Senior 1 Math Problems Urgent Seeking solutions to senior 1 math problems

1.(1) Add left and subtract right, so choose

2) a≠0, =1+4=5>0, 2 intersections.

3) The axis of symmetry is x=1 and the vertex is (1,5).

2.(1) The axis of symmetry is x=0, m=a

2) The vertex is on the x-axis, (4ac-b 2) 4a = 0, 8m-12 = (m-a) 2, solution.

3) Over the origin, that is, over the (0, 0) point, 2m-3=0, m=3 23axis of symmetry x = -1

The minimum value is taken at -1, and 2-4-3 = -5

So the range is [-5,+

4.The first one is not to be said.

Say the second. axis of symmetry x=-a

When a=0.

The maximum value y=27

When a>0, y=27+10a

When a=0.

y=27-10a

axis of symmetry x=-a

When a=-1 2.

y(2)=4+1+2≠4, rounded.

When a=-1 2.

The maximum value y(2) = 5 + 4 a = 4, a = -1 4 when a < -1 2.

The maximum value y(-1)=2-2a=4, a=-1

In summary, a = -1 or -1 4

1) The fourth.

m=am=a+4+2 times rigid (2a+1) or a+4-2 times rigid (2a+1).

m=3/2y>=-5

1) When a=-1, y=x -2x+2=(x-1) 2+1, so the maximum value is 37 when x=-5, and the minimum value is 12)y=(x+a) 2+2-a 2

When a>=0, when x=5, y takes the maximum value of 27+10a, when a<0, when x=-5, y takes the maximum value of 27-10ay=(x+a) 2-a 2+1

When -a<=1 2, when x=2, y takes the maximum value of 5+4a, so a=-1 4

When -a>1 2, when x=-1, y takes the maximum value of 2-2a, so a=-1

So a=-1 or -1 4

It's so tiring to write, don't forget to add points.

m=am=a+4+2 times rigid (2a+1) or a+4-2 times rigid (2a+1).

m = 3 2 axis of symmetry x = -1

The minimum value is taken at -1, and 2-4-3 = -5

So the range is [-5,+

1) When a=-1, y=x -2x+2=(x-1) 2+1, so the maximum value is 37 when x=-5, and the minimum value is 12)y=(x+a) 2+2-a 2

When a>=0, when x=5, y takes the maximum value of 27+10a, when a<0, when x=-5, y takes the maximum value of 27-10ay=(x+a) 2-a 2+1

When -a<=1 2, when x=2, y takes the maximum value of 5+4a, so a=-1 4

When -a>1 2, when x=-1, y takes the maximum value of 2-2a, so a=-1

So a=-1 or -1 4

(1) Let x=y=1 bring in f(xy)=f(x)+f(y) and the answer will come out directly.

2) From the formula f(xy)=f(x)+f(y), it is easy to get the function as a logarithmic function, because only the logarithmic function has such a property, and then you know that f(6)=1, so you can know that the base of the logarithmic function is 6, and then the rest of the solution inequality is very simple. It's a simple solution to the inequality problem, and the final answer is (0,3 35).

Hope it helps!

Solution: (1) Because f(xy)=f(x)+f(y), let x=1, y=1 get:

f（1）=f（1）+f（1)

So f(1)=0

2) Because f(6)=1, f(6)+f(6)=f(6*6)=f(36)=2

So the original inequality can be reduced to: f(x+3)-f(x) f(36)f(x+3) f(36)+f(x).

f（x+3）＞f（36x）

and f(x) is a monotonic increasing function defined on (0, positive infinity), so:

x+3＞36x>0

0

Solution: From the problem, f(1+y 2)=3y+5y-1 makes 1+y 2=2x-1 then y=4x-4 substitution.

obtained: f(2x-1)=48x-76x+272f(x)+f(1 x)=2x.1

2f（1/x）+f（x)=2/x ..2

Synoptic f(x) = 4x 3-2 (3x).

Hope it works for you!

It is added that f(5)=f(3+2)=-1 f(3), f(1+2)=-1 f(1), so f(1)=f(5)=-5, it can be seen that the function is a function with a period of 4.

f(f(1))=f(-5)=f(3)=-1/f(1)=1/5

1) Let t=1+x2, then x=2t-2

f(t)=3(2t-2)²+5(2t-2)-112t²-14t+1

f(2x-1)=12(2x-1)²-14(2x-1)+148x²-76²+27

2) Replace x in a known equation with 1 x to obtain.

2f(1/x)+f(x)=2/x

f(x) and f(1 x) are regarded as unknowns, and they are solved in conjunction with the known equations.

f(x)=(4/3)x-2/(3x)

x²-100x+49

I'm substituting 1+x 2 with x as a whole, then replacing 2x-1 with an equation containing x, and then solving it)

2. There is nothing that can be done. Sweat...

From the known x 2+y 2-4x-2y-4=0, that is, (x-2) 2+(y-1) 2=3 2, let the required center coordinates (a,b), the equation is (x-a) 2+(y-b) 2=r 2, tangent to the straight line y=0 (i.e., x-axis), so the radius of the circle r= b =4, the two circles are tangent, we can know that the distance between the two circles (a,b) to (2,1) is equal to the sum of the radius, that is, (a-2) 2+(b-1) 2=(3+4) 2, where b=4, a=2 3 15, b=-4, get a=2 119

So the equation for the circle is.

1](x-2+3√15)^2+(y-4)^2=16

2](x-2-3√15)^2+(y-4)^2=16

3](x-2+√119)^2+(y+4)^2=16

4](x-2-√119)^2+(y+4)^2=16

A total of four circles. Let the equation for the circle be (x-a) 2 + y-b) 2 = c (c is the square of the radius).

y = 0,x-a)^2 = c - b^2,x1 = a + c-b^2)^(1/2),x2 = a - c-b^2)^(1/2).

2(c-b^2)^(1/2) = 6,c - b^2 = 9,c = b^2 + 9.

Substitute p q.

2-a)^2 + 4-b)^2 = c,3-a)^2 + 1-b)^2 = c.

Combined simplification: 4 + 4a + a 2 + 16 - 8b + b 2 = 9 - 6a + a 2 + 1 + 2b + b 2, 10a - 10b + 10 = 0, a = b - 1

3-(b-1)] 2 + 1-b) 2 = b 2 + 9,4-b) 2 + 1 + b) 2 = b 2 + 9,16 - 8b + b 2 + 1 + 2b + b 2 = b 2 + 9, b 2 - 6b + 8 = 0, b-2) (b-4) = 0, b = 2, or, b = 4

a = 1, or, a = 3

c = 13, or, c = 25

The equation for the circle is .

1] (x-1)^2 + y-2)^2 = 13

2] (x-3)^2 + y-4)^2 = 25.

There are 2 circles in total.

Classmates, how can you not listen to class in class?

If you don't listen to it, you have to read it.

I don't understand it.

If you want to learn. I don't want to learn.

I also want to learn that my brother is a person who has come over.

Student, the solution expression for this problem is easy to obtain, let's say, the time for A to arrive is 6 o'clock x minutes, and the time for B to arrive is 6 o'clock y;

Then the probability of A and B getting together is equal to p(|x-y|<10) where x<=60 and y<=60

However, the solution is not a classical generalization, and it needs to use a geometric generalization, use a Cartesian coordinate system, and convert it into a linear programming.

then p(|x-y|<10) = hexagonal area Square area = 11 36

Draw the coordinate axis x-axis is the time when A arrives and the y-axis is the time when B arrives When the absolute value of x-y is less than 10, you can meet Then, on the coordinate axis, draw the area of the encounter The area sandwiched in the middle is the area of the encounter The total area is 1 The area of non-encounter is 5 6 * 5 6 = 25 36 So the area of the encounter is 11 36 This is the probability of 11 36

bc·1?

Obviously, bb1 is perpendicular to the face a1b1c1d1

Therefore, the angle BC1B1 is the angle of 45° between BC·1 and the plane A1B1C1D1

.It seems that the problem is the angle between de and the plane a1b1c1d1, right? Otherwise, the condition of e is useless, [solution] over e as ef bc, hand over bc to f, connect df

EF Plane ABCD, EDF is the angle between the straight line DE and the plane ABCD.

The title means that EF is the median line of BC1C, so EF=C1C 2=1, CF=CB 2=1

df = 5 ef df, tan edf = 5 5 so the magnitude of the angle between the straight line de and the plane abcd is arctan 5 5

1.Build a space Cartesian coordinate system, let the coordinate origin be a, then the plane composed of two pairs of the three coordinate axes in the coordinate system is described in the condition, imagine for yourself). Then the three data given by the problem are the three coordinate values of point c in the coordinate system.

The required east and east is the distance from point c (coordinates may be written as (3,4,5), and the number order reversal does not affect the answer) to the origin a. i.e. ac=sqr(3 2+4 2+5 2). sqr stands for the root number, and the result is calculated by itself.

2.This is the original question of the 5th question of the 09 Zhejiang Mathematics College Entrance Examination Paper (Science), the answer is by yourself, and you will not ask again.