IIT JEE , AIEEE Forum - Mathematics , Algebra

shirishpathak

f(n+1)>f(n) , f[f(n)]=3n,

then find f(1458).

a)0

b)1458

c)1458*3/2

d)none

shirishpathak

anyone????????????

please

sachin_gupta1991

As f{f(n)}=3n , f(n+1)>f(n)

f(n)=

3 n

Thus,

f(1458)=1458

3

Therefore,

d)none option is correct.

iitkgp_bipin

Basic idea behind this :

Since f{f(n)} = 3n

so, f(n) should be a linear function of n.

Let f(n) = an+b

f{f(n)} = a.f(n)+b = a(an+b)+b = a²n + b(1+a)

But f{f(n)} = 3n

Hence a²n + ab+b = 3n

Equating the coefficient of n and constant term :

a² = 3 and b(1+a) = 0

a =

3 and b = 0

But since f(n+1) > f(n), f(n) is an increasing function.
Hence a should be positive.

so, a =

3

so, f(n) = (

3)n

f(1458) = (

3)(1458)

correct option is none of these.

karthik2007

I didnt follow your first step.. that linear function thing

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