2 high school math problems to find the process, help to do 2 high school math problems The second q

1。(x,y) represents the point on the circle, and (y-2) (x-1) can be understood as the slope of the line connecting (x,y) and (1,2).

kpa = + infinity, kpb = 3 4Let's talk about the calculation of 3 4: the tangent of the angle opa is 1 2, and the tangent of the angle apb is calculated by the formula of the double angle: 4 3, and the inclination angle of pb is congruent with the angle apb, so kpb = 3 4.

The latter question uses the parametric equation of the circle, or trigonometric commutation. x=cosa,y=sina.

x 3 + y 4 = cosa 3 + sina 4 = 1 12 (4 cosa + 3 sina) = 5 12 sin (a + b), here the two angles and the sine formula are used inversely.

Therefore, the maximum is 5 12.

2 The function defines the domain r, and the denominator can be associated with trigonometric formulas. Let x=tana and the denominator becomes the square of seca. The molecule is atana+b

So y=(atana+b)cosa2. i.e. y=asinacosa+bcosa 2, y=1 2asin2a+1 2b(1+cos2a).

y=1 2(asin2a+bcos2a)+1 2b, y=1 2 root number(a 2+b 2)sin(a+b)+1 2b, here the two angles and the sine formula are used inversely.

by sin between -1,1.

According to the title: -1 2 root number (a 2 + b 2) + 1 2b = -1

1 2 root number (a 2 + b 2) + 1 2b = 4

The solution yields b = 3 root number (a 2 + b 2) = 5

That is, a=+-4, b=3

In the first question, the geometric meaning of (y-2) (x-1) is the slope of the straight line connecting a point on a circle with (1,2), and it is easy to get that when it is taken as the minimum value, the straight line is tangent to the circle, and the equation y=k(x-1)+2 (of course, k can also be meaningless, but it is not what we need).

The distance from the center of the circle to the straight line d=|-k+2|sqrt(k +1)=1, the solution is k=3 4, that is, the minimum value is 3 4

In the second question, directly let x=cos and y=sin

then x 3 + y 4 = (4 x + 3y) 12

Using the auxiliary angle formula for 4cos +3sin, we get 4cos +3sin =5 (4 5cos +3 5sin) = 5sin( +arctan(4 3)).

Its maximum value is 5, so the maximum value of x 3+y 4 is 5 12

By the title, (ax+b) (x 2+1) -1, (ax+b) (x 2+1) 4

Dissolve, got. x^2+ax+b+1≥0

4x^2-ax-b+4≥0

Of course, for any x r, both of these inequalities must hold.

Obtained by discriminant.

x+a/2)^2+b+1-a^2/4≥0

2x+a/4)^2-b+4-a^2/16≥0

Sorted out. a^2/4-(b+1)=0

a^2/16-(b-4)=0

The solution yields b = 3 a = 4

1.(y-2) (x-1) can be regarded as the slope of any point (x,y) and the point (1,2) on the circle, so when drawing a figure, it is easy to get the minimum value, just the tangent of the point (1,2) is tangent to the circle, set to y=k(x-1)+2, the distance from the center of the circle to the straight line d=|-k+2|(k +1)=1, the solution is k=3 4, that is, the minimum value of 3 4

2.-1=< y =<4

i.e. -1=< (ax+b) (x 2+1) =<4 after decomposition. ax+b >= -x^2-1

4x^2+4 >= ax+b

And then get. x^2+ax+b+1 >= 0

4x^2-ax-b+4 >= 0

Since it is greater than or equal to, it is required that the above two must be matched into a completely flat mode.

x+a 2) 2+b+1-a 2 4 >= 0 gives b+1-a 2 4=0

2x+a 4) 2-b+4-a 2 16 >= 0 gives -b+4-a 2 16=0

Solve the system of binary quadratic equations.

a=4 b=3

Or. a=-4 b=3

1) The sum of the maximum and minimum values on y=a x [0,1] is the 3 exponential function a>0 and is not equal to 1 (defined).

Since the exponential function is monotonic, the endpoint of [0,1] is taken regardless of the maximum or minimum value.

So a 0 + a 1 = 3, then a = 2

2）f(x)=2^2(x-1/2)-3*2^x+5=2^(2x-1)-3*2^x+5

1/2*2^2x-3*2^x+5

1/2(2^2x-6*2^x+10)

1/2[(2^x)^2-6*2^x+10]=1/2[(2^x)^2-6*2^x+9+1]=1/2[(2^x-3)^2+1]

Because 2 x > 0, when 2 x = 3, f(x) has a minimum value of 1 2.

I was originally on the first floor, but now I want to supplement the first question.

Note: Indicates the square.

1) Is it square or power? If it is squared, the answer is plus or minus root number 3, and if it is power, there is no maximum!

2) f(x)=2 2(x-1, 2)-3*2 x+5 (the process of the first floor is given).

Because 2 x > 0, when 2 x = 3, f(x) has a minimum value of 1 2.

I think the first question seems a little wrong, if a is not equal to 1 or -1 or 0, and it does not limit the range of x, this function has no maximum or minimum value, but if a is 1 or -1 or 0, it is not a sum of 3, I don't know how to do it.

2，f(x)=4^(x-1/2)-3*2^x+5

2 [2*(x-1 2)]-3*2 x+5 (replace 4 with 2 squared).

2^（2x-1)-3*2^x +5

2 (2x)] 2-3*2 x +5 (replace the -1 power of 2 with 1 2).

Let y=2x, then 2 (2x)=y2

So f(x)=y 2-3 y+5

And because x is in the range between 0 and 2, 2 x is in the range between 1 and 4, that is, y is greater than or equal to 1 and less than or equal to 4.

So the range of f(x) is 1 2 (when y=3) and when y=1).

That is, the maximum value of f(x) when x=0 is.

x=log2 3, (2 is the base) the minimum value is 1 2

In the first question, if a is a repetition, the function y decreases monotonically in the defined domain, and the maximum value is a=0 and y=1. So a is a positive number. The function y increases monotonically in the definition domain, i.e. the power of 0 of a is the minimum value of 1, and the power of a is the maximum value of a=2

c 2 = 36 - 27 = 9, so, c = 3If the hyperbola to the focal point is m(0,3) and n(0,-3), and a(x,4) is the intersection point, then x 2 27 + 16 36 = 1, and x 2 = 15 2a=an-am=8-4=4 .

a=2 a^2=4

So b 2 = c 2 - a 2 = 5 so the hyperbolic equation is: y 2 4-x 2 5 = 1

Because there is a common focus, let y 2 a 2-x 2 b 2 = 1 , a 2 + b 2 = c 2 = 9

Since y=4, substituting the elliptic equation gives x substituting x y into the given equation and substituting b 2=9-a 2 gives a 2

Therefore, the equation can be found.

1.The definition domain is the value range of x, and the value range of f( ) should be found first, and the value range of f( ) is unchanged.

The domain of the function f(2x-1) is defined as [0,1).

0≤x＜1-1≤2x-1＜1

1≤1-3x＜1

0＜x≤2/3

The domain of f(1-3x) is (0,2,3].

2.First, find the domain of the function y = 3rd root ax to the 2nd power + 4ax + 3rd ax-1.

The solution set for ax 2+4ax+3≠0 is r

16a^2-12a＜0

0＜a＜3/4

If you want to think of an image of y=ax 2+4ax+3, you need to make the solution set r, whether a 0 or a 0, it must be 0

belongs to [0,1], then 2x-1 belongs to [-1,1], (1-3x) belongs to [-1,1], and x belongs to (0,2 3].

2.As long as f(x)=ax to the power of 2 + 4ax + 3 = 0 has no solution, when a>0, <0, a<3 4

When a=0, it is compliant.

When a<0, a>3 4, there is no solution.

In summary: 0 a< 3 4

1: Simplification yields c(-1,1). The straight line passes the fixed point a(0,-3) and connects to ac,k(ac)=-4, so when the straight line is perpendicular to ac, the maximum k=1 4

2: ABC three-point collinear, o is one point outside the circle, so 2m+3n=1 (theorem). Therefore, 1 m+2 n=(2m+3n) m+2*(2m+3n) n=2+3n m+4m n+6=8+3n m+4m n, and the min obtained from the basic inequality is 8+4 root number 3 (but the premise m and n must be greater than 0).