Sophomore math, desperate for answers!!

The derivative of y=x 2+1 is y'=2x

Let p(a,a 2+1), then the tangent equation is y-a 2-1=2a(x-a) and y is tangent to y=-2x 2-1, so y'=-4x=2a, x=-a 2, so y=-2x 2-1=-a 2 2-1

Substitute y-a 2-1=2a(x-a).

A 2 = 4 3 and a = 2 3 3

So p(2 3 3,7 3) or p(-2 3 3,7 3) is wrong to assume that the tangent equation is y=ax+b, which is tangent to both curves, i.e., the system of equations formed by both curves has only a unique solution.

Bring the tangent equation into the first quadratic equation:

ax+b=x 2+1 i.e. x 2-ax+1-b=0 It has a unique solution and the sufficient and necessary condition is a 2-4(1-b)=0, i.e., a 2+4b-4=0

Bring the tangent equation into the second curve equation.

ax+b=-2x 2-1 i.e. 2x 2+ax+b+1=0 It has a unique solution, and the sufficient and necessary conditions are a 2-8(b+1)=0, i.e., a 2-8b-8=0

Solving these two equations for a and b together yields:

b=-1 3, a=+ - 4 sqrt(3) and then invert a,b to the above equation to find the coordinates of point p. (omitted...)

First find the common tangent, and then the intersection of the straight line with the curve y=x +1 is p

1/2^2+1/3^2+..1/n^2>1/(2*3)+1/(3*4)+.1/[n*(n+1)]

And: 1 (2*3)+1 (3*4)+1/[n*(n+1)]

1/2-1/3+1/3-1/4+..1/n-1/(n+1)

1/2-1/(n+1)

1/2^2+1/3^2+..1/n^2<1/(1*2)+1/(2*3)+.1/[(n-1)*n]

And: 1 (1*2)+1 (2*3)+1/[(n-1)*n]

1-1/2+1/2-1/3+..1/(n-1)-1/n

1-1 n So, it is proven.

Note: The left side of the certificate you want to prove is"1/2-1/(n+1)"bar.

Since the bright posture is 2x+a x squire 1, and x is a positive number, 2x2 -x+a 0, for this inequality to hold, then δ=1-2a old key cavity 0, a 1 8

Therefore, by the title, p q, and q cannot deduce p, i.e., p is a sufficient but not necessary condition for q.

Scribbling downstairs, don't mislead people's children, the two questions are actually very simple, classified discussion, and the axis of symmetry is in the known range of left, middle and right.

Let the height of the OCD be X, then the height of the OAB will be 17-X.

5²+x²=12²+(17-x)²

Solve out x=12

The radius of the circle r = 5 +12 , r = 13

When these four points are the key points of the corresponding edge, the maximum area is taken.

My own answer: The roots of x 2-2x + k=0 are a, b has a+b=2, ab=k because |a-b|=2√2(a-b)^2=(a+b)^2-4ab=4-4k=8k=-1

Let Red's shadow length at point A be x

Since OA = 40 meters, the distance between the top of the shadow and the distance from the O point to remove the tassel = 40 + X is 14 meters, and the height of the little red is 14 meters

14, i.e.:

14x=, x=60/

That is, the shadow length is meters.

Senior, are you sure it's a math problem?

The domain of the odd function f(x) is defined as [-1,1] and monotonically reduced at [-1,0], f(x) is monotonically reduced at [-1,1], and is determined by f(a)+f(a-1).