1 high school math problem. Speed!!

y=sin²x+2sinxcosx+3cos²x1+sin2x+2cos²x

sin2x+cos2x+2

2sin(2x+π/4)+2

The minimum value is 2-2 at this time.

2x+ 4=k, which is x=- 8+k 2

y=sin x+2sinxcosx+3cos xy=1+sin2x+2cos x【Exploit sin x+cos x=1】y=sin2x+cos2x+2【utilize cos2x=2cos x-1】y= 2sin(2x+ 4)+2 [utilize sin2xcos 4 + sin 4cos2x== 2sin(2x+ 4)+2].

When 2x+4=k, i.e., x=- 8+k2, y takes the minimum value of 2-2

y=sin²x+2sinxcosx+3cos²x(sin²+cos²x)+sin2x+2cos²x1+sin2x+cos2x+1

2+2√2sin(2x+∏/4）

The value range is [2-2 2, 2+2 2] (2 2 means 2 to the 2nd power).

(1) The equation for a circle is (x-a) +y = a +4, then the center of the circle c(a,0) and the radius = a +4

k=1, then the equation for the line l is x-y+2=0, and the chord length obtained by the truncated line is 2 2 c to the distance from the straight line = la+2l 2

a +4 = 2 + (a + 2) 2, a -4a = 0, solution a = 0 or a = 4

The equation for a circle is x + y -4 = 0 or x -8x + y -4 = 0 (2) substituting y=kx+2 into the equation for a circle gives x -2ax + (kx + 2) -4 = 0

1-k²)x²-(2a-4k)x=0

There must be x=0 as the solution of the equation, in this case y=2, the coordinates of the change point are (0,2) No matter what real number k and a take, there is always an intersection point between a line and a circle.

(1) The equation of the circle c is arranged to obtain (x-a) 2+y 2=a 2+4 the center of the circle c(a,0) with a radius of (a 2+4) when k=1, l is x-y+2=0

l distance to the center of the circle d = a+2 2

d. The radius and the half-chord length form a triangle at a straight angle.

a+2 2) 2+2=a 2+4 gives a=0 or 4

The equation for a circle is x 2 + y 2 = 4 or (x-4) 2 + y 2 = 20 (2) The equation of simultaneous l and c is simplified.

k 2 + 1) x 2 + (4k-2a) x = 0 Obviously, no matter what value k and a take, the equation always has roots.

That is, there is always an intersection point between a straight line and a circle.

Let t=x 2-2, then x=plus and minus root signs (t+2).The inverse of both sides of the original equal sign at the same time gives f(t)=f inverse (positive and negative root numbers t+2).then x 2-2 = plus and minus root signs (x + 2); then x is equal to 0 or 2

1.Proof: Take any 10, get it!

2.I guess: the height of the cone and cylinder is 2, otherwise there is ambiguity. Next, please flip through the book to find the formula and calculate the result: s=(5+ 5), v=8 3(Word limit 100, can't say more).

1 f(x)=-x 2+2x=-(x-1) 2+1 axis of symmetry x=1 with the opening facing downward.

In (1,+poor) subtract function.

2 cones, column height 2

Cone: Female = 5, S = RL = 5

Body v 1 3 * r 2 * 2 = 2 3

Cylindrical surface: 2*2 r+ r 2=5, body v 2* r 2=2 plane(5+ 5), body v=8 3

take x1,x2 on (1, positive infinity) and x1 x2f(x1)-f(x2)=(x2-x1)(x1+x2-2)x2-x1 0 x1+x2 2

f(x1)-f(x2)=(x2-x1)(x1+x2-2) 0f(x)=-x's square + 2x is a subtraction function on (1, positive infinity), so I don't know what it means.

The first question is the same as above (originally the first asana was written, but you can only enter 100 words at most) 2, s = 1 2l * r (cone surface area) + 2 r * h (cylindrical side area) + r * r (cylindrical bottom area) = 1 2 * 5 * 1 + 2 * 1 * 2 +

v=1/3sh+πr*r*h=1/3*π*1*1*2+π*1*1*2=8/3π

So cos( +30) <0

sin(α+30)=3/5

Because sin ( 30) + cos ( 30) = 1 so cos( +30) = -4 5

So cos

cos[(α30)-30]

cos(α+30)cos30+sin(α+30)sin30

(1) From a(n+1)=sn 3, we can bring in and obtain:

n=1,a2=s1 /3=1/3 (s1=a1)n=2,a3=s2 /3=4/9 (s2=a1+a2)n=3,a4=s3 /3=16/27 (s3=a1+a2+a3)n=4,a5=s4 /3=64/81 (s4=a1+a2+a3+a4)

an=sn-s(n-1)=3a(n+1)-3an (n>1, n is a natural number).

Find a(n+1)=4 3*an

then = 1 (n=1);

1/3 (n=2);

4 (n-1)] 3 n (n>=3, n is a natural number) (2)a2+a4+a6+.a2n=s2n -s1=3a(2n+1) -a1

4/3)^2n -1

a1 = 1, a(n+1) = one-third sn

So a2=1 3, a2=1 9, a3=1 27an=(1 3) (n-1).

Original = 1 3 + (1 3) 3 + (1 3) 5 + ...1/3)^(2n-1)

1/3(1-(1/9)^n]/(8/9)=3(1-(1/9)^n)/8

a(n+1)=s(n)/3 s(n)=3a(n+1)

a(1)=s(1)=1

a(2)=s(1)/3=1/3

a(3)=s(2)/3=[a(1)+a(2)]/3=4/9

a(4)=s(3)/3=[a(1)+a(2)+a(3)]/3=16/27

a(n)=s(n)-s(n-1)=3a(n+1)-3a(n) (n>=2)

a(n+1)/a(n)=4/3

Therefore, the number series is an proportional series with 1 3 as the first term and 4 3 as the common ratio.

So a(n)=4 (n-2) 3 (n-1) n>=2 a(1)=1

2)a(2n)=4^(2n-2)/3^(2n-1) a(2n+2)/a(2n)=4^2n/3^(2n+1)/[4^(2n-2)/3^(2n-1)]=16/9

a2+a4+ +a2n

a2(1-(16/9)^n)/(1-16/9)

3/7*[1-(16/9)^n}

Ellipse x 2 a 2 + y 2 b 2 = 1 (a>b>0), e = 3 3, the straight line l passing through the right focal point intersects the ellipse at a and b two points, and when the slope of the straight line l is 1, the distance from the quiet origin of the high trembling cavity to the straight line l is 2 2.

1) Find the values of a, b.

2) Whether there is a point p on c, when l rotates a certain position around f, there is a vector op=oa+ob is true, if it exists, find the p coordinates and l equation, if not, please explain the reason.

1) Analysis: e= 3 3, a= 3c

Let the equation for the line l be x=y+c

The distance from the origin to the straight line l is 2 2, c|/√2=√2/2==>c=1

a=√3, b=√2

2) Analysis: Let the equation of the straight line l be x=my+1==> x 2=m 2y 2+2my+1

Ellipse: 2x 2+3y 2=6

2m^2+3)y^2+4my-4=0

y1+y2=-4m/(2m^2+3), y1y2=-4/(2m^2+3)

x1+x2=m(y1+y2)+2

Let p(x,y).

Vector op=oa+ob

x=x1+x2=-4m^2/(2m^2+3)+2=6/(2m^2+3)

y= y1+y2=-4m/(2m^2+3)

Substitute shirt into the ellipse: 72 (2m 2+3) 2+48m 2 (2m 2+3) 2=6

24(3+2m^2)/(2m^2+3)^2=6

4=(2m^2+3)==m1=-√2/2, m1=√2/2

There is a point p(3 2, -2 2) or p(3 2, 2 2).

The equation for a straight line l is x=- 2 2y+1 or x= 2 2y+1

1, (1 + root number 3) sin2 + (1 - root number 3) cos2 = root number (1 + 3 + 1 + 3) sin (2 -15) = 2 root number 2 sin (2 -15).

The maximum value is 2 root number 2

2,tana=1/3

sina = root number 10 10

a/sina=c/sinc

1 (root number 10 10) = ab (1 2).

ab = root number 10 2

3,tan(a+b)=(1/4+3/5)/(1-3/20)=17/17=1

tan(a+b)=tan(180-c)=180-c=45c=135°

The angle A is the smallest and the A side is the smallest.

tana=1/4

sina = root number 17 17

a/sina=c/sinc

a = (root number 17 17 * root number 17) sin135a = root number 2

The edge length of the smallest side is 2