Let f: R - {frac{alpha}{6}} rightarrow R be d | vedclass

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(B) Given $f(x) = \frac{5x + 3}{6x - \alpha}$.For $(f \circ f)(x) = x$,the function $f$ must be its own inverse,i.e.,$f(x) = f^{-1}(x)$.Let $y = f(x)

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$5$

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(B) Given $f(x) = \frac{5x + 3}{6x - \alpha}$.For $(f \circ f)(x) = x$,the function $f$ must be its own inverse,i.e.,$f(x) = f^{-1}(x)$.Let $y = f(x) = \frac{5x + 3}{6x - \alpha}$.Then $y(6x - \alpha) = 5x + 3$.$6xy - \alpha y = 5x + 3$.$6xy - 5x = \alpha y + 3$.$x(6y - 5) = \alpha y + 3$.$x = \frac{\alpha y + 3}{6y - 5}$.Thus,$f^{-1}(x) = \frac{\alpha x + 3}{6x - 5}$.Since $f(x) = f^{-1}(x)$,we have $\frac{5x + 3}{6x - \alpha} = \frac{\alpha x + 3}{6x - 5}$.Comparing the coefficients,we get $\alpha = 5$.

Let $f: R - \{\frac{\alpha}{6}\} \rightarrow R$ be defined by $f(x) = \frac{5x + 3}{6x - \alpha}$. Then the value of $\alpha$ for which $(f \circ f)(x) = x$,for all $x \in R - \{\frac{\alpha}{6}\}$,is:

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