1and subs x as x-2;
so f(x-2)=f(x)+f(x-4);
and add both so.....
f(x)+f(x-2)=f(x+2)+f(x-2)+f(x)+f(x-4)
that is f(x+2)+f(x-4)=0
subs x as x-2......
f(x)+f(x-6)=0
so f(x)=-f(x-6)......... so i think period is 6.....may be
f:R->[0,infinity) is such dat f(x-1)+f(x+1)=√3 f(x) then find the period of f(x).
do we really have a method to find it, my sir told me to do it by hit and trial !!
and it was a MSQ but i m nt givn da opt
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6 Answers
1ya v hav a method.............substitute x as x+1..........and ull get sumthing..........then substitute x as x-1.........and ull get sumthing else...............and subtract if u want................and at last ull get an eqn with f(x)..............from which u can find the period........
1f(x-1)+f(x+1)=√3f(x)
subs x as x+1
f(x)+f(x+2)=√3f(x+1)
and subs x as x-1 in d first eqn...
f(x-2)+f(x)=√3f(x-1)
add these
2f(x)+f(x+2)+f(x-2)=√3(f(x+1)+f(x-1))
so......f(x)=f(x+2)+f(x-2).....
1
1not 6 akand
as f(x)=-f(x-6)
f(x-6)=-f(x-6-6)=-f(x-12)
hence f(x)=f(x-12)
period =12
1well !!!! thanks for the method
but ans=12
you got
f(x)=-f(x-1) remove -ve sign to get ans
thanks again!!!!
and rohan too for correcting