Ask two high school math questions and two sophomore math questions

1.Because a=1, c=0, so f(x)=x 2+bx 1, that is, f(x)-1 0, that is, x 2+bx-1 0, and then the main dimension is reversed, and b is regarded as the main element, and x is regarded as the dimension, that is, x is known, so it becomes a one-dimensional inequality about b, because x (0, 1, so the inequality is brought in, -1 0 is constant, 1 2+1 b-1 0, and b 0, in summary, b 0 2That is, 4 x + m (2 x) + 1 = 0 holds, and the equal sign shifts both sides, that is, m=-(2 x+2 -x), that is, find the range of f(x) = -(2 x+2 -x), because x r, so (2 x) (0, + commutation, so that 2 x=t, t (0, + i.e., the original formula is y=-(t+1 t), and y (-2) is obtained from t, that is, m (-2).

1.From [f(x)]<=1 to obtain -1<=x 2+bx<=1, i.e., 1-x 2>=bx>=-x 2-1, 1 x-x>=b>=-x-1 x.

x-1 x increments, b>=decreasing, b<=0In summary, -2<=b<=0

1. f(x) denotes the absolute value, and the result is [-2,0]2, p is a true proposition.

2 x=t,t 2+tm+1=0 is constant on the defined domain.

Because t>0, the axis of symmetry is 0, 0

m≤-2

Agree with the view of the Seven Elements of the World, and use derivatives when judging increments. The answer is the same as his, and so is the process.

Let (m,n) be the point on the curve, then there is n=m 3-m ......The slope of the tangent line passing through this point is k=3m 2-1 (derivative of f(x)), so the equation for the tangent line passing this point is: y-n=(3m 2-1)(x-m), this tangent has to pass through the point (a, b), so b-n=(3m 2-1)(a-m), and because , it is sorted out: g(m)=2m 3-3am 2+a+b=0, the problem requires g(m)=0 to be derived from three solutions, and the derivative of g(m) is obtained, and two extreme points m=0, m=a, due to a 0, so judging by the image, the sufficient and necessary conditions for the three solutions of this equation are:

g(0) 0, g(a) 0, from g(0) 0 to get a+b 0, i.e., -a b, from g(a) 0 to get 2a 3-3a 3+a+b 0, i.e., b a 3-a=f(a), i.e., there is -a b f(a).

f(x)=(ax+b)/(x^2+1)=ttx^2-ax+t-b=0

Discriminant a 2-4t(t-b) >=0

i.e. 4t 2-4bt-a 2<=0

Discriminant (4b) 2-4*4(-a*2)=16(a2+b2)>0 The maximum value is 1, and the minimum value is -1

1,-1 is 4t 2-4bt-a 2=0 two roots.

4-4b-a 2=0,4+4b-a 2=0b=0,a=2,a=-2 (rounding).

ab=2*0=0

The second question is: write the derivative form of f(x) first, and then make them equal to 1 and -1 respectively, so that a system of equations can be formed, and the equations can be solved.

1.Let the coordinates of the q point be (3 2cosx, 2sinx) and replace them with triangulations.

Point a(3,1), point p(4,4),3).（3√2cosx-3，√2sinx-1）=3√2(sinx+cosx)-6=6sin(x+π/4)-6。

sin(x+4) is between [-1,1].

It should be at [-12,0].

Of course, you may wonder why there is no positive value, but in fact, the slope k1 between the two points of a and p and the slope of the elliptic curve k2 at point a satisfy k1*k2=-1, that is, the straight line ap and the elliptic curve are perpendicular to the tangent at point a. Specific Proofs:

Take the upper part of the ellipse of y 0, and then the original elliptic equation is converted to f(x)=y= (2-x 9), and the derivative of this function is obtained f(x).'=x (18-x), then the elliptic curve is tangent at point a, k1=f(3).'=1, and it is easy to get k2=1, which can be proved in summary.

Solution: Let q be labeled as (xo,yo), then the vector ap=(1,3), aq vector=(xo-3,yo-1), then the vector ap multiplies the vector aq=xo-3+3yo-3=xo+3yo-6

I can use trigonometric commutation to calculate the range of values of p(sin,cos) and then columnar sphere trigonometric formulas.

Let the point chord intersect at the two points a and b in the lease imitation ellipse, a(x1,y1) b(x2,y2) is obtained by x 36 + y 9 = 1

9 χ1² +36y1²=324……Cons 9 2 +36y2 =324......②

From — obtained, Zheng liquid x2-x1) 4(y2-y1) =0y2-y1) x2-x1)= x2+x1) y2+y1)