A few high school math problems, solve! Solve a high school math problem!!

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

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

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.

(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.

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

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

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

You should be a freshman in high school!

Swap xy, subtract it, and pull it out.

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

sinx) 2 (cosx) 2=1 and cosx=7 13-sinx, so bring in sinx=12 13 cosx=-5 13 so tanx=-12 5

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

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

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.

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

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.

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.

1 a=-2 b=-8

2m is less than or equal to -5 (derivative).

3 exists.