Solve a high 2 math problem and a high 2 math problem

Solution:1Because f(x)=x2(ax+b)=ax 3+bx 2 then f'(x)=3ax^2+2bx

And because f(x) has a limit value at x=2.

then f'(2)=12a+4b=0………1)

And because the tangent of f(x) at (1, f(1)) is parallel to the straight line 3x+y=0.

The slope of the straight line 3x+y=0 k=-3

then f'(1)=3a+2b=k=-3………2) Simultaneous (1) and (2) are solved to a=1, b=-3

2.There is 1 to get a = 1 and b = -3

So f'(x)=3x^2-6x

When f'(x) >0, i.e., when x>2 and x<0, f(x) is monotonically increasing.

When f'(x) <0, i.e. when 0When x<2 is <, f(x) is monotonically decreasing.

f'(x)=3ax^2+b

f at x = 2'(x)=0

12a+b=0

Because the tangent of the image at the point (1, f(1)) is parallel to the straight line 3x+y=0.

So f'(1)=-3

3a+b=-3

So: a=1 3

b=-4f(x)=1/3x^3-4x

f'(x)=x^2-4

f'(x) x>2 or x<-2 at >0

f'(x) -22 x<-2 at <0

Monotically decreasing -2

The process of solving the problem is correct, the minimum slope is -3 2, the minimum slope point is (1 2, -1 2), and the tangent equation using the point slope is (y+1 2)=-3 2*(x-1 2).

b -2a is the abscissa, substituting y'The slope of the curve at x=b -2a is obtained, which can be substituted into the original equation to obtain the ordinate of the curve at x=b -2a.

Maybe it's better to relate this problem to physics, where you think of x as time, y as the equation of the trajectory of the object's motion (i.e., the equation of position), and the derivative of it as the velocity of the object at each position, y', this problem is equivalent to the trajectory equation of a known object, finding the point with the smallest velocity. In this way, the velocity y is derived from y', by y'Get the moment when the point with the smallest velocity is x=-b 2a, substituting y'Find the minimum velocity, substitute y to find the position at the moment when the velocity is the smallest.

Let the radius of the circumscribed circle be r, then b sinb=2r c sinc=2r a sina=2r

sinb+sinc+sina) times (sinb+sinc-sinina0=3sinbsinc,b 2r+c 2r+a 2r)(b 2r+c 2r-a 2r) = 3bc 4r 2

b^2+c^2-a^2=bc (1)

1) Substituting the cosine theorem cosa=(b 2+c 2-a 2) 2bc=bc 2bc=1 2

a=60 b, c are the two roots of the equation x-square-3x+4cosa=0 b》c, so b+c=3

bc=4cosa=4*1/2=2

b^2+c^2-a^2=bc……》b+c)^2-a^2=3bc……》9-a^2=6

a = 3 x square - 3 x + 4cosa = 0......》x^2-3x+2=0

x-2)(x-1)=0 two x1=2 x2=1 b>c b=2 c=1

The degree of angle a and the value of abc a= 60 a= 3 b=2 c=1

Since a 2+c 2 = b 2, the triangle abc is a right triangle r=1 (because the midpoint of the hypotenuse of the right triangle is the center of the outer garden).

1)y=2/（ t+1）

t=(2/y)-1

x=((2/y)-3)/(2/y)

x=(2-3y)/2

2x+3y=2

2) a n+1 3an+1an 4a n=0(an+1-4an) (an+1+an)=0an+1=4an or an+1=-an (rounded off because it is a positive sequence) an=4 n 2

Question 1: t=2 y-1 is obtained from y=2 (t+1), which can be obtained by substituting x=(t-2) (t+1).

The second question is: (an+1-4an) (an+1+an)=0 from the equation, so that an+1=4an or an+1=-an; When an+1=4an, an=2*4 (n-1), when an+1=-an, an=2*(-1) (n-1);

3y/2=3/(t+1),x+3y/2=(t-2)/(t+1)+3/(t+1)=1

Simplification yields 2x+3y-2=0

The original formula can be reduced to (an+1-4an) (an+1+an)=0 because it is a positive sequence, so an+1=4an

So an=2*4 (n-1).

As for the fact that x is not equal to 1, the reason is that the equation for x can be reduced to 1-3 (t 1) and only when x approaches infinity there is x 1

Looking at the parametric equation, we can see that it is impossible for x to take 1

1) First of all, the image of y=absolute value x is in one or two quadrants, you draw the angular bisector of one or two quadrants, which is the image of this function, m is a point on this image, and the perpendicular line is made in the direction of m to the x-axis y-axis, because the angle is 45°, so it is hooked into a square with the xy-axis respectively, so the distance from the point m to the two coordinate axes is equal, 2) first, to ensure the existence of circle c, use the discriminant formula d 2+e 2-4f of the circle, that is, (-a) 2+(2a) 2-4(2a+1)>0, From this to get a range, as for the process of finding the range of a, you should be, I won't say, find a< a>2, in addition a also has a range, because it is through p(2,1) to do tangents, so this straight line through p, there are countless straight lines through p, and each of them has 2 tangent points with the circle c, so to ensure that p is not in the circle or in the circle, p has only one in the circle, p in the circle then any straight line through p has no tangent point with the circle c, so at this time use the radius formula of the circle, d 2 + e 2-4F is squared, multiplied by 1 2, which is the radius of the circle C, the coordinates of the center of the circle C are (A 2, -A), and the distance between the point P and the center of the circle is (2-A 2) 2+(1+A) 2

To ensure that p is not on or in a circle, we can list an inequality (2-a 2) 2+(1+a) 2 >12 (5a 2-8a-4) solution a>3, which is combined with a< a>2, and the final solution set of a can be obtained.

So the end result is a>3

Tired。。

1.When the distance from the point m to the two coordinate axes is equal, the point m passes through the straight line y=|x|and perpendicular to the plane of the plane where the two axes are located.

Because the curve is in the quadrant (including the origin), the quadrant angle is bisected, so the adequacy is satisfied; But point m can also be in the first.

Three and four quadrants, so the necessity is not satisfied.

2)(x- a 2) 2+(y+a) 2=5 4a 2-2a-1, the square of the distance between the center of the circle and p is.

2-A 2) 2+(1+a) 2>5 4a 2-2a-1 to get a>-3

1.Because y=-|x| 2.Equation thinking, list it ten equations to solve together, here I will not solve,

1.Knowable.

Point m is in the curve y=|x|Above" => m, the distance to the two axes is equal, then the distance from m to the two axes is equal => |y|=|x|m may be below the x-axis and thus not in y=|x|Above.

In summary2Point m is in the curve y=|x|"Up" is a sufficient and unnecessary condition for "the distance from the point m to the two axes is equal".

x²+y²-ax+2ay+2a+1=0

x-a 2) 2+(y-a) 2=5a 2 4-2a-1 then 5a 2 4-2a-1>0 and.

OP 2>5A2 4-2A-1 i.e.

0<5a 2 4-2a-1<(2-a 2) 2+(1-a) 2 to get a<-2 5 or 2

Because the distance from the point m on the curve y=-x to the two coordinate axes is also equal;

(1): The solution is determined by x 2-x+1 =>(x-1 2) 2+(3 4) 3 4, so two equations are obtained.

3x 2+px-6<6*(x 2-x+1)3x 2+px-6>-9*(x 2-x+1)i.e.: 3x 2-(p+6)x+12>0

12x^2+(p-9)x+3>0

(p+6) 2-144<0, (p-9) 2-144<0, i.e., -12=0 gives (x-a)(x-b) 0 and (x-c) 0 a2 The equation is (x-1)(x-3)(x+2)<0 is (x-1)(x-3)>0 by 1, so (negative infinity, 1) and (3, positive infinity).

x+2)<0 so x<-2

x (negative infinity, -2) and (3, positive infinity).

Friend, give it to me, and thank me for my hard work. I didn't do the math, but I'll leave it to you!

The first question: the denominator is (x-1 2) 2+3 4, which is Evergrande in 0 (this kind of problem, generally the denominator is Evergrande in 0), so multiply the denominator directly into two formulas without changing the sign, 3x 2-(p+6)x+12>0, and 12x 2+(p-9)x+3>0, both are unary quadratic inequalities, to be everbright in 0, the discriminant formula can be less than 0 (you can't take =), (p+6) 2-144<0, he (p-9) 2-144< 0, we get -12=0 to get (x-a)(x-b) 0 and (x-c) 0, which is the result of not getting it, and at most it is only one of the cases.

For this type of problem, the needle threading method is the fastest, and almost all of them can be solved by oral arithmetic. Of course, if you want to be rigorous in the process of argumentation, you can discuss it in categories:

You can ignore the equal sign first, and add it at the end, or you can do it in one step, and each step is strictly written with the equal sign and the unequal sign), the numerator denominator must be divided into two categories of discussion with the same sign, because the numerator has two, so it needs to be discussed again, a total of 4 situations, each case is an inequality group with three inequalities, and the solution can be done separately.

Although it is a bit troublesome, it is basically just a matter of distinguishing and writing directly.

You can do this for b, x 2-2ax+(a+2)=x 2-2ax+a 2-a 2+a+2=(x-a) 2+(-a 2+a+2)=0

So a 2-a-2 = (x-a) 2> = 0

a+1)(a-2)>=0 is solved to a=<-1 or a>=2b, and the two roots of the equation are a a+1)(a-2), and for a solution is x<1orx>4

When a=<-1 is a solution that is not empty.

When a>=2, a large root a+ a+1)(a-2)>4 or a small root a-a+1)(a-2)<1 is required

Need to solve the range of a can be found, I calculated that it should be a>18 7 Let's see. The answer to the possible correct solution should be a=<-1or a>18 7 There is no detailed answer, it may be wrong to solve it tomorrow, hehe.