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Let $f : R \rightarrow R$ be a continuous function such that $f(3 x)-f(x)=x$. If $f(8)=7$, then $f(14)$ is equal to.
$4$
$10$
$11$
$16$
Solution
$f(x)-f(x / 3)=x / 3$
$f(x / 3)-f\left(x / 3^{2}\right)=x / 3^{2}$
$f(x)-\lim _{n \rightarrow \infty} f\left(\frac{x}{3^{n}}\right)=x\left(\frac{1}{3}+\frac{1}{3^{2}} \ldots \infty\right)$
$f(x)-f(0)=\frac{x}{2}$
$f(8)=7 ; f(0)=3$
$f(x)=x / 2+3$
$f(14)=10$
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: