Let $f:R - \left\{ {{\alpha \over 6}} \right\} \to R$ be defined by $f(x) = {{5x + 3} \over {6x - \alpha }}$. Then the value of $\alpha$ for which (fof)(x) = x, for all $x \in R - \left\{ {{\alpha \over 6}} \right\}$, is :
Solution
$f(x) = {{5x + 3} \over {6x - \alpha }} = y$ ..... (i)<br><br>$5x + 3 = 6xy - \alpha y$<br><br>$x(6y - 5) = \alpha y + 3$<br><br>$x = {{\alpha y + 3} \over {6y - 5}}$<br><br>${f^{ - 1}}(x) = {{\alpha x + 3} \over {6x - 5}}$ ...... (ii)<br><br>fo $f(x) = x$<br><br>$f(x) = {f^{ - 1}}(x)$<br><br>From eq<sup>n</sup> (i) & (ii)<br><br>Clearly $(\alpha = 5)$
About this question
Subject: Mathematics · Chapter: Sets, Relations and Functions · Topic: Sets and Operations
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