High school math problems. Desperate for expert answers. Much appreciated!!

1)f(x)=-a(x^2-x/a)=-a[x-1/(2a)]^2+1/(4a)

r, the maximum point is x=1 (2a)>1, so the maximum value in the interval is obtained at the endpoint:

f(-1)=-a-1, f(1)=-a+1

f(1)>f(-1)

f(1)=-a+1=2 3 is the maximum value, and a=1 3

2) |sinx|<=1, f(sinx) is equivalent to f(t) and takes the value of t in the interval [-1,1], a>0, and its maximum value is t=1 (2a).

If its <=1, then a>1 2, then the maximum value f=1 (4a) 2, does not match.

Its minimum is at the endpoint f(-1)=-a-1<=-2-->a>=1, which is compliant.

If it is 1, then the maximum value is at the endpoint f(1)=-a+1<=-2-->a>=1Accord with.

a=0, f(t)=t, the maximum and minimum values are 1, and -1 does not match.

a<0, its maximum value is f(1)=-a+1>=2-->a<=-1, which is in accordance with the f(1)=-a+1.

Its minimum value is f(-1)=-a-1<=-2---a>=1, which does not match.

Therefore, the range of the combined a is: a>=1 or a<=-1

3)a=-1, f(x)=x^2+x=x(x+1)

Note b(k+1)=k [k(k+1)]=1 (k+1).

k<=n.

tk=b(2n+k)+b(2n-k)=1/(2n+k)+1/(2n-k)=4n/(4n^2-k^2)>1/n

tk<4n/(4n^2-n^2)=4/(3n)<2/n

t= (i f(i)) (up 3n down i = n) = t1 + t2 + .tn>1/n*n=1

t<2/n*n=2

It should be late to find the equation when the radius of the circle m is minimal...

Common chord AB equation: (x y 2x volt Zheng 2y 2) - (x y 2m x 2ny m 1).

2+2m)x+(2+2n)y-m -1=0 Because A and B score circle n, AB is the diameter of circle N, and through the center of the circle n (-1, -1) so bring in the straight line Fang Qin Li Chengde: -2-2m-2-2n-m -1=0 to get it: m +2m + 2n + 5 = 0

Since the equation for the circle m can be written as: (x-m) 2+(y-n) = n 2+1 so the square of the radius is: n 2+1

The radius is minimal, i.e., n is minimum, i.e., 2n minimum.

Since 2n=m +2m+5, so when m=-1, 2n is the smallest, and the n=-2 circle n equation is x y +2x+4y=0

Solution: Do AC parallel line MN over D, as shown in the figure above.

It is easy to know that MDA=45 degrees ADB=30 degrees CDb=30 degrees CDN=75 degrees ADC=60 degrees DAC=45 degrees.

There is also cd=30km, so the land beam mill has a sinusoidal definite understanding triangle ADC to get AC=30*V3 Morning Bucket 2* (V2 2)=15*V6

In the triangle abc, ab 2=ac 2+bc 2-2*ac*bc*cos45 yields ab 2=750 ab=5v30 (v represents the root number).

This problem can also be understood with scumbag cosine, but the operation is more complicated, and the above is the simplest algorithm).

From the theorem of the relationship between roots and coefficients, it can be seen that the proposition p is equivalent to the discriminant formula 16-16m 0, that is, m 1, and 4 m is an integer, so that m = -4, -2, -1 and 1 are obtained, but to ensure that both roots are integers, m can only be 1Thus m=1 is the only condition for the proposition p to be true. When it is verified that m=1, the proposition q is also true.

Therefore, m=1 is a sufficient condition for p q to be true. Because it is unique, it is also a sufficient and necessary condition for a proposition that p q is true.

1. When m=0, q is not true.

2. When m ≠ 0, the first two equations should be δ 0, and solving these two inequalities yields -4 5 m 1, so the possible values of m are only -1 and 1, and the original equation is brought back to verify respectively, and we can know that only when m=1, the root of the two equations is an integer.

Therefore, the sufficient and necessary condition for p q is m=1

Solution: From the meaning of the question, it can be seen that the point p is the upper part of the ellipse x 2 4 + y 2=1, and the analysis problem shows that y-2 x-4 is the minimum value of the slope of the straight line passing through the point p and the point (4,2), and the simultaneous y-2=k(x-4).

x^2/4+y^2=1

Let =0 solve k=1 2

The minimum value of Y-2 X-4 is 1 2 satisfied].

In this problem, it can be seen as a system of equations x 2 4 + y 2 = 1 (y 0), y-2 x-4 = a, there is a real number solution, and the range of values of a is solved.

The graph of y-2 x-4=a is the inverse function y=2 x upward translation a+4 (or downward translation (-4-a)). At x>0 and close to zero, a can be taken as an infinitesimal negative. At x<0 and close to zero, a can be taken as a positive number of infinity.

There is a big problem with this question.

f(2)=2 (2a+b)=1, so 2a+b=2x (ax+b)-x=0, x[1 (ax+b)-1]=0, the only solution with a cluster should be x=0, the solution in the middle brackets gives x=(1-b) a, this formula should omit Zheng Chun's return = 0, so 1-b=0, b=1

Substituting the previous formula, a=1 2

So f(x)=2x(x+2).