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If $f(x) = \frac{{\alpha x}}{{x + 1}},x \ne - 1$, for what value of $\alpha $ is $f(f(x)) = x$
$\sqrt 2 $
$ - \sqrt 2 $
$1$
$-1$
Solution
(d) $f(x) = \frac{{\alpha x}}{{x + 1}}$;
$f(f(x)) = f\left( {\frac{{\alpha x}}{{x + 1}}} \right) = \frac{{\alpha \left( {\frac{{\alpha x}}{{x + 1}}} \right)}}{{\frac{{\alpha x}}{{x + 1}} + 1}}$
But $f(f(x)) = x$,
$\therefore$ $\frac{{{\alpha ^2}x}}{{\alpha x + x + 1}} = x$
Put $\alpha = – 1$, $\frac{{{{( – 1)}^2}x}}{{( – 1)x + x + 1}} = \frac{x}{{ – x + x + 1}} = x$;
$\therefore$ $\alpha = – 1$.
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:
