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F(x)=f(f(x))=f(x^2).....how many such functions with degree>2 exist?

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See here f(f(x))=f(x^2).. well, what is inside there within f( )??
so.. we may say f(x)=x^2
now F(x)=f(f(x))=f(x^2)=(x^2)^2= x^4 (using the definition of f(x) above)
so.. ans is 1 such function exists.. i guess this is the answer.. comments are welcome
11 years agoHelpfull: Yes(5) No(3)

the answer is 0
11 years agoHelpfull: Yes(4) No(5)
plz explain
11 years agoHelpfull: Yes(2) No(2)

TCS Other Question

how many digit of xeros at the end of product of the number from 1 to 100? That's an interesting question! In a normal year - that is, a year
with no leap day - there are 365 days, which comes to 52 weeks of
7 days, plus one more day.

So if a normal year starts on a Monday, then the following year has to
start on a Tuesday. Does that make sense?

However, if a leap year starts on a Monday, then the following year
has to start on a Wednesday, because of the extra day in February.

And if two years start on the same day, and if they're both normal
years or both leap years, then they have to have the same calendar. Do
you see why this is true?

1990 is a normal year, and in fact, it _does_ start on a Monday. So
1991 must start on a Tuesday, and 1992 must start on a Wednesday. But
1992 is a leap year, so 1993 must start on a Friday (instead of a
Thursday).

I've started a table below:

Year Starts on Leap? Add
---- --------- ----- ---
1990 Monday N 1
1991 Tuesday N 1
1992 Wednesday Y 2
1993 Friday N 1
1994 Saturday N 1

If you continue the table, you'll eventually find a year that starts
on Monday and isn't a leap year. That's the year you're looking for.