Let a function f(x) defined on the set of integers, satisfy the conditions f(92+x)=f(92-x), f(1748 + x) = f(1748 - x) and f(1992 + x) = f(1992-x). Then maximum number of elements that can be present in the range of f(x), is:
As per mine analysis answer should be 5...but in the book the answer given is 8...plz share the solution...
If $f(a-x)=f(a+x)$ and $f(b-x)=f(b+x)$ then $f(x)$ can be proved to be periodic with period as $2|b-a|$.
Proof of the claim:
$f(x+2b-2a)=f(b+(x+b-2a))=f(b-(x+b-2a))=f(2a-x)=f(a+(a-x))=f(a-(a-x))=f(x)$
Taking into account all possible three pairs we obtain the possible period as $488,3800$ and $3312$. The HCF of these is $8$.
So $8$ is the period of $f(x)$
Since $f(x)$ is symmetric about $x=92$,we may have at most five distinct integers $f(91)=f(93)=k_1,f(90)=f(94)=k_2,f(89)=f(95)=k_3,f(88)=f(96)=k_4,f(92)=k_5$
After this $f(87)=f(87+8)=f(95)=k_3$ and on seeing further we realize that same five integers keep popping up because of periodicity.
So the answer is indeed $5$ and NOT $8$