A question about senior one mathematics an equation has a solution in a certain interval and the ra

x 2 + (m-1) x + 1 = 0, discriminant = (m-1) 2-4 = m 2-2m-3> = 0, m< = -1 or m> = 3.

Let f(x)=x 2+(m-1)x+1, and there are three cases:

1. There is only one solution to the equation.

If m=-1, then the equation is x 2-2x+1 = 0 and x = 1 in the interval [0,2].

If m=3, then the equation is x 2 + 2x + 1 = 0 and x = -1 is not in the interval [0,2].

2, the discriminant formula is greater than 0, but there is only one solution in the interval [0,2].

then f(0)=1>0,f(2)=4+2(m-1)+1<0,m<-3 2.

3, there are two solutions in the interval [0,2] with the axis of symmetry x=(1-m) 2.

then 0<(1-m) 2<2,f[(1-m) 2]=(1-m) 2 4-(1-m) 2 2+1<0, f(0)=1>0, f(2)=4+2(m-1)+1>=0.

3/2<=m<-1。

To sum up, take the union of m=-1, m<-3 2, -3 2<=m<-1, then m<=-1.

Let y=x2+(m-1)x+1

Because: the equation has a solution on [0,2].

So: when x is equal to 0 and 2 respectively, the value of y should be a different sign.

When x=0, y=1

When x=2, y=4+2m-2+1<=0

2m<=-3

m<=

Therefore, m takes the maximum value at 1 2, and in the interval [-1 ,1] 1 2 is farthest from -1, so the minimum value at -1 is substituted by x=-1 to obtain the minimum value of -2, and the maximum value is 1 4, so that the value range of the real number m.

2 + (m-1) 1 = 0, discriminant.

(m-1) 2-4 = square meters 2-2 m-3> = 0, m = 3.

Let f(x) = x 2 + (m-1) x +1, in three cases:

1. The equation is the only solution.

If m = 1, then the equation x 2-2x of 1 = 0, x = 1 when in the interval [0, 2].

If m = 3, then the equation x 2 2 1 = 0, x = 1, not in the interval [0, 2].

2, the discriminant is greater than 0, but there is only one solution in the interval [0,2].

f(0) = 1>0, and f(2) = 4 + 2 (m-1) 1 < 0, m <-3 2.

In 3, there are two solutions, in the interval [0,2], the axis of symmetry.

of = (1-m) of 2.

0 < (1-m) of 2 < 2, f [(1-m) of 2] = 1-m) of 2 4-(1-m) of 2 2 1 0, f(2) = 4 + 2 (m-1) 1> = 0.

3 2 < = m<-1. “

For summary, take m = 1 and m <-3 2, -3 2 < = m<-1, set, m< = 1.

All you need to do is prove that the function values at both endpoints of the interval are outliers.

f=(x+1)(x-2)(x-3)-1, substituting x=-1 and x=0, one greater than zero and less than zero, according to the continuity of the function, there must be a 0 solution in the interval.

Find the values of f(-1) and f(0), and if you have a different sign, you have a solution.

To make f(x)=ln(x+8-a x) an increment on [1,+, just g(x)=x+8-a x is an increment on [1,+, i.e., g'(x)=1+a x =(x +a) x 0 Constant Formation Known x 1

Then x +a 0 is constant.

It only takes x=1 to be true when the condition is satisfied.

So 1+a 0

Solve a -1

Hope it helps you o(o

Solution: lnx is an increasing function, so if this problem is transformed, it will be x+8-a x is an increasing function on [1,+, and further x-a x is an increasing function on [1,+, and then classified and discussed.

When a<0, on x[1,+, when a2=-a, i.e., x= (a), take the minimum value, so = (a) 1,-1 a<0

When a=0, the primitive =x is an increasing function on [1,+, and when a>0, it is also an increasing function on [1,+.

In summary, a(-1,+

From f(2)=4a+2 b, i.e., -3<=4a+2 b<=6, 6a <=3 b<=9-6a

f(3)=9a+3 b, then from above, it can be obtained, < = 9a+3 b<=(9-6a)+9a, i.e., <=f(3)<= 9+3a

This kind of topic is just a matching number, you can do this in the future, count it yourself, study hard, go up every day, hehe.

f(3)=2 3f(2) knows the range of f(2) and f(3) comes out naturally.

1) Set the coordinates of point p to (x,y).

Then the slope of the pa is y (x-2).

x≠2), the slope of pb is y (x+2).

x≠-2) is the product of the slope of the straight line pa,pb.

Get y 2 (x 2-4) = -3 4

x≠ 2) Organize the trajectory of the moving point p. c

The equation is: 3x 2+4y 2=12

At 20, t 1 m+m 2 (m*(1 m)) 2, and the equal sign is taken if and only if m 1 m, i.e., m 1 (m=-1 rounded).

When m<0, t 1 m+m (1 m+(-m)) 2 (-m*(-1 m))=2

Take the equal sign if and only if -m -1 m, i.e., m -1 (m = 1 rounded).

So t 2 or t -2

1/2≤1/t≤1/2

i.e. -1 8 k 1 8

Since -2 in the curve c is rounded off at this point.

When the banquet segment m is , the linear l equation is x, then the two points e and f are symmetrical with respect to the x axis, and the midpoint m coordinate is (,0).

The straight line am obviously coincides with the x-axis, i.e. y 0, and the slope is 0, indicating that k 0 exists.

Therefore, the value of k is: -1 8 k 1 8

1) Let p(x,y) then kpa=y (x-2) kpb=y (x+2).

kpa*kpb=-3/4

substitution. x2 4 + y2 3 = 1 is an ellipse.

2) Let the equation of the straight hail line l: y=k1 (and the elliptic equation are coupled to obtain the equation failure:

4k*k+3)

x*x-4k*kx+4k*k-12=0

x1+x2=4k*k/(4k*k+3)

In the same way, y1+y2=-3k (4k*k+3).

m(2k*k (4k*k+3),-3k 2(4k*k+3)) can be read to find k=k 4(1+k*k) to apply the important inequality k+1 k>=1 to get k<=1 8

Let the coordinates of point p be (x,y).

Rule. y (x-2)]*y (x+2)]=3 4 Solving the equation yields 3x 2+4y 2=12

That is, the square ratio of x is 4 plus the square ratio of y is 3 equal to 1

It's an ellipse.

Let the coordinates of e and f be (x1,y1), respectively, and (x2,y2), then the coordinates of point m are ((x1+x2) 2,(y1+y2) 2) with the equation buried in 1.

My heart is not too good at doing it, so I don't count it down, I'm sorry.

Three steps:1Because there is a zero point, the discriminant must be greater than 0 (this can determine the range of a for the first time);

2.Use the solution equation to find the solution of x (including a), so that the solution is greater than or equal to 1 and less than or equal to 1 next to the rubber scattering side of the town to obtain the second range of a (note: the two ranges are combined and operated);

3.Just hand over the first and second ranges.

According to the defined domain, a x+4a -x k>0

From the mean inequality we get the original formula 4-k>0 to get k<4

And because the original formula ≠ 1 (the denominator is not 0), k≠3

In summary: k<4 and k≠3

The condition for f(x) to define the domain r is .

a x+4a -x k>0 is always true for any x.

Let a x=t>0

t+4/t>k t+4/t≥4

To make this inequality true for any t constant.

then k<4

Consider k=4 as long as t=2. The inequality does not hold up to k<4

sin bai+sin =1, then , must be the angle du of one or two quadrants, therefore: when zhi= =150 degrees, dao, cos +cos obtains the minimum internal value -

capacity 3; When = =30 degrees, cos +cos achieves a maximum value of 3. Therefore, the value range of cos +cos is: [-3, 3].