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1.Relation and Function
hard
If $f(x)$ is a function satisfying $f(x + y) = f(x)f(y)$ for all $x,\;y \in N$ such that $f(1) = 3$ and $\sum\limits_{x = 1}^n {f(x) = 120} $. Then the value of $n$ is
A
$4$
B
$5$
C
$6$
D
None of these
(IIT-1992)
Solution
(a) Since $f(x + y) = f(x)f(y)$ for all $x,\;y \in N$, therefore for any $x \in N$
$f(x) = f(x – 1 + 1) = f(x – 1)f(1)$
$ = f(x – 2){[f(1)]^2} = ……. = {[f(1)]^x}$
$ \Rightarrow $$f(x) = {3^x}$, $(\;f(1) = 3)$
Now $\sum\limits_{x = 1}^n {f(x) = 120} $
$ \Rightarrow $$\sum\limits_{x = 1}^n {{3^x} = 120} $
$ \Rightarrow $$\frac{{3({3^n} – 1)}}{{(3 – 1)}} = 120$
$ \Rightarrow $$n = 4$.
Standard 12
Mathematics
