Let [f:R - 6 R] be defined by [f(x) = 5x + 3 6x - ].

Question

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 :

Accepted Answer

$$f(x) = {{5x + 3} \over {6x - \alpha }} = y$$ ..... (i)

$$5x + 3 = 6xy - \alpha y$$

$$x(6y - 5) = \alpha y + 3$$

$$x = {{\alpha y + 3} \over {6y - 5}}$$

$${f^{ - 1}}(x) = {{\alpha x + 3} \over {6x - 5}}$$ ...... (ii)

fo $f(x) = x$

$f(x) = {f^{ - 1}}(x)$

From eqn (i) \& (ii)

Clearly $$(\alpha  = 5)$$

Solutions · 1