1. If a(0) = x , a (n+1) = f( a ( n ) ) , n= 0 , 1 , 2 , ............................., find a(n) when
(i) f(x) = root of (mod x )
(ii) f(x) = 1 / 1-x
2. Let f(x) be a real valued funcn defined for all real nos x such that for some fixed real no. a>0 , f(x+a) = 1/2 + root of ( f(x) - square of f(x) ) and 1/2 less than equal to f(x) less than equal to 1 for all x . Show that f(x) is periodic with period '2a'.
3. Let f(x) = ( 2x^3 - 9x^2 + 12x +3) (x-a) /(x-a) . If range of f(x) is a proper subset of real nos. , then what is the most exhaustive set in which a lies??
392)
i dont know if i am being absolutely rubbish
but for f to have period 2a , we must have f(x)=f(x+2a)
putting x=x-a
f(x)=1/2+√[f(x-a)-√f(x-a)]
putting x=x+a
f(x+2a)=1/2+√[f(x+a)-√f(x+a)]
now these 2 are equal only when
f(x-a)=f(x+a)
so f is periodic with period 2a
391)
f(x)=1/(1-x)
a(0)=x and a(n+1)=f(a(n))
n=0 gives us a(1)=f(x)
so a(2)=f(a(1))=f(f(x))
and a(3)=f(f(f(x)))
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a(n)=f(f(f(.......f(x))))))))))) (n times f occurs)
a(1)=f(x)=1/1-x
a(2)=f(f(x))=f(1/(1-x))=(x-1)/x
a(3)=f(f(f(x)))=f((x-a)/x)=x
a(4) will similarly be 1/(1-x)
and series follows
1/(1-x) , (x-1)/x , x
these keep on occuring one after another
so a(n) willd epend on what remainder n leaves when divided by 3
if its 1 , then a(n)=1/(1-x)
if its 2, then a(n)=(x-1)/x
if its 0, then a(n)=x
393rd question , please look ta it again , i think u have made some mistake in typing the question
