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Let f be a real valued function defined for all real x such that for some fixed real number a > 0, such that..
\huge f(x+a)=\frac{1}{2}+\sqrt{f(x)-(f(x))^2}
and
\huge \frac{1}{2}\leq f(x)\leq 1, \forall x
Show that f(x) is periodic and find period of f(x)
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29 Answers
341The algebraic way to go about it would be:
We are given f(x+a) - \frac{1}{2} = \sqrt{f(x) - f^2(x)} ...............1
By squaring and manipulation we get f(x) - \frac{1}{2} = \sqrt{ f(x+a) - f^2(x+a)}
We immediately see from this that f(x+a) - \frac{1}{2} = \sqrt{ f(x+2a) - f^2(x+2a)}..........2
From 1 and 2 we have
\sqrt{f(x) - f^2(x)} = \sqrt{ f(x+2a) - f^2(x+2a)} \\ \\ \Rightarrow f(x) - f^2(x) = f(x+2a) - f^2(x+2a)
which means either f(x) - f(x+2a) = 0 or f(x) + f(x+2a) = 1
The first gives f(x) = f(x+2a) while the second gives f(x) = f(x+2a) = 1/2 (since 1/2 <= f(x) <= 1)
In either case f(x) = f(x+2a) for all x
Hence 2a is a period of f(x)
33Take any f(x)'s curve... (simple one )
transform it to -(f(x))2
then other graph transformations....
add f(x) to it...
take it root add 1/2 to it...
u will get the final curve.. now period can be seen easily...
33lol...
don worry...
that technique is not 100% right... it is giving correct answer only because f(x) was periodic because of given condition.. if it would not have been periodic then also it would have given some answer... But since it is periodic then only it can give correct answer.. so it can't check if its periodic or not..
13i m not gettin dis graph wali technique
plzzz explain.........
21PL. NOTE : ONLY EXPERTS AT THIS MUST USE IT..... PL. DONT TRY THIS AT HOME!!!
LOL
33
bahut log nahi manenge.. isko.. :P
par mere ko answer aata hai is sab se..
Note:This is not to test if function is periodic or not(in such questions...) just to find period if its periodic
21CAN U TELL GRAF SE TUME KAISE KIYA THA???
HOW DID U RANDOMLY PLOT A GRAF?
MAYB USIN LIMITS OF Y - AXIS AS 1 AND 1/2 AND THWEN APPLYIN DA KEY CONDITION???
KAISE KIYA??
21[12]
UMMMMM.......
IF f(k) = 1;
f(k+a) = 1/2;
so the range of the function is [f(a),f(k+a)]
..... aage lemme [12]
btw priyam do u know how 2 solv it or is it ur dbt as well?? [7]
21f(k)=1; ==> f(k+a)=1/2
==> f(k-a) = 1/2;
iska matlab f(k-a) = f(k+a)
ther4 T = 2a
HOPE ITS RITE
39theoory se banana kuch nahiin, bas manipulate karte baito diya gaya condition ko.
agar luck accha hain toh ban jaayega, nahiin toh laat marega.
waise priyam aur tapan i dont know about u, but luck never goes with me.
33Isse tumko period approximately dikhne lagega.. phir check kar lena x ko x+T put kar ke... :)
Pata nahi acah se theory likh likh kar kaise banega...
maine to graph se kiya... :)
33Ek andaj se khicho f(x) ka curve (kaisa bhi..) phir RHS ka sara transformation karo.. aur phir dekho..
u have done main things..
f(k)=1 f(k+a)=1/2
21wat wud ur approacjh b while solvin this kinda sums??
wat i thot was 2 keep in mind dat i need to fid a T such that f(x+T) = f(x) and then manipulate upon the given condtn to get this!!!
can u pl. tell me how wud u hav gon abt it?
21k k...
LOL.......
mere ko thooda periodicity find karne vale que me gadbad hai, magar lemme try!!!
If i dont get, I'll ask....
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33Do do tha likhne ka isme isliye post kiye question :P
easy hai... :)
33Solved this last year... abhi bhi ban jaega... bas latex me likhne ka man kar raha tha...[3]