You will laugh, once you see the solution
Find the number of solutions of the equation :
8 ⌊ x 2 − x ⌋ + 4 ⌊ x ⌋ = 1 3 + 1 2 ⌊ sin x ⌋
where ⌊ . ⌋ represents the Greatest Integer Function
Note: This problem appeared in our AITS (All India Test Series) -7
The answer is 0.
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3 solutions
Discussions for this problem are now closed
Another way to interpret it is that the LHS is always even whereas the RHS is always odd.
i followed the same method
I followed the same one.
Oh yes, thats true. For a moment, I thought you were talking about even and odd functions .
I realised something else. Take ⌊ sin x ⌋ . Now, this must equal either 0 or − 1 , since − 1 ≤ sin x ≤ 1 . If it is equal to 0 , let ⌊ x 2 − x ⌋ be a , and ⌊ x ⌋ be b . Now, a , b ∈ Z . However, 8 a + 4 b = 1 3 has no integer solutions. Now, if ⌊ sin x ⌋ = − 1 , then 8 a + 4 b = 1 , which also has no integer solutions. Thus, there are no solutions to the equation.
I guessed the answer right from your caption ;)
that seems to be so logical....sometimes you have to apply plane logic rather than thinking too much
I really laughed! Thanks for the problem!
If you increase or decrease the value of x, changes in polynomial forms is drastic in comparison to sine function. So, logically, there can't be any solution.
what are the LHS & the RHS , please ?
Left Hand Side and Right Hand Side. :P
The LHS is an even term while the RHS is all Odd so there are no possible solution !!
Very nice observation!
what are the LHS & the RHS , PLEASE ?
8[x^2 - x] + 4[x] = 1 or 13 or 25
4{ 2[x^2 - x] + [x] } = 1,13,25
Zero solutions ; None multiple of 4
Since ⌊ . ⌋ (Greatest Integer function) always returns an integral value, let us assume,
⌊ x 2 − x ⌋ = a
⌊ x ⌋ = b
⌊ sin x ⌋ = c where a , b , c ∈ I
Hence, the equation becomes,
8 a + 4 b = 1 3 + 1 2 c
Rearrange the equation to get,
4 ( 2 a − b − 3 c ) = 1 3
2 a − b − 3 c = 4 1 3
Note that the above equation is a linear combination of a , b , c
Since a , b , c ∈ I , there will be no set of a , b , c , the above linear combination of which, will give us a fraction.
In other words, no matter what integers a , b and c are, 2 a − b − 3 c will always be an integer. So, hence there are no solutions.