High School Freshman Math Problems Senior 1 Math Problems.

Test Points: Trajectory Equations; the positional relationship between a line and a circle; The Positional Relationship between Circles and Circles and Their Determination Topic: Calculation Problems Analysis:

Let the coordinates of the p point be (x,y),c1(-2,0) and the radius of the moving circle is r, then the distance from p(x,y) to c1(-2,0) and the straight line x=4 can be obtained according to the properties of the two circles tangent and the tangent of the straight line and the circle, which can be simplified.

Answer: Solve the center of the circle (x+2)2+y2=4 c1(-2,0), the center of the circle p(x,y) of the moving circle p, and the radius is r, as .

x=4,x=2,pq The straight line x=4,q is perpendicular, because the circle p is tangent to x=2, so the distance from the circle p to the straight line x=4 pq=r+2, and PC1=r+2, so the distance from p(x,y) to c1(-2,0) and the straight line x=4 is equal, the trajectory of p is parabolic, the focus is c1(-2,0), the quasi-line x=4, the vertex is (1,0), the opening is to the right, the focal parameter p=6, and the equation is: y2=-12(x-1).

Let the coordinates of the center of the moving circle be (x,y) and the radius be r;

1) Because the moving circle and the circle (x+2) +y =4 are inscribed, (x+2) 2+y 2=2+r;

2) Because the straight line x=2 is tangent, r=2-x;

(x+2) 2+y 2=2+2-x is obtained from (1) and (2);

Simplification yields: y 2-12x-12=0.

See figure. <>

The moving circle c is (x-a) +y-b) =r

Tangent to the straight line x=2.

then a=2 r

x+2) +y =4 center of the circle (-2,0).

take a=2-r

r=2-a moving circle and circle(x+2) +y =4 inscribed.

then (a+2) +b = (r+2).

a+2)²+b²=(2-a+2)²

a²+4a+4+b²=16-8a+a²

12a-12+b²=0

a=-b²/12+1

The equation for the center of the moving circle is.

x=-y²/12+1

The intersection of 3x+4y+1=0 and 5x+12y-1=0 is (-1,1 2). Let the angle division equation be: y=k(x+1)+1 2=kx+k+1 2

The angular equinox crosses the dot (0, k+1 2). The distance from this point to the straight line 3x+4y+1=0 is equal to 5x+12y-1=0.

4k+2+1] 5=[12k+6-1] 13 k=7 4 or k=-4 7

The angular division equation is: y=(7 4)x+7 4+1 2 4x-7y+9=0 or y=-(4 7)x-4 7+1 2 8x+14y+1=0

The slope of l1 l2 is k1=tana k2=tanb, the slope of the bisector k=tan(a+b) 2, and there is a condition that crosses the intersection of l1 l2. You can find the analytic formula of the bisector.

First find the coordinates of the intersection of the two straight lines, and then find it, let the inclination angle of the first straight line be a, and the inclination angle of the second straight line is b, then the inclination angle of the angle bisector is tan(a+b) 2, so you can find the slope k, and then substitute the coordinates of the intersection point, you can find bAt this point, the straight line is determined.

Since f(x+1) and f(x-1) are both odd functions, from the definition, we can know that f(x+1)=-f(-x+1), and there is f(x-1)=-f(-x-1), so from the definition of symmetry, we can know that (1,0) and (-1,0) are the two symmetry centers of the function. Thus there is a function that is periodic, and the period t=4. Therefore, (3,0) is also the center of symmetry, so d is chosen.

1.Proof (1) |a|²=a²=1，|b|²=b²=1

a+b)●（a-b）= a²-b²=1-1=0

a+b) (a-b) (Note: The above are vector operations).

2）∵|a+b|=|κa-b| ∴a+b|²=|κa-b|²

4ka●b=0, κ0 ∴a●b=0

cosacosβ+ sinαsinβ=0

i.e. cos( =0

2.The "2 x" in the title should be "x 2", and the "2 root number 3" should be "3 2".

1).f(x) = (1 2) sin x (cosx sinx) + (3 2) cos2x (intermediate application "cut string").

1/2）sin2x+( 3/2) cos2x=sin(2x+π/3)

From 2k - 2 2x + 3 2k + 2, k z, get k -5 x 12 x k + 12, k z

From 2k + 2 2x + 3 2k +3 2 , k z, get k + 12 x k +7 12, k z

The increasing interval of f(x) is [k -5 12 , k + 12] and the decreasing interval is [k + 12, k +7 12], k z

2) From f(x) = 3 2 to sin(2x+ 3) = 3 2

2x+3=2k+3 or 2x+3=2k+2 3,kz

x=k or x=k+6

0∴x=π/6

1.Evidence: a+b=(cos + cos, sin + sin), a-b = (cos -cos, sin -sin).

a+b)· (a-b)=(cos) 2-(cos) 2+(sin) 2-(sin) 2=(cos) 2+(sin) 2-[(cos) 2+(sin) 2]=1-1=0 a+b is perpendicular to a-b.

ka+b=(kcosα+cosβ,ksinα+sinβ),a-kb=(cosα-kcosβ,sinα-ksinβ)

Ka+b is equal in length to a-kb.

kcosα+cosβ)^2+(ksinα+sinβ)^2=(cosα-kcosβ)^2+(sinα-ksinβ)^2

It is simplified as: 4k(cos·cos +sin·sin)=4k·cos( -=0

k is a non-0 constant

cos( -=0, -=t + 2 (t is a non-negative integer).

t=0,β-=π/2

2.I didn't get it. 2 root number 3 This is 2/3 ?.. root

Landlord, this is the third year of high school, it's so difficult.

(1) Square plus round = 3

2) Circle minus square = 1 to get square equal to 1 circle equal to 2 (3) square plus triangle = 6

4) Trigonometric minus square = 4 to get trigonometric equal to 5 Finally: trigonometric minus circle = 5-2=3

Circle + Square = 3

Circle-square = 1

Then, circle = 2 and square = 1

Square + triangle = 6

Triangle - square = 4

then triangle = 5

So triangle + circle = 7

According to the topic. f（1-a）＜-f（1-a^2）

f(x) is an odd function, so -f(x) = f(x)f(1-a) f(a2-1).

It is also a subtraction function that defines the domain -1 to 1

So 1-a is between -1,1.

1-a2 is between -1,1.

And there is 1-a 1-a 2

The range of the simultaneous solution a is (1, root number 2).

The answer is wrong!

Because f(x) is an odd function, -f(x) = f(-x) and because f(1-a) -f(1-a) 2

i.e. f(1-a) f(a2-1).

So: 1>1-a>a 2-1>-1

Seeking.

I guess the problem is that the true number of logarithms is an absolute value added, so there are two intervals, the quadratic function f(x)=x 2-x-12=(x-4)(x+3), and the axis of symmetry is x=1 2, which is graphed by the function of f(x), |f（x）|The image is a function image on the interval (-3,4) with respect to the x-axis symmetry (in layman's terms, it is about the x-axis flipping), and the image on the other intervals is retained. According to the image and the law of same increase and difference, the easy to obtain monotonic increase interval is (-infinity, -3) and on (1 2, 4) Thank you, you don't understand it if you don't understand it, thank you.

It is to function |x^2-x-12|of the subtract interval.

x^2-x-12=0

x1 = 4, x2 = -3 at this time |x^2-x-12|The function value is the smallest, which is 0, and the function y=x2-x-12 has a symmetry axis of x=1 2, so at x-3, |x^2-x-12|Decreases as x increases, is |x^2-x-12|of the subtract interval.

1 2 x 4, |x^2-x-12|It also decreases as x increases, which is |x^2-x-12|of the subtract interval.

Therefore y=log1 2 |x^2-x-12|The increase interval is (- 3) 1 2, 4).

It's not wrong to increase the same and subtract the same. The outer function log1 2t (let t=|.)x^2-x-12|For the subtraction function, the monotonic interval of your inner function is wrong.

t=|x^2-x-12|Because there is an absolute value, what was originally only one subtraction interval has become two:

infinity, -3) and (1 2, 4).

As for why, you can see if you fold the image below the x-axis.

Don't forget the absolute value, |x^2-x-12|In (3) is monotonically decreasing, |x^2-x-12|At (1 2,4) is also monotonically decreasing, but since the base is 1 2 1, the main y is monotonically increasing in these two intervals.

|x^2-x-12|=|(x-4)(x+3)||x-4)(x+3)|On (- 3) and (1 2, 4) is subtracted, and on (-3, 1, 2) and (4, ) is increased.

log1 2 t is minus.

According to the law of same increase and difference decrease.

y is increased on (- 3) and (1, 2, 4).

On (-3, 1, 2) and (4, ) is minus.

The slope of the straight line l: x+ 3y=0 is -1 3, so, the slope of the perpendicular line of the straight line l is 3, the equation of the perpendicular line m of the straight line l passing the point q is y+ 3= 3(x-3), y = 3(x-4), the circle p is tangent to the line l with the point q, so p is on the straight line m. Let p(x, 3(x-4)).

The radius of the circle p is equal to pq= [(x-3) 2+3(x-3) 2]=2|x-3|.

Circle x 2 + y 2-2x = 0, (x-1) 2 + y 2 = 1, center of circle a(1,0), radius is 1

Circle p and circle x 2+y 2-2x=0 inscribed, pa=1+pq, x-1) 2+3(x-4) 2=1+4(x-3) 2+4|x-3|,12-2x=4|x-3|, from 12-2x=4(x-3), we get x=4, where p(4,0), pq=2, the equation of the circle p.

x-4)^2+y^2=4;

From 12-2x=-4(x-3), x=0, where p(0,-4 3), pq=6, the equation of the circle p x 2+(y+4 3) 2=36

**Unclear welcome to continue to add, thank you for adopting!