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The right side of the first course becomes the ratio of sin, and then you continue to do it, otherwise you won't be admitted to college.

The monotonicity of the field number of the quadratic letter bureau mainly depends on the symmetry of the Tong brigade shouting axis and the opening square town to do the symmetry axis.

1 .The axis of symmetry x=1 opening upwards increments on x>=1 there must be a>=1

2.The axis of symmetry x=-a 4 opening upwards increments on x>=1 must have -a 4<=1 a>= 4

13. The primary function y=ax+b

f(f(x))=a(ax+b)+b=a^2x+ab+b=4x+8a^2=4 ab+b=8

a=2 b=8/3

a=-2 b=-8

14、y=a(x+2)(x-4)=a(x^2-2x-8)=a(x-1)^2-9a

There are maximum values a<0, -9a=9 a=-1<0 y=-(x+2)(x-4).

1) If b>0, x>0, then g(x)=x+b x+a.

g(x)≥2√x·b/x+a=2√b+a

x=b x.

0, b) on g(x) is the increasing function, (b, on g(x) is the decreasing function.

So b=1, b=1

If b<0,x>0, then g(x)=x+b x+ay=x is the function of increasing in x (0, + is the increasing function, y=b x is the increasing function in x (0, + is the increasing bureau function.

Then g(x)=x+b x+a in x (0,+ is also an increasing function, which does not satisfy the topic, so b 0 is not true.

If b = 0 and g(x) = x + a, it is obviously not true.

In summary, b=1

2) Let the concession be under the condition of .

g(x)≥2√x·1/x+a=2+a=1

a=-1g(x)=x²-x+1/x

Don't you have a step? Substituting k=2 into the solution gives a=2b -b+1=a 2 a=1 b=1 2

f（x）=sin（2x+π/3） g（x）=1/2tan（2x-π/3）

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f(1) =f(-1)

(-m-1) power of m*m-5*m+7)*1 = (-m-1) power of m*m-5*m+7)*(1).

Because of the digging of the socks is (m*m-5*m+7) >0

So: 1 to the power (-m-1) = 1) to the power (-m-1).

m-1 = even.

m = the number of laughs in the judgment of filial piety.

I have time to write about it below.

1 is very simple, and the answer can be clearly obtained based on the function image.

The x power of y=2 has only one intersection point with y=-x, in the second quadrant (draw your own picture).