A review paper high school math A high school math problem

Solution: (1) The function y=lg(1-x 2)+1 2x+1,1-x 2>0,2x+1>0, the solution: -1-1 2, -1 20

The domain of the 00, -1 function y is (-1,1).

1-x^2≠0,∴y∈r.

5) Let f(x) = log(1 4)x, it is easy to know that f(x) is a subtraction function.

y=[f(x)] 2+5, when f(x) is the minimum value, y has a minimum value, that is, f(4)=log(1 4)4=-1, y min=(-1) 2+5=4, when f(x) is the maximum value, y has a maximum value.

That is, f(2)=log(1 4)2=-1 2,, y max=(-1 2) 2+5=21 4, and the range of the function y is [4,21 4].

6) Functions y=x-1 x+1,x (0,1),y=(x+1-2) (x+1)=1-2 (x+1),x (0,1).

2 (x+1)≠-1,2 (x+1)≠1, the range of the function y is (-1,1)

x is less than 1 and is greater than -1 x is greater than minus half (first question) Hey, without a pen by your side, it's really hard to calculate.

Solution: (1) p(x0,y0)(x0≠ a) is the hyperbolic e: x a -y b =1(a 0, b 0) on the previous point, x0 a -y0 b =1, from the meaning of the title there is y0 (x0-a) y0 (x0+a)=1 5, a = 5b, c = a +b, then e= c a = 30 5

a, 2-a square = b square -2 so b square + a square = 4, according to the fundamental inequality theorem, a square + b square = 2ab, if and only if a square = b square, the equal sign holds, because a = 2 root sign ab, a > 0 and b > 0This conditional question has been given.

2-a^2=b^2-2

a^2+b^2=4

A 2 + B 2 > = 2ab (the equal sign needs to be removed in this question) ab<2

There is an ab>0 to choose a according to the question

That is, when a 2 + b 2 = 4.

The problem of finding the maximum value of ab.

There is a maximum value of 2 only when a=b=2

So, the answer is a

It's best not to look at their answers, if you want to go, you really can't think of it, ask your teachers to go!

It is estimated that you either haven't learned math and geometry well, or you are good at going online, and you can use the time you ask questions, find answers, and send answers on the Internet to read books, and you will already get the answers, and maybe there will be unexpected gains. I wish you success.

Connect A'c', because it is a cube, so a'c'⊥b'd', because of CC'⊥b'd', so b'd'Surface A'cc', so b'd'⊥a'c。

Connect A'b or a'd, the process is the same as above.

Don't blame me for saving steps, the principle is the same, if you want a complete answer you don't have a little bit of thinking.

on=1 2(ob+oc) om=1 2oa so mn=on-om=1 2(ob+oc)-1 2oa

and oq=om+2 3mn op=om+1 3mn will be brought in , respectively: oq=1 6oa+1 3(ob+oc) op=1 3(oa+ob+oc).

Pay attention to the writing format, and add the vector symbols yourself.

2) The vector A + vector B = (2 into -3, into +4) and because the vector A + vector B is collinear with vector C.

So -7 (2 into -3) + 4 (into +4) = 0

So into = 37 10

17.(1) f(x)=2sinx-2 root number: 3cosx=4sin(x-3).

Because -1<=sin(x- 3)<=1

So -4<=f(x)<=4

2) The period of f(x) t=2 w=2

Monotonic Transmission Source 2k + 2<=x- 3<=2k +3 2 So 2k +5 6 <=x<=2k +11 6 So the monotonically decreasing interval of x is [2k +5 6 ,2k +11 6 ]15(1) t=(4 9 - 9)x2=2 3 2) from the figure we can see a=2

w=2π/t=3

i.e. y=2sin(3x+?)

Substitute the point (9,0) into the function y

Then the analytic formula is y=2sin(3x- 3).

18.(1) Vector ab=(8,-8).

modulo of vector ab = 8 under the root number 2 + (-8) 2 = 8 times the root number 22) vector ca = (-8, -8).

The inner product of the vector ab and the vector hail scatter ca = 0, so the vector ab is perpendicular to the vector ca.

So this triangle is a right-angled triangle.

Science class (1) because vector A is parallel to vector B

So -9sinx-3(1-sinx)=0

So sinx=-1 2

Again, x belongs to - 2 to 2

So x=- 6

2) Analytic f(x)=9 2sinx 2-9 2sinx+6 I didn't see fill-in-the-blank and selection.

1)a'b'c'd'The midpoint of o', even ao'then ao'Parallel c'o，ao'On the surface ab'd'Inside.

So c'o Plane ab'd'

2) Not established. Because of AC'ab' is not true, ac'ad' is not true.

Even A'c'Handed over b'd'with the point o'

Connect the AO'Because o'c'AO so the quadrilateral is a parallelogram.

So c'o//ao

So c'o Plane ab'd'

The second question should be A'C Plane AB'd'

Cube ABCD A'b'c'd'Medium b'd'Face AA'c'c∴b'd'⊥a'c

a'C in plane AA'b'The projection on b is a'b,a'b⊥ab'

ab'⊥a'c

then a'C Plane AB'd'

Have you ever learned about space vectors, multiplying coordinates equals 0 is perpendicular?

Very common college entrance examination questions:

1) Use mathematical induction:

Suppose 00 so f(x) is subtracted.

Let f(x)=(x-sinx)-1 6x 3 then f'(x)contains=1-cosx-1 2x 2 list ......When x=0, f(x) has a maximum value f(x)=0, so f(x) < 0

So (x-sinx)<1 6x 3

Substituting an into x is established!!

I've got a relatively complete understanding of Kai's blindness," ...Part of yourself is sure you can.

Because of the grinding blob f'(x)=1-cosx>0

So f(x) is subtracted.

Let f(x)=(x-sinx)-1 6x 3 then f'(x)=1-cosx-1 be ...... with 2x 2 listsWhen x=0, f(x) has a maximum value f(x)=0, so f(x) < 0

So (x-sinx) "Blind Orange 1 6x 3

Substituting an into x is established!!