Doubts from a test

![q2|690x233]

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  1. 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 :no_mouth:

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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

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Me too.

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

For the first one I am getting AD.Is it correct?
@Cannizzaro

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

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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.

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

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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.

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yeah that number is 1/2
p=0 and q=1/2

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bro what is the answer for 41 @Cannizzaro??

41 cd

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

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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

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53- Is the answer 5/4?

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