If m=2, the solution of f(x)=0 is x=-0.25, which is consistent with the meaning of the question.

If m<& gt2, the discriminant = 8 (m 2+5m-6) > =0, m.

If the image intersects the negative semi-axis of x, only the smaller root is less than 0.

The smaller root is {2m- radical sign [2 (m 2+5m-6)]}/(m-2).

If m < =-6, the denominator of the root is less than 0, just 2m- root sign [2 (m 2+5m-6)] > 0, and there is no solution.

If m>= 1, there are two cases:

If 1

If m>2, denominator is greater than 0, and 2m- radical sign [2 (M 2+5m-6)] is less than 0, the solution is 2.

To sum up, the value range of m is [2,3].

the second question

f(x)=[2x+(2p- 1)][2x-(p+ 1)]

The function has two zeros, one is 1-2p and the other is p+ 1. As long as one of these two zeros is in [- 1, 1], then the title condition holds.

So there is-1

The solution is 0

Now consider whether the boundaries can be equal. After testing, -2 and 1 are not desirable, and 0 is desirable, so:

-2 & lt； p & lt 1。

Third question

The first question:

(y-2)/(x-1) = (2-y)/(1-x) = [2 radicals (1-x 2)]/( 1-x).

The numerator and denominator are both greater than 0, so take the negative sign, the numerator is the subtraction function of X, and the denominator is the subtraction function of X, so the integral fraction is the increasing function of X, so when x=0, the minimum value is 1.

The second question: θ φ

Let x=sinθ, y = cos θ, and θ belongs to [0,2pi].

Using the auxiliary angle formula, then x/3+y/4=sinθ/3+cosθ/4= root sign (1/9+116) sin (θ+φ), where tanφ=3/4.

Because sin (θ+φ) < = 1, the maximum value of the above formula is the root sign (1/9+116), which is 5/ 12.

The fourth question

It's basically painting. These two functions are even functions. So we only need to consider the positive semi-axis.

At (0, 1), two function images intersect. At (1, infinity), since 2 x is above y=x and lgx is below y=x, there is no intersection between them.

So there are only 1 intersections. Considering the symmetry of even function, the equation has two solutions.

It should be pointed out that not all logarithmic functions and exponential functions do not intersect with y = X.

The fifth question

Let t = (1/3) | x | > 0, and the original equation is t 2-4t-m = 0.

The equation has a real number solution, that is, the equation has a solution greater than zero.

Then the discriminant 16+4m >: =0, that is, m & gt=-4.

[4+ number of roots (16+4m)]/2 > as long as the larger root is greater than 0; 0。 Obviously established. Therefore, m & gt=-4.