Senior 3 math problems. Finding a solution, senior three, mathematics, finding a solution

Do it with the ** method.

First, draw a unit circle in the Cartesian coordinate system, with the center of the circle at the coordinate origin.

Then let the vectors a and b correspond to the points a and b on the unit circle, respectively. It can be found that vector OA = vector A, vector ob = vector B.

From the vector a*vector b=, it can be obtained from the angle aob=60. (This should be very simple, don't explain in detail, you can't ask).

Let the point corresponding to vector C in the coordinate system be point C, and the vectors A-C correspond to ray CA, and vector B-C correspond to ray CB.

Then the angle between the vector a-c and b-c is 30 degrees, and the angle acb is equal to 30 degrees.

It can be found that the angle AOB is equal to twice the angle ACB, and it is known from the garden theorem that the circumferential angle of an arc is equal to half of the angle of the center of the circle it opposes, and the point c is on the circumference.

At this time, you can move point C on the circumference of the unit at will, and it can be found on the graph that when the points c, o, and a are collinear, the maximum AC is 2That is, the modulus value of vector a-c is the largest, which is 2

In terms of a person's learning experience, when doing multiple-choice fill-in-the-blank questions, it will be simpler and clearer to do it with the ** method

<> OA and OB represent a unit vector with an angle of 60 degrees, and the possible position of C can only be the point on the two equal circles in the figure, (both are the size of the unit circle), so the maximum modulus of OA-OC should be 2 in diameter, ** is better to understand, and the construction of algebraic methods may be cumbersome.

CD is definitely not right, it should be A.

(1) an=3n-2, (2) tn=n (3n+1) analysis: sn-sn-1=an can be found; Finding the general term of an shows that the series a1=1, and the equal difference series with a tolerance of 3 is 1 anan+1=(1 an-1 an+1) 1 3

then find tn.

First, write out the Cartesian coordinate system equations for circles C1 and C2.

c1：(x-2)^2+y^2=4

c2：x^2+y^2=4y

That is, x 2+(y-2) 2=4

Draw two circles on a Cartesian coordinate system, which can be seen.

Two circles intersect at two points, one is the coordinate origin, and the other is (2,2) The straight line formed by the connection of the two points is the straight line where the common chord is located.

The equation for the straight line is y=x, and the conversion to the polar coordinate system is rsin =rcos, and the solution is = 4 or =3 4

The second question is to rotate the straight line 30 degrees clockwise, and intersect the two circles at two points ab, |ab|=|oa|-|ob|= 4cos15 degrees - 4sin15 degrees = 4 * (cos.)

45 degrees - 30 degrees) - sin (45 degrees - 30 degrees)) = 4 * root number 2 2 = 2 times root number 2

Answer: 2 times the root number 2

log16x log4 to the 2nd power.

x is known by the previous step to the 3rd power of x 4.

So the problem is equivalent to the 2nd power of 4, and the number of times is equal to the 3rd power of 4.

So it should be 3/2

p(x,y)

then pf= (x-1) +y].

p to x = -1 distance = |x-(-1)|=|x+1|√(x-1)²+y²]=|x+1|

square x -2x+1+y =x +2x+1

So c1 is y = 4x

f(x)=sin2x+cos2x

2(√2/2*sin2x+√2/2cos2x)=√2(sin2xcosπ/4+cos2xsinπ/4)=√2sin(2x+π/4)

So t=2 2=

Maximum = 2

1. The trajectory of the moving point is a parabola with f(1,0) as the focus and the straight line x= 1 as the line, p=2, then the trajectory equation of the moving point is y = 4x. Since the point t is on the curve C1, let T(t, 2T), the radius of the circle C2 is R, using the perpendicular diameter theorem, r = (t) 4, the distance from the center of the circle to the straight line x= 1 is d = t 1, r d = [(t) 4] (t 1) =3 2t, according to the value of t, the relationship between r and d can be judged, so as to judge the position relationship between the line x= 1 and the circle c2.

2、f(x)=2sinxcosx＋cos2x=sin2x＋cos2x

2sin(2x＋π/4)

The minimum positive period t=2 |ω|=2 2= , the maximum value of the function is 2, obtained if and only if 2x 4=2k 2, i.e., x=k 8, where k is an integer.

1.Let the coordinates of the moving point p be (x,y).

By the title, got.

(x-1)^2+(y-0)^2]=|x+1|I tidy it up and get it.

x=y2 4, this is the equation for the curve c1.

2.Every time the question is written, it should be f(x)=2sinxcosx+cos(2x).

f(x)=2sinxcosx+cos(2x)=sin(2x)+cos(2x)

2sin[2(x+π/8)]

The minimum positive period tmin=2 2=

When 2(x+8)=2k+2(kz), there is f(x)max=2

1: Let p(x,y).

x+1 = under the root number (x-1) 2+y 2

The two sides are squared and simplified to C1: y 2=4x

2: I don't know if you made a mistake in the title.

f(x)=2sinxcosx+cos2=sin2x+cos2 has a minimum positive period and a maximum value of 1+cos2

Also: f(x)=2sinxcosx+cos2x=sin2x+cos2x=sin(2x+4) twice the root number

The minimum positive period is , and the maximum is root number 2

is a parabola and the equation is y 2=4x

Root number 2 x sin(2x+ 4) has a minimum positive period of , and a maximum value of 1 root number 2

You have to ask for the C2 equation.

Split the latter into the previous form of the final unary equation with a formula, and then the graph or anything can be solved, very simple, send a proposition.

The second question seems problematic to me, but that's the way it works