1. Apparently, there are no solutions for x>1 and x<-1.

Since 97~=30,8Pi,

the value of cos(97)<0

and graphically it also decreases in this point (that is also possible to check using first derivative test). Using the fact that cos(x) is the even function we could sketch out graphs for both y=cos(97x) and y=x and evaluate the number of intersetion points, which would be equal to 61.

2. If you try to put 4 balls of one color and 4 balls of another color in 7 bins, that each bin does not contain two balls of the same color, in that case at least one bin would contain two balls of different colors. Therefore, 0 is the answer.

3. Let points of this set S be the points of the square with vertices (0,0), (1,0), (1,1), (0,1).

Consider subset A of this set that

A={(x,0), where x is irrational}.

Apparently, the set A is both open and closed.

Therefore, S is

disconnected. (

Warning! Mistake! See further posts for more information)

Consider set family B(n) of such points in squares with side length 1-1/n (n>1). If we "insert" such "squares" in our square and add the set C of border points:

C={(x,0),(x,1),(0,y),(1,y), x,y are irrational}

we would have the open covering of our set S that has no finite subcovering.

Therefore, S is

not compact.

4. I couldn't solve this integral so far.

Please confirm whether it looks like this one.

5. To get the answer we should actually calculate the number of all possible 2x2 non-invertible matrixes in field q.

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