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If $f$ 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\left( x \right) = 120,} $ find the value of $n$
$4$
$4$
$4$
$4$
Solution
It is given that,
$f(x+y)=f(x) \times f(y)$ for all $x, y \in N$ …..$(1)$
$f(1)=3$
Taking $x=y=1$ in $(1)$
We obtain $f(1+1)=f(2)=f(1) f(1)=3 \times 3=9$
Similarly,
$f(1+1+1)=f(3)=f(1+2)=f(1) f(2)=3 \times 9=27$
$f(4)=f(1+4)=f(1) f(3)=3 \times 27=81$
$\therefore f(1), f(2), f(3), \ldots \ldots,$ that is $3,9,27, \ldots \ldots,$ forms a $G.P.$ with both the first term and common ratio equal to $3 .$
It is known that, $S_{n}=\frac{a\left(r^{n}-1\right)}{r-1}$
It is given that, $\sum\limits_{x = 1}^n {f\left( x \right) = 120,} $
$\therefore 120=\frac{3\left(3^{n}-1\right)}{3-1}$
$\Rightarrow 120=\frac{3}{2}\left(3^{n}-1\right)$
$\Rightarrow 3^{n}-1=80$
$\Rightarrow 3^{n}=81=3^{4}$
$\therefore n=4$
Thus, the value of $n$ is $4$
Similar Questions
Let $\quad E_1=\left\{x \in R : x \neq 1\right.$ and $\left.\frac{x}{x-1}>0\right\}$ and $\quad E_2=\left\{x \in E_1: \sin ^{-1}\left(\log _e\left(\frac{x}{x-1}\right)\right)\right.$ is a real number $\}$.
(Here, the inverse trigonometric function $\sin ^{-1} x$ assumes values in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ )
Let $f : E _1 \rightarrow R$ be the function defined by $f(x)=\log _c\left(\frac{x}{x-1}\right)$ and $g: E_2 \rightarrow R$ be the function defined by $g(x)=\sin ^{-1}\left(\log _e\left(\frac{x}{x-1}\right)\right)$
| $LIST I$ | $LIST II$ |
| $P$ The range of $f$ is | $1$ $\left(-\infty, \frac{1}{1- e }\right] \cup\left[\frac{ e }{ e -1}, \infty\right)$ |
| $Q$ The range of $g$ contains | $2$ $(0,1)$ |
| $R$ The domain of $f$ contains | $3$ $\left[-\frac{1}{2}, \frac{1}{2}\right]$ |
| $S$ The domain of $g$ is | $4$ $(-\infty, 0) \cup(0, \infty)$ |
| $5$ $\left(-\infty, \frac{ e }{ e -1}\right]$ | |
| $6$ $(-\infty, 0) \cup\left(\frac{1}{2}, \frac{ e }{ e -1}\right]$ |
The correct option is:
