=x+1
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sinx) 2 (cosx) 2=1 and cosx=7 13-sinx, so bring in sinx=12 13 cosx=-5 13 so tanx=-12 5
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It's too laborious, I won't write it for you, it's to use the addition and split method to convert to tanx*tanx=sinx*sinx+cosx*cosx
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Because sina 2+cosa 2=1
So 1-sin2 2=cos2 2 is cos2 2=-cos2
Get cos2 =-1 or 0
When cos2 =-1.
Get 2 = ==> = 2 when cos2 = 0.
Get 2 = 2 or 3 2 ==> = 4 or 3 4
Sum = 2 or 4 or 3 4
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Choose a(a) a is the set of positive numbers, b is the set of real numbers, and for f:a, the squares of the numbers in the set all correspond one-to-one in the set b, and the preforms are a subset of b.
b) The 1 in the set of a does not conform to the one-to-many principle under the correspondence of f, and the 0 in a does not conform to the law f
d) 0 in a does not find a corresponding in b.
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Do you choose B?
A is missing a negative number.
There are 2 3 and other scores missing in c.
What should I do if I have 0 in d?
So consider choosing B again
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0 in ba does not have b so b cannot be a set of real numbers.
There is no reciprocal 0 in c.
D is also a 0 issue.
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1)a=-4 b=-16
Drawing method Draw an image of g(x) first: If f(x)<=g(x) (both are absolute values), then the expression of f g must be the same.
2) f(x)>=m+2)x-m-15] to 2x 2-(m+6)x+m-1> scramble=0
The coefficient of the quadratic term is 2, so the opening is upward, there is a minimum value, which is the axis of symmetry, and the corresponding chain of values is the minimum value, and the minimum value at this time is calculated, and the other "=0" solves a range of m.
2x 2-(m+6)x+m-1>=0 is reduced to the equation (1-x)m>=-2x 2+6x-1 about m
m>=-2(1-x)-1+3 (1-x) is the checkpoint function, because x is greater than 2, so -2(1-x)-1+3 (1-x)>=2*3-1=-7 is m>=-7
Compound m>=-7 with the range of one m found above to find the range of m.
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1 a=-2 b=-8
2m is less than or equal to -5 (derivative).
3 exists.
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