- OptionA: consider two tetrahedrons of the cube, not having any two vertices among them in common. Each tetrahedron must have numbers of same parity. Hence the ans is 2*(4!)².

Option B: This is division of 4even and 4odd numbers into pairs. ({4 \choose 2})^2= (4!)²/16. Then we can distribute these in 2^4 * \frac{4!}{2!*2!}= 6*16 ways.

So final answer is 6*4!².

Option C: answer is 0.

Option D: number of tetrahedron’s in a cube=2. Each tetrahedron has 4C3=4 equilateral triangles. So answer is 8.

Other questions are my doubts as well

Is answer 4 Buddy @Cannizzaro for 53.

If Yes then What I just did is at first ignore the Box Funxn and Solve for x you will get x=0 and x=1/2 now just apply the GIF Property so 0 will be going to 1 and 1/2 remains as it is. So (1+0.5+0.5)^2=2^2=4

If for 41 Answer is B and C then,

Just Define Cases,

x=0 then from two equations

y<= 3 and y^2+y<=3 solve both and Apply [GIF] coz here x and y both are integers. so

I did that and got

y belongs to[-4,2]

similarly put y=0

x belongs to [-4,2]

and then carry on with x=1 but make sure you eliminate the repeated solution sets.

In Total I got different 17 solution Sets’

Aand 14 vertexes (Combined all 4) to make a Square

Me too.

answer is 1,25, in solution they directly wrote interval as (0,(root(5)-1)/2)

ac, it has been solved in another thread already by the name shm soap bubble

Ok.I might have missed a factor of 2 somewhere.

Oh got it.I need to take the mean height.

no, its not just a factor of 2. check out danish javed’s solution in the aforementioned thread, rather it’s something that normally eludes thought

Which ones are remaining?

I have done a bit different.In my solution I need to take mean height.

41,53 (41 has been solved in this thread, but the method gets really lengthy)

Find range of second function, and critical points.

x=-1 is asymptote of RHS. Now, see where they are close to each other and find intersection points there.

I’m getting from 0 to some number between 0 and 1, didn’t calculate.

yeah that number is 1/2

p=0 and q=1/2

41 cd

as per answer, p is 0, q is (root(5)-1)/2

for the 41th…

we can add both the inequality to get

x^2+y^2+2x+2y<6

then adding 2 both side

(x+1)^2 + (y+1)^2\le6 now this looks more solvable

53- Is the answer 5/4?